851 lines
18 KiB
TeX
851 lines
18 KiB
TeX
\chapter{Matrices}
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\index{matrix}
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A \key{matrix} is a mathematical concept
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that corresponds to a two-dimensional array
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in programming. For example,
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\[
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A =
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\begin{bmatrix}
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6 & 13 & 7 & 4 \\
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7 & 0 & 8 & 2 \\
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9 & 5 & 4 & 18 \\
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\end{bmatrix}
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\]
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is a matrix of size $3 \times 4$, i.e.,
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it has 3 rows and 4 columns.
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The notation $[i,j]$ refers to
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the element in row $i$ and column $j$
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in a matrix.
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For example, in the above matrix,
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$A[2,3]=8$ and $A[3,1]=9$.
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\index{vector}
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A special case of a matrix is a \key{vector}
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that is a one-dimensional matrix of size $n \times 1$.
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For example,
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\[
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V =
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\begin{bmatrix}
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4 \\
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7 \\
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5 \\
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\end{bmatrix}
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\]
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is a vector that contains three elements.
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\index{transpose}
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The \key{transpose} $A^T$ of a matrix $A$
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is obtained when the rows and columns in $A$
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are swapped, i.e., $A^T[i,j]=A[j,i]$:
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\[
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A^T =
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\begin{bmatrix}
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6 & 7 & 9 \\
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13 & 0 & 5 \\
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7 & 8 & 4 \\
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4 & 2 & 18 \\
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\end{bmatrix}
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\]
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\index{square matrix}
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A matrix is a \key{square matrix} if it
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has the same number of rows and columns.
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For example, the following matrix is a
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square matrix:
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\[
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S =
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\begin{bmatrix}
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3 & 12 & 4 \\
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5 & 9 & 15 \\
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0 & 2 & 4 \\
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\end{bmatrix}
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\]
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\section{Operations}
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The sum $A+B$ of matrices $A$ and $B$
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is defined if the matrices are of the same size.
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The result is a matrix where each element
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is the sum of the corresponding elements
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in matrices $A$ and $B$.
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For example,
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\[
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\begin{bmatrix}
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6 & 1 & 4 \\
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3 & 9 & 2 \\
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\end{bmatrix}
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+
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\begin{bmatrix}
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4 & 9 & 3 \\
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8 & 1 & 3 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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6+4 & 1+9 & 4+3 \\
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3+8 & 9+1 & 2+3 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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10 & 10 & 7 \\
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11 & 10 & 5 \\
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\end{bmatrix}.
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\]
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Multiplying a matrix $A$ by a value $x$ means
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that we multiply each element in $A$ by $x$.
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For example,
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\[
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2 \cdot \begin{bmatrix}
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6 & 1 & 4 \\
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3 & 9 & 2 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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2 \cdot 6 & 2\cdot1 & 2\cdot4 \\
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2\cdot3 & 2\cdot9 & 2\cdot2 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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12 & 2 & 8 \\
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6 & 18 & 4 \\
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\end{bmatrix}.
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\]
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\subsubsection{Matrix multiplication}
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\index{matrix multiplication}
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The product $AB$ of matrices $A$ and $B$
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is defined if $A$ is of size $a \times n$
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and $B$ is of size $n \times b$, i.e.,
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the width of $A$ equals the height of $B$.
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The result is a matrix of size $a \times b$
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whose elements are calculated using the formula
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\[
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AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j].
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\]
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The idea is that each element in $AB$
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is a sum of products of elements in $A$ and $B$
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according to the following picture:
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\begin{center}
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\begin{tikzpicture}[scale=0.5]
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\draw (0,0) grid (4,3);
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\draw (5,0) grid (10,3);
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\draw (5,4) grid (10,8);
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\node at (2,-1) {$A$};
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\node at (7.5,-1) {$AB$};
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\node at (11,6) {$B$};
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\draw[thick,->,red,line width=2pt] (0,1.5) -- (4,1.5);
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\draw[thick,->,red,line width=2pt] (6.5,8) -- (6.5,4);
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\draw[thick,red,line width=2pt] (6.5,1.5) circle (0.4);
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\end{tikzpicture}
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\end{center}
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For example,
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\[
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 6 \\
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2 & 9 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 \cdot 1 + 4 \cdot 2 & 1 \cdot 6 + 4 \cdot 9 \\
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3 \cdot 1 + 9 \cdot 2 & 3 \cdot 6 + 9 \cdot 9 \\
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8 \cdot 1 + 6 \cdot 2 & 8 \cdot 6 + 6 \cdot 9 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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9 & 42 \\
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21 & 99 \\
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20 & 102 \\
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\end{bmatrix}.
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\]
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Matrix multiplication is not commutative,
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so $AB = BA$ doesn't hold.
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However, it is associative,
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so $A(BC)=(AB)C$ holds.
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\index{identity matrix}
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An \key{identity matrix} is a square matrix
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where each element on the diagonal is 1,
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and all other elements are 0.
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For example, the $3 \times 3$ identity matrix
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is as follows:
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\[
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I = \begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{bmatrix}
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\]
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\begin{samepage}
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Multiplying a matrix by an identity matrix
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doesn't change it. For example,
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\[
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix} \hspace{10px} \textrm{ja} \hspace{10px}
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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1 & 0 \\
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0 & 1 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1 & 4 \\
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3 & 9 \\
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8 & 6 \\
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\end{bmatrix}.
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\]
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\end{samepage}
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Using a straightforward algorithm,
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we can calculate the product of
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two $n \times n$ matrices
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in $O(n^3)$ time.
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There are also more efficient algorithms
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for matrix multiplication:
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at the moment, the best known time complexity
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is $O(n^{2.37})$.
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However, such special algorithms are not needed
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in competitive programming.
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\subsubsection{Matrix power}
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\index{matrix power}
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The power $A^k$ of a matrix $A$ is defined
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if $A$ is a square matrix.
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The definition is based on matrix multiplication:
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\[ A^k = \underbrace{A \cdot A \cdot A \cdots A}_{\textrm{$k$ times}} \]
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For example,
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\[
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix}^3 =
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix} \cdot
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix} \cdot
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix} =
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\begin{bmatrix}
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48 & 165 \\
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33 & 114 \\
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\end{bmatrix}.
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\]
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In addition, $A^0$ is an identity matrix. For example,
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\[
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix}^0 =
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\begin{bmatrix}
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1 & 0 \\
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0 & 1 \\
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\end{bmatrix}.
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\]
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The matrix $A^k$ can be efficiently calculated
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in $O(n^3 \log k)$ time using the
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algorithm in Chapter 21.2. For example,
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\[
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix}^8 =
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix}^4 \cdot
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\begin{bmatrix}
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2 & 5 \\
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1 & 4 \\
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\end{bmatrix}^4.
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\]
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\subsubsection{Determinant}
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\index{determinant}
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The \key{determinant} $\det(A)$ of a matrix $A$
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is defined if $A$ is a square matrix.
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If $A$ is of size $1 \times 1$,
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then $\det(A)=A[1,1]$.
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The determinant of a larger matrix is
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calculated recursively using the formula \index{cofactor}
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\[\det(A)=\sum_{j=1}^n A[1,j] C[1,j],\]
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where $C[i,j]$ is the \key{cofactor} of $A$
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at $[i,j]$.
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The cofactor is calculated using the formula
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\[C[i,j] = (-1)^{i+j} \det(M[i,j]),\]
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where $M[i,j]$ is a copy of matrix $A$
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where row $i$ and column $j$ are removed.
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Because of the multiplier $(-1)^{i+j}$ in the cofactor,
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every other determinant is positive
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and negative.
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\begin{samepage}
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For example,
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\[
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\det(
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\begin{bmatrix}
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3 & 4 \\
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1 & 6 \\
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\end{bmatrix}
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) = 3 \cdot 6 - 4 \cdot 1 = 14
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\]
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\end{samepage}
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and
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\[
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\det(
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\begin{bmatrix}
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2 & 4 & 3 \\
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5 & 1 & 6 \\
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7 & 2 & 4 \\
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\end{bmatrix}
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) =
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2 \cdot
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\det(
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\begin{bmatrix}
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1 & 6 \\
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2 & 4 \\
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\end{bmatrix}
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)
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-4 \cdot
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\det(
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\begin{bmatrix}
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5 & 6 \\
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7 & 4 \\
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\end{bmatrix}
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)
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+3 \cdot
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\det(
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\begin{bmatrix}
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5 & 1 \\
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7 & 2 \\
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\end{bmatrix}
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) = 81.
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\]
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\index{inverse matrix}
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The determinant indicates if matrix $A$
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has an \key{inverse matrix}
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$A^{-1}$ for which $A \cdot A^{-1} = I$,
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where $I$ is an identity matrix.
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It turns out that $A^{-1}$ exists
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exactly when $\det(A) \neq 0$,
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and it can be calculated using the formula
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\[A^{-1}[i,j] = \frac{C[j,i]}{det(A)}.\]
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For example,
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\[
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\underbrace{
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\begin{bmatrix}
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2 & 4 & 3\\
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5 & 1 & 6\\
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7 & 2 & 4\\
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\end{bmatrix}
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}_{A}
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\cdot
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\underbrace{
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\frac{1}{81}
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\begin{bmatrix}
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-8 & -10 & 21 \\
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22 & -13 & 3 \\
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3 & 24 & -18 \\
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\end{bmatrix}
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}_{A^{-1}}
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=
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\underbrace{
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\begin{bmatrix}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{bmatrix}
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}_{I}.
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\]
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\section{Linear recurrences}
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\index{linear recurrence}
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A \key{linear recurrence}
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can be represented as a function $f(n)$
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with initial values
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$f(0),f(1),\ldots,f(k-1)$,
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whose values for $k$ and larger parameters
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are calculated recursively using a formula
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\[f(n) = c_1 f(n-1) + c_2 f(n-2) + \ldots + c_k f (n-k),\]
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where $c_1,c_2,\ldots,c_k$ are constant multipliers.
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We can use dynamic programming to calculate
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any value $f(n)$ in $O(kn)$ time by calculating
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all values $f(0),f(1),\ldots,f(n)$ one after another.
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However, if $k$ is small, it is possible to calculate
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$f(n)$ much more efficiently in $O(k^3 \log n)$
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time using matrix operations.
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\subsubsection{Fibonacci numbers}
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\index{Fibonacci number}
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A simple example of a linear recurrence is the
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function that calculates Fibonacci numbers:
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\[
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\begin{array}{lcl}
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f(0) & = & 0 \\
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f(1) & = & 1 \\
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f(n) & = & f(n-1)+f(n-2) \\
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\end{array}
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\]
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In this case, $k=2$ and $c_1=c_2=1$.
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\begin{samepage}
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The idea is to represent the formula for
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calculating Fibonacci numbers as a
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square matrix $X$ of size $2 \times 2$
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for which the following holds:
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\[ X \cdot
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\begin{bmatrix}
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f(i) \\
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f(i+1) \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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f(i+1) \\
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f(i+2) \\
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\end{bmatrix}
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\]
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Thus, values $f(i)$ and $f(i+1)$ are given as
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''input'' for $X$,
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and $X$ constructs values $f(i+1)$ and $f(i+2)$
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from them.
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It turns out that such a matrix is
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\[ X =
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\begin{bmatrix}
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0 & 1 \\
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1 & 1 \\
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\end{bmatrix}.
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\]
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\end{samepage}
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\noindent
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For example,
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\[
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\begin{bmatrix}
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0 & 1 \\
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1 & 1 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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f(5) \\
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f(6) \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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0 & 1 \\
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1 & 1 \\
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\end{bmatrix}
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\cdot
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\begin{bmatrix}
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5 \\
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8 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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8 \\
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13 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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f(6) \\
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f(7) \\
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\end{bmatrix}.
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\]
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Thus, we can calculate $f(n)$ using the formula
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\[
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\begin{bmatrix}
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f(n) \\
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f(n+1) \\
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\end{bmatrix}
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=
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X^n \cdot
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\begin{bmatrix}
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f(0) \\
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f(1) \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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0 & 1 \\
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1 & 1 \\
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\end{bmatrix}^n
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\cdot
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\begin{bmatrix}
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0 \\
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1 \\
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\end{bmatrix}.
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\]
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The power $X^n$ on can be calculated in
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$O(k^3 \log n)$ time,
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so the value $f(n)$ can also be calculated
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in $O(k^3 \log n)$ time.
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\subsubsection{General case}
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Let's now consider a general case where
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$f(n)$ is any linear recurrence.
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Again, our goal is to construct a matrix $X$
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for which
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\[ X \cdot
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\begin{bmatrix}
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f(i) \\
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f(i+1) \\
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\vdots \\
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f(i+k-1) \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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f(i+1) \\
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f(i+2) \\
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\vdots \\
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f(i+k) \\
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\end{bmatrix}.
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\]
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Such a matrix is
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\[
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X =
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\begin{bmatrix}
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0 & 1 & 0 & 0 & \cdots & 0 \\
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0 & 0 & 1 & 0 & \cdots & 0 \\
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0 & 0 & 0 & 1 & \cdots & 0 \\
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\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
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0 & 0 & 0 & 0 & \cdots & 1 \\
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c_k & c_{k-1} & c_{k-2} & c_{k-3} & \cdots & c_1 \\
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\end{bmatrix}.
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\]
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In the first $k-1$ rows, each element is 0
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except that one element is 1.
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These rows replace $f(i)$ with $f(i+1)$,
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$f(i+1)$ with $f(i+2)$, etc.
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The last row contains the multipliers in the recurrence,
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and it calculates the new value $f(i+k)$.
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\begin{samepage}
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Now, $f(n)$ can be calculated in
|
|
$O(k^3 \log n)$ time using the formula
|
|
\[
|
|
\begin{bmatrix}
|
|
f(n) \\
|
|
f(n+1) \\
|
|
\vdots \\
|
|
f(n+k-1) \\
|
|
\end{bmatrix}
|
|
=
|
|
X^n \cdot
|
|
\begin{bmatrix}
|
|
f(0) \\
|
|
f(1) \\
|
|
\vdots \\
|
|
f(k-1) \\
|
|
\end{bmatrix}.
|
|
\]
|
|
\end{samepage}
|
|
|
|
\section{Graphs and matrices}
|
|
|
|
\subsubsection{Counting paths}
|
|
|
|
The powers of an adjacency matrix of a graph
|
|
have an interesting property.
|
|
When $V$ is an adjacency matrix of an unweighted graph,
|
|
the matrix $V^n$ contains the numbers of paths of
|
|
$n$ edges between the nodes in the graph.
|
|
|
|
For example, for the graph
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (1,1) {$4$};
|
|
\node[draw, circle] (3) at (3,3) {$2$};
|
|
\node[draw, circle] (4) at (5,3) {$3$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (5,1) {$6$};
|
|
|
|
\path[draw,thick,->,>=latex] (1) -- (2);
|
|
\path[draw,thick,->,>=latex] (2) -- (3);
|
|
\path[draw,thick,->,>=latex] (3) -- (1);
|
|
\path[draw,thick,->,>=latex] (4) -- (3);
|
|
\path[draw,thick,->,>=latex] (3) -- (5);
|
|
\path[draw,thick,->,>=latex] (3) -- (6);
|
|
\path[draw,thick,->,>=latex] (6) -- (4);
|
|
\path[draw,thick,->,>=latex] (6) -- (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
the adjacency matrix is
|
|
\[
|
|
V= \begin{bmatrix}
|
|
0 & 0 & 0 & 1 & 0 & 0 \\
|
|
1 & 0 & 0 & 0 & 1 & 1 \\
|
|
0 & 1 & 0 & 0 & 0 & 0 \\
|
|
0 & 1 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 1 & 0 & 1 & 0 \\
|
|
\end{bmatrix}.
|
|
\]
|
|
Now, for example, the matrix
|
|
\[
|
|
V^4= \begin{bmatrix}
|
|
0 & 0 & 1 & 1 & 1 & 0 \\
|
|
2 & 0 & 0 & 0 & 2 & 2 \\
|
|
0 & 2 & 0 & 0 & 0 & 0 \\
|
|
0 & 2 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 0 & 0 & 0 & 0 \\
|
|
0 & 0 & 1 & 1 & 1 & 0 \\
|
|
\end{bmatrix}
|
|
\]
|
|
contains the numbers of paths of 4 edges
|
|
between the nodes.
|
|
For example, $V^4[2,5]=2$,
|
|
because there are two paths of 4 edges
|
|
from node 2 to node 5:
|
|
$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$
|
|
and
|
|
$2 \rightarrow 6 \rightarrow 3 \rightarrow 2 \rightarrow 5$.
|
|
|
|
\subsubsection{Shortest paths}
|
|
|
|
Using a similar idea in a weighted graph,
|
|
we can calculate for each pair of nodes the shortest
|
|
path between them that contains exactly $n$ edges.
|
|
To calculate this, we have to define matrix multiplication
|
|
in another way, so that we don't calculate the number
|
|
of paths but minimize the length of a path.
|
|
|
|
\begin{samepage}
|
|
As an example, consider the following graph:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (1,1) {$4$};
|
|
\node[draw, circle] (3) at (3,3) {$2$};
|
|
\node[draw, circle] (4) at (5,3) {$3$};
|
|
\node[draw, circle] (5) at (3,1) {$5$};
|
|
\node[draw, circle] (6) at (5,1) {$6$};
|
|
|
|
\path[draw,thick,->,>=latex] (1) -- node[font=\small,label=left:4] {} (2);
|
|
\path[draw,thick,->,>=latex] (2) -- node[font=\small,label=left:1] {} (3);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=north:2] {} (1);
|
|
\path[draw,thick,->,>=latex] (4) -- node[font=\small,label=north:4] {} (3);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=left:1] {} (5);
|
|
\path[draw,thick,->,>=latex] (3) -- node[font=\small,label=left:2] {} (6);
|
|
\path[draw,thick,->,>=latex] (6) -- node[font=\small,label=right:3] {} (4);
|
|
\path[draw,thick,->,>=latex] (6) -- node[font=\small,label=below:2] {} (5);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
Let's construct an adjacency matrix where
|
|
$\infty$ means that an edge doesn't exist,
|
|
and other values correspond to edge weights.
|
|
The matrix is
|
|
\[
|
|
V= \begin{bmatrix}
|
|
\infty & \infty & \infty & 4 & \infty & \infty \\
|
|
2 & \infty & \infty & \infty & 1 & 2 \\
|
|
\infty & 4 & \infty & \infty & \infty & \infty \\
|
|
\infty & 1 & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & 3 & \infty & 2 & \infty \\
|
|
\end{bmatrix}.
|
|
\]
|
|
|
|
Instead of the formula
|
|
\[
|
|
AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j]
|
|
\]
|
|
we now use the formula
|
|
\[
|
|
AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j],
|
|
\]
|
|
for matrix multiplication, so we calculate
|
|
a minimum instead of a sum,
|
|
and a sum of elements instead of a product.
|
|
After this modification,
|
|
matrix powers can be used for calculating
|
|
shortest paths in the graph:
|
|
|
|
\[
|
|
V^4= \begin{bmatrix}
|
|
\infty & \infty & 10 & 11 & 9 & \infty \\
|
|
9 & \infty & \infty & \infty & 8 & 9 \\
|
|
\infty & 11 & \infty & \infty & \infty & \infty \\
|
|
\infty & 8 & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & \infty & \infty & \infty & \infty \\
|
|
\infty & \infty & 12 & 13 & 11 & \infty \\
|
|
\end{bmatrix}
|
|
\]
|
|
For example, the shortest path of 4 edges
|
|
from node 2 to node 5 has length 8.
|
|
This path is
|
|
$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
|
|
|
|
\subsubsection{Kirchhoff's theorem}
|
|
|
|
\index{Kirchhoff's theorem}
|
|
\index{spanning tree}
|
|
|
|
\key{Kirchhoff's theorem} provides us a way
|
|
to calculate the number of spanning trees
|
|
in a graph as a determinant of a special matrix.
|
|
For example, the graph
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1) at (1,3) {$1$};
|
|
\node[draw, circle] (2) at (3,3) {$2$};
|
|
\node[draw, circle] (3) at (1,1) {$3$};
|
|
\node[draw, circle] (4) at (3,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1) -- (2);
|
|
\path[draw,thick,-] (1) -- (3);
|
|
\path[draw,thick,-] (3) -- (4);
|
|
\path[draw,thick,-] (1) -- (4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
has three spanning trees:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\node[draw, circle] (1a) at (1,3) {$1$};
|
|
\node[draw, circle] (2a) at (3,3) {$2$};
|
|
\node[draw, circle] (3a) at (1,1) {$3$};
|
|
\node[draw, circle] (4a) at (3,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1a) -- (2a);
|
|
%\path[draw,thick,-] (1a) -- (3a);
|
|
\path[draw,thick,-] (3a) -- (4a);
|
|
\path[draw,thick,-] (1a) -- (4a);
|
|
|
|
\node[draw, circle] (1b) at (1+4,3) {$1$};
|
|
\node[draw, circle] (2b) at (3+4,3) {$2$};
|
|
\node[draw, circle] (3b) at (1+4,1) {$3$};
|
|
\node[draw, circle] (4b) at (3+4,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1b) -- (2b);
|
|
\path[draw,thick,-] (1b) -- (3b);
|
|
%\path[draw,thick,-] (3b) -- (4b);
|
|
\path[draw,thick,-] (1b) -- (4b);
|
|
|
|
\node[draw, circle] (1c) at (1+8,3) {$1$};
|
|
\node[draw, circle] (2c) at (3+8,3) {$2$};
|
|
\node[draw, circle] (3c) at (1+8,1) {$3$};
|
|
\node[draw, circle] (4c) at (3+8,1) {$4$};
|
|
|
|
\path[draw,thick,-] (1c) -- (2c);
|
|
\path[draw,thick,-] (1c) -- (3c);
|
|
\path[draw,thick,-] (3c) -- (4c);
|
|
%\path[draw,thick,-] (1c) -- (4c);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\index{Laplacean matrix}
|
|
We construct a \key{Laplacean matrix} $L$,
|
|
where $L[i,i]$ is the degree of node $i$
|
|
and $L[i,j]=-1$ if there is an edge between
|
|
nodes $i$ and $j$, and otherwise $L[i,j]=0$.
|
|
In this graph, the matrix is as follows:
|
|
|
|
\[
|
|
L= \begin{bmatrix}
|
|
3 & -1 & -1 & -1 \\
|
|
-1 & 1 & 0 & 0 \\
|
|
-1 & 0 & 2 & -1 \\
|
|
-1 & 0 & -1 & 2 \\
|
|
\end{bmatrix}
|
|
\]
|
|
|
|
The number of spanning trees is the same as
|
|
the determinant of a matrix that is obtained
|
|
when we remove any row and any column from $L$.
|
|
For example, if we remove the first row
|
|
and column, the result is
|
|
|
|
\[ \det(
|
|
\begin{bmatrix}
|
|
1 & 0 & 0 \\
|
|
0 & 2 & -1 \\
|
|
0 & -1 & 2 \\
|
|
\end{bmatrix}
|
|
) =3.\]
|
|
The determinant is always the same,
|
|
regardless of which row and column we remove from $L$.
|
|
|
|
Note that a special case of Kirchhoff's theorem
|
|
is Cayley's formula in Chapter 22.5,
|
|
because in a complete graph of $n$ nodes
|
|
|
|
\[ \det(
|
|
\begin{bmatrix}
|
|
n-1 & -1 & \cdots & -1 \\
|
|
-1 & n-1 & \cdots & -1 \\
|
|
\vdots & \vdots & \ddots & \vdots \\
|
|
-1 & -1 & \cdots & n-1 \\
|
|
\end{bmatrix}
|
|
) =n^{n-2}.\]
|
|
|
|
|
|
|