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@ -73,7 +73,7 @@ The sum $A+B$ of matrices $A$ and $B$
is defined if the matrices are of the same size. is defined if the matrices are of the same size.
The result is a matrix where each element The result is a matrix where each element
is the sum of the corresponding elements is the sum of the corresponding elements
in matrices $A$ and $B$. in $A$ and $B$.
For example, For example,
\[ \[
@ -99,8 +99,7 @@ For example,
\] \]
Multiplying a matrix $A$ by a value $x$ means Multiplying a matrix $A$ by a value $x$ means
that we multiply each element in $A$ by $x$. that each element of $A$ is multiplied by $x$.
For example, For example,
\[ \[
2 \cdot \begin{bmatrix} 2 \cdot \begin{bmatrix}
@ -180,18 +179,18 @@ For example,
\end{bmatrix}. \end{bmatrix}.
\] \]
Matrix multiplication is not commutative, Matrix multiplication is associative,
so $AB = BA$ doesn't hold. so $A(BC)=(AB)C$ holds,
However, it is associative, but it is not commutative,
so $A(BC)=(AB)C$ holds. so $AB = BA$ does not hold.
\index{identity matrix} \index{identity matrix}
An \key{identity matrix} is a square matrix An \key{identity matrix} is a square matrix
where each element on the diagonal is 1, where each element on the diagonal is 1
and all other elements are 0. and all other elements are 0.
For example, the $3 \times 3$ identity matrix For example, the following matrix
is as follows: is the $3 \times 3$ identity matrix:
\[ \[
I = \begin{bmatrix} I = \begin{bmatrix}
1 & 0 & 0 \\ 1 & 0 & 0 \\
@ -202,7 +201,7 @@ is as follows:
\begin{samepage} \begin{samepage}
Multiplying a matrix by an identity matrix Multiplying a matrix by an identity matrix
doesn't change it. For example, does not change it. For example,
\[ \[
\begin{bmatrix} \begin{bmatrix}
1 & 0 & 0 \\ 1 & 0 & 0 \\
@ -220,7 +219,7 @@ doesn't change it. For example,
1 & 4 \\ 1 & 4 \\
3 & 9 \\ 3 & 9 \\
8 & 6 \\ 8 & 6 \\
\end{bmatrix} \hspace{10px} \textrm{ja} \hspace{10px} \end{bmatrix} \hspace{10px} \textrm{and} \hspace{10px}
\begin{bmatrix} \begin{bmatrix}
1 & 4 \\ 1 & 4 \\
3 & 9 \\ 3 & 9 \\
@ -328,13 +327,11 @@ where $C[i,j]$ is the \key{cofactor} of $A$
at $[i,j]$. at $[i,j]$.
The cofactor is calculated using the formula The cofactor is calculated using the formula
\[C[i,j] = (-1)^{i+j} \det(M[i,j]),\] \[C[i,j] = (-1)^{i+j} \det(M[i,j]),\]
where $M[i,j]$ is a copy of matrix $A$ where $M[i,j]$ is obtained by removing
where row $i$ and column $j$ are removed. row $i$ and column $j$ from $A$.
Because of the multiplier $(-1)^{i+j}$ in the cofactor, Due to the multiplier $(-1)^{i+j}$ in the cofactor,
every other determinant is positive every other determinant is positive
and negative. and negative.
\begin{samepage}
For example, For example,
\[ \[
\det( \det(
@ -344,10 +341,7 @@ For example,
\end{bmatrix} \end{bmatrix}
) = 3 \cdot 6 - 4 \cdot 1 = 14 ) = 3 \cdot 6 - 4 \cdot 1 = 14
\] \]
\end{samepage}
and and
\[ \[
\det( \det(
\begin{bmatrix} \begin{bmatrix}
@ -381,9 +375,9 @@ and
\index{inverse matrix} \index{inverse matrix}
The determinant indicates if matrix $A$ The determinant of $A$ tells us
has an \key{inverse matrix} whether there is an \key{inverse matrix}
$A^{-1}$ for which $A \cdot A^{-1} = I$, $A^{-1}$ such that $A \cdot A^{-1} = I$,
where $I$ is an identity matrix. where $I$ is an identity matrix.
It turns out that $A^{-1}$ exists It turns out that $A^{-1}$ exists
exactly when $\det(A) \neq 0$, exactly when $\det(A) \neq 0$,
@ -426,16 +420,16 @@ For example,
A \key{linear recurrence} A \key{linear recurrence}
can be represented as a function $f(n)$ can be represented as a function $f(n)$
with initial values such that the initial values are
$f(0),f(1),\ldots,f(k-1)$, $f(0),f(1),\ldots,f(k-1)$
whose values for $k$ and larger parameters and the larger values
are calculated recursively using a formula are calculated recursively using the formula
\[f(n) = c_1 f(n-1) + c_2 f(n-2) + \ldots + c_k f (n-k),\] \[f(n) = c_1 f(n-1) + c_2 f(n-2) + \ldots + c_k f (n-k),\]
where $c_1,c_2,\ldots,c_k$ are constant multipliers. where $c_1,c_2,\ldots,c_k$ are constant multipliers.
We can use dynamic programming to calculate We can use dynamic programming to calculate
any value $f(n)$ in $O(kn)$ time by calculating any value of $f(n)$ in $O(kn)$ time by calculating
all values $f(0),f(1),\ldots,f(n)$ one after another. all values of $f(0),f(1),\ldots,f(n)$ one after another.
However, if $k$ is small, it is possible to calculate However, if $k$ is small, it is possible to calculate
$f(n)$ much more efficiently in $O(k^3 \log n)$ $f(n)$ much more efficiently in $O(k^3 \log n)$
time using matrix operations. time using matrix operations.
@ -445,7 +439,7 @@ time using matrix operations.
\index{Fibonacci number} \index{Fibonacci number}
A simple example of a linear recurrence is the A simple example of a linear recurrence is the
function that calculates Fibonacci numbers: following function that defines the Fibonacci numbers:
\[ \[
\begin{array}{lcl} \begin{array}{lcl}
f(0) & = & 0 \\ f(0) & = & 0 \\
@ -456,9 +450,9 @@ f(n) & = & f(n-1)+f(n-2) \\
In this case, $k=2$ and $c_1=c_2=1$. In this case, $k=2$ and $c_1=c_2=1$.
\begin{samepage} \begin{samepage}
The idea is to represent the formula for The idea is to represent the
calculating Fibonacci numbers as a Fibonacci formula as a
square matrix $X$ of size $2 \times 2$ square matrix $X$ of size $2 \times 2$,
for which the following holds: for which the following holds:
\[ X \cdot \[ X \cdot
\begin{bmatrix} \begin{bmatrix}
@ -473,7 +467,7 @@ for which the following holds:
\] \]
Thus, values $f(i)$ and $f(i+1)$ are given as Thus, values $f(i)$ and $f(i+1)$ are given as
''input'' for $X$, ''input'' for $X$,
and $X$ constructs values $f(i+1)$ and $f(i+2)$ and $X$ calculates values $f(i+1)$ and $f(i+2)$
from them. from them.
It turns out that such a matrix is It turns out that such a matrix is
@ -540,14 +534,14 @@ X^n \cdot
1 \\ 1 \\
\end{bmatrix}. \end{bmatrix}.
\] \]
The power $X^n$ on can be calculated in The power $X^n$ can be calculated in
$O(k^3 \log n)$ time, $O(k^3 \log n)$ time,
so the value $f(n)$ can also be calculated so the value of $f(n)$ can also be calculated
in $O(k^3 \log n)$ time. in $O(k^3 \log n)$ time.
\subsubsection{General case} \subsubsection{General case}
Let's now consider a general case where Let us now consider the general case where
$f(n)$ is any linear recurrence. $f(n)$ is any linear recurrence.
Again, our goal is to construct a matrix $X$ Again, our goal is to construct a matrix $X$
for which for which
@ -583,8 +577,8 @@ In the first $k-1$ rows, each element is 0
except that one element is 1. except that one element is 1.
These rows replace $f(i)$ with $f(i+1)$, These rows replace $f(i)$ with $f(i+1)$,
$f(i+1)$ with $f(i+2)$, etc. $f(i+1)$ with $f(i+2)$, etc.
The last row contains the multipliers in the recurrence, The last row contains the multipliers of the recurrence
and it calculates the new value $f(i+k)$. to calculate the new value $f(i+k)$.
\begin{samepage} \begin{samepage}
Now, $f(n)$ can be calculated in Now, $f(n)$ can be calculated in
@ -674,8 +668,8 @@ Using a similar idea in a weighted graph,
we can calculate for each pair of nodes the shortest we can calculate for each pair of nodes the shortest
path between them that contains exactly $n$ edges. path between them that contains exactly $n$ edges.
To calculate this, we have to define matrix multiplication To calculate this, we have to define matrix multiplication
in another way, so that we don't calculate the number in a new way, so that we do not calculate the numbers
of paths but minimize the length of a path. of paths but minimize the lengths of paths.
\begin{samepage} \begin{samepage}
As an example, consider the following graph: As an example, consider the following graph:
@ -700,8 +694,8 @@ As an example, consider the following graph:
\end{center} \end{center}
\end{samepage} \end{samepage}
Let's construct an adjacency matrix where Let us construct an adjacency matrix where
$\infty$ means that an edge doesn't exist, $\infty$ means that an edge does not exist,
and other values correspond to edge weights. and other values correspond to edge weights.
The matrix is The matrix is
\[ \[
@ -721,15 +715,16 @@ AB[i,j] = \sum_{k=1}^n A[i,k] \cdot B[k,j]
\] \]
we now use the formula we now use the formula
\[ \[
AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j], AB[i,j] = \min_{k=1}^n A[i,k] + B[k,j]
\] \]
for matrix multiplication, so we calculate for matrix multiplication, so we calculate
a minimum instead of a sum, a minimum instead of a sum,
and a sum of elements instead of a product. and a sum of elements instead of a product.
After this modification, After this modification,
matrix powers can be used for calculating matrix powers correspond to
shortest paths in the graph: shortest paths in the graph.
For example, as
\[ \[
V^4= \begin{bmatrix} V^4= \begin{bmatrix}
\infty & \infty & 10 & 11 & 9 & \infty \\ \infty & \infty & 10 & 11 & 9 & \infty \\
@ -738,9 +733,9 @@ V^4= \begin{bmatrix}
\infty & 8 & \infty & \infty & \infty & \infty \\ \infty & 8 & \infty & \infty & \infty & \infty \\
\infty & \infty & \infty & \infty & \infty & \infty \\ \infty & \infty & \infty & \infty & \infty & \infty \\
\infty & \infty & 12 & 13 & 11 & \infty \\ \infty & \infty & 12 & 13 & 11 & \infty \\
\end{bmatrix} \end{bmatrix},
\] \]
For example, the shortest path of 4 edges we can conclude that the shortest path of 4 edges
from node 2 to node 5 has length 8. from node 2 to node 5 has length 8.
This path is This path is
$2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$. $2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
@ -750,9 +745,9 @@ $2 \rightarrow 1 \rightarrow 4 \rightarrow 2 \rightarrow 5$.
\index{Kirchhoff's theorem} \index{Kirchhoff's theorem}
\index{spanning tree} \index{spanning tree}
\key{Kirchhoff's theorem} provides us a way \key{Kirchhoff's theorem} provides a way
to calculate the number of spanning trees to calculate the number of spanning trees
in a graph as a determinant of a special matrix. of a graph as a determinant of a special matrix.
For example, the graph For example, the graph
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.9] \begin{tikzpicture}[scale=0.9]
@ -802,12 +797,12 @@ has three spanning trees:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
\index{Laplacean matrix} \index{Laplacean matrix}
We construct a \key{Laplacean matrix} $L$, To calculate the number of spanning trees,
we construct a \key{Laplacean matrix} $L$,
where $L[i,i]$ is the degree of node $i$ where $L[i,i]$ is the degree of node $i$
and $L[i,j]=-1$ if there is an edge between and $L[i,j]=-1$ if there is an edge between
nodes $i$ and $j$, and otherwise $L[i,j]=0$. nodes $i$ and $j$, and otherwise $L[i,j]=0$.
In this graph, the matrix is as follows: The Laplacean matrix for the above graph is as follows:
\[ \[
L= \begin{bmatrix} L= \begin{bmatrix}
3 & -1 & -1 & -1 \\ 3 & -1 & -1 & -1 \\
@ -817,7 +812,7 @@ L= \begin{bmatrix}
\end{bmatrix} \end{bmatrix}
\] \]
The number of spanning trees is the same as The number of spanning trees equals
the determinant of a matrix that is obtained the determinant of a matrix that is obtained
when we remove any row and any column from $L$. when we remove any row and any column from $L$.
For example, if we remove the first row For example, if we remove the first row