On the computation of the nth power of a matrix

TL;DR

This paper explores methods for computing the nth power of a matrix, emphasizing the Cayley-Hamilton theorem and Gauss elimination to find the minimum polynomial, relevant to Markov chains and other areas.

Contribution

It highlights the importance of the Cayley-Hamilton theorem and demonstrates Gauss elimination as a practical tool for matrix power computation.

Findings

Cayley-Hamilton theorem is crucial for matrix power calculation.

Gauss elimination can determine the minimum polynomial without the characteristic polynomial.

The approach is applicable to Markov chains and related mathematical topics.

Abstract

In this note we discuss the problem of finding the nth power of a matrix which is strongly connected to the study of Markov chains and others mathematical topics. We observe the known fact (but maybe not well known) that the Cayley-Hamilton theorem is of key importance to this goal. We also demonstrate the classical Gauss elimination technique as a tool to compute the minimum polynomial of a matrix without necessarily know the characteristic polynomial.

Equations

$$k^n={\delta }_A(k)\pi (k)+v(k)$$

$$\displaystyle P=\left(\begin{array}[]{rrr}-3&6&0\\ 2&1&0\\ 0&0&3\\ \end{array}\right)$$

$$\displaystyle P^{n}=aP^{2}+bP+c=\left(\begin{array}[]{ccc}21a-3a+c&-12a+6b&0\\ -4a+2b&13a+b+c&0\\ 0&0&9a+3b+c\\ \end{array}\right)$$

$$\displaystyle P=\left(\begin{array}[]{ccc}0&1&0\\[5.69046pt] 0&\frac{2}{3}&\frac{1}{3}\\[5.69046pt] \frac{1}{3}&0&\frac{2}{3}\\ \end{array}\right)$$

$$k_2$$

$$k_3$$

$$1^n$$

$$\frac{1}{3^n}\left(\cos \frac{n\pi }{3}+i\sin \frac{n\pi }{3}\right)$$

$$\frac{1}{3^n}\left(\cos \frac{n\pi }{3}-i\sin \frac{n\pi }{3}\right)$$

$$\frac{1}{3^n}\cos \frac{n\pi }{3}=a\frac{1}{3^2}\cos \frac{2\pi }{3}+b\frac{1}{3}\cos \frac{\pi }{3}+c$$

$$\frac{1}{3^n}\sin \frac{n\pi }{3}=a\frac{1}{3^2}\sin \frac{2\pi }{3}+b\frac{1}{3}\sin \frac{\pi }{3}$$

$$1^n$$

$$\frac{1}{3^n}\cos \frac{n\pi }{3}$$

$$\frac{1}{3^n}\sin \frac{n\pi }{3}$$

$$a$$

$$b$$

$$c$$

$$q(k)=k^r+a_{r-1}{k}^{r-1}+\cdots +a_0$$

$$b_{r-1}{A}^{r-1}+\cdots +b_0{I}_{m\times m}=0_{m\times m}$$

$$I_{m\times m},A,\cdots ,A^{r-1},A^r,A^{r+1},\cdots ,A^m$$

$$\displaystyle\hat{C}=\left(\begin{array}[]{cccc}1&0&0&\cdots\\ 0&1&0&\cdots\\ 0&0&1&\cdots\\ \vdots&\vdots&\vdots&\vdots\\ 0&0&0&0\\ 0&0&0&0\\ \end{array}\right)$$

$$a_n{A}^n+a_{n-1}{A}^{n-1}+\cdots +a_0{I}_{n\times n}=0_{n\times n}$$

$$q(k)=a_0+a_{1}k+\cdots +a_{r-1}{k}^{r-1}+k^r$$

$$q(k)=k^r-{\hat{B}}_{r,r+1}{k}^{r-1}-{\hat{B}}_{r-1,r+1}{k}^{r-2}-\cdots -{\hat{B}}_{1,r+1}$$

$$\displaystyle A=\left(\begin{array}[]{rrr}-4&2&0\\ -2&-1&0\\ 0&0&1\\ \end{array}\right)$$

$$\displaystyle B=\left(\begin{array}[]{rrrrr}1&-4&12&-28\\ 0&-2&10&-34\\ 0&0&0&0\\ 0&2&-10&34\\ 1&-1&-3&23\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 1&1&1&1\\ \end{array}\right)$$

$$\displaystyle\hat{B}=\left(\begin{array}[]{rrrr}1&0&0&8\\ 0&1&0&-3\\ 0&0&1&-4\\ 0&0&0&0\\ \vdots&\vdots&\vdots&\vdots\\ \end{array}\right)$$

$$q(k)=\delta (k)=k^3+4k^2+3k-8$$

$$\displaystyle A=\left(\begin{array}[]{rrr}-3&6&0\\ 2&1&0\\ 0&0&3\\ \end{array}\right)$$

$$\displaystyle B=\left(\begin{array}[]{rrrrr}1&-3&21&-87\\ 0&2&-4&38\\ 0&0&0&0\\ 0&6&-12&114\\ 1&1&13&-11\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 1&3&9&27\\ \end{array}\right)$$

$$\displaystyle\hat{B}=\left(\begin{array}[]{rrrr}1&0&15&-30\\ 0&1&-2&19\\ 0&0&0&0\\ \vdots&\vdots&\vdots&\vdots\\ \end{array}\right)$$

$$q(k)=k^2+2k-15=(k+5)(k-3)$$

Full text

Keywords matrix, nth power, Cayley-Hamilton theorem, minimum polynomial

2010 Mathematics Subject Classification

1 Using Cayley-Hamilton theorem to find the nth power of a matrix

Let $A_{m\times m}$ be a real matrix and suppose that we want to compute the nth power. Let us denote by $\delta(k)$ the characteristic polynomial of $A$. Then it holds that

[TABLE]

where $v(k)$ is a polynomial of degree less or equal to $m-1$. Using the Cayley-Hamilton theorem we see that $A^{n}=v(A)$ since $\delta(A)=0$. Therefore, in order to find the form of $A^{n}$ we have to find the coefficients of $v(k)$ which are at most $m$. In this direction we will use the eigenvalues $k_{1},\cdots,k_{n}$ (not necessarily distinct) of the matrix $A$ since $\delta(k_{i})=0$. Setting $k=k_{i}$ we produce some equations involving the unknown coefficients of $v$. Of course if some of the eigenvalues are of multiplicity two or more then we have to produce some more equations. This can be done by differentiating equation 1 and setting $k=k_{i}$. We differentiate this equation $m_{i}-1$ times (where $m_{i}$ is the multiplicity of $k_{i}$ eigenvalue) and each time we set $k=k_{i}$ in order to produce one more equation. We do this for any eigenvalue $k_{j}$ with multiplicity $m_{j}$. In this fashion we produce $m$ different equations in order to determine the $m$ unknown coefficients of $v$. One can easily observe that the computation of the coefficients of $v(k)$ is in fact an interpolation problem which has a unique solution (see [2], Thm. 2.2.2). Therefore, we can always find the nth power of matrix by using the Cayley-Hamilton theorem even if the matrix is not diagonizable. Below, we give some examples.

Example 1

We will compute the nth power of the matrix

[TABLE]

The eigenvalues are $k_{1}=3$ with multiplicity two and $k_{2}=-5$ that is the characteristic polynomial is $\delta_{P}(k)=(k-3)^{2}(k+5)$.

Therefore, it holds that $k^{n}=t(k)\delta_{P}(k)+v(k)$ where $v(k)=ak^{2}+bk+c$. We have to produce three equations in order to determine the unknown coefficients. Setting $k_{1}=3$ we obtain the first equation $3^{n}=9a+3b+c$ and setting $k=-5$ we produce another one which is $(-1)^{n}5=25a-5b+c$. In order to have one more equation we differentiate the equality $k^{n}=t(k)\delta_{P}(k)+v(k)$ and then we set again $k=3$ because this eigenvalue has multiplicity two. Therefore, we obtain the third equation which is $3^{n-1}n=6a+b$. Next we solve for $a,b,c$ and thus

[TABLE]

Finally, one can verify the result by induction.

Example 2

We will compute the nth power of the following matrix

[TABLE]

This matrix has the following eigenvalues, $k_{1}=1$, $k_{2}=\frac{1+i\sqrt{3}}{6}$ and $k_{3}=\frac{1-i\sqrt{3}}{6}$. We have the following equation $k^{n}=\delta_{P}(k)t(k)+v(k)$ where $v(k)=ak^{2}+bk+c$. Next, we will produce three equations in order to evaluate the unknown coefficients. We transform every complex number in the polar form, that is $x+iy=r(\cos\phi+i\sin\phi)$ where $r=\sqrt{x^{2}+y^{2}}$ and $\phi=\arctan\frac{y}{x}$. Therefore we have that

[TABLE]

We obtain the following equations by setting $k=k_{i}$ in the equation $k^{n}=\delta_{P}(k)t(k)+v(k)$

[TABLE]

Adding the second equation to the third we produce the equality

[TABLE]

while subtracting the third equation from the second we produce the following

[TABLE]

Finally, we have the following system of equations

[TABLE]

Solving for $a,b,c$ we obatin

[TABLE]

and therefore $P^{n}=aP^{2}+bP+cI_{2\times 2}$. One can verify the result by induction.

2 Minimum polynomial

Recall that the minimum polynomial of $A$ is the polynomial $q(k)$ of less degree such that $q(A)=0$. Therefore one can use the minimum polynomial rather than the characteristic one in order to compute the nth power of the matrix $A$.

We will demonstrate the classical Gauss elimination procedure in order to find the minimum polynomial.

Let the minimum polynomial has the form

[TABLE]

$r\leq m$.

Obviously the relation

[TABLE]

drive us to the conclusion that $b_{r-1}=b_{r-2}=\cdots=b_{0}=0$ otherwise there will be a polynomial $\hat{q}$ such that $\hat{q}(A)=0$ with less degree than the minimum polynomial which is a contradiction. That means that the matrices $I_{m\times m},A,\cdots,A^{r-1}$ are linearly independent while the matrices $I_{n\times n},A,\cdots,A^{r-1},A^{r}$ are linearly dependent. Obviously the matrices

[TABLE]

are also linearly dependent.

Let $C_{m^{2}\times r}$ the matrix that has in its first column the identity matrix $I$, that is at the first $m$ places of the first column of $C$ we put the first column of $I$, next at the next $m$ places of the first column of $C$ we put the second column of $I$ and so on. We do the same for the matrices $A,A^{2},\cdots,A^{r-1}$. This matrix is the matrix of the system 5. Since the matrices $I,A,A^{2},\cdots,A^{r-1}$ are linearly dependent then the reduced row echelon form of $C$ will have the following form

[TABLE]

where the number of leading 1 equals to $r$.

Let now $D_{m^{2}\times(m+1-r)}$ the matrix that has in its columns the matrices $A^{r},\cdots,A^{m}$ with the same fashion as with $C$. We construct next the matrix $B=(C|D)$ (which is in fact the matrix of system 7 below) and we compute the reduced row echelon form. Obviously, the number of the leading 1 equals $r$ again and these are located at the $r$ first columns of $B$.

That means that the system

[TABLE]

has infinite many solutions with $n+1-r$ free parameters. Setting $a_{r}=1$, $a_{r+1}=a_{r+2}=\cdots=a_{n}=0$ and solving for the others we obtain the minimum polynomial

[TABLE]

One can easily verify that the minimum polynomial is as follows

[TABLE]

Example 3

Let the matrix

[TABLE]

We will compute the minimum polynomial.

We construct the matrix $B$ by using the matrices $I,A,A^{2},A^{3}$. Then

[TABLE]

end the reduced row echelon matrix is the following

[TABLE]

The number of the leading 1 is 3 and that means that the minimum polynomial if of third degree. Thus,

[TABLE]

and in this case coincides with the characteristic polynomial.

Example 4

Let the matrix

[TABLE]

We construct the matrix $B$ as before and therefore we have

[TABLE]

The reduced row echelon matrix of $B$ is

[TABLE]

The number of the leading 1 is 2 therefore the degree of the minimum polynomial is 2. Setting $a_{2}=1$, $a_{3}=0$ and solving for the other we get

[TABLE]

That means that the matrix $A$ has the $k_{1}=-5$ and $k_{2}=3$ as eigenvalues.