Math 511: Linear Algebra

The Cayley-Hamilton Theorem

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The Cayley-Hamilton Theorem

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Definition

The characteristic polynomial of a square matrix $A$ is the determinant $|A-\lambda I|$. If matrix $A$ is an $n\times n$ matrix, then the characteristic polynomial $p(\lambda) = |A - \lambda I|$ is a polynomial of degree $n$ in variable $\lambda$.

$$ p(\lambda) = \det(A - \lambda I) = (-1)^n\lambda^n + c_{n-1}\lambda^{n-1} + \cdots + c_1\lambda + c_0 $$

As matrix $A$ is square, we may raise matrix $A$ to positive integer powers. So we can write

$$ p(A) = (-1)^nA^n + c_{n-1}A^{n-1} + \cdots + c_1A + c_0I_n $$

The Cayley-Hamilton Theorem says that every square matrix $A$ satisfies its characteristic polynomial. That means that for an $n\times n$ matrix $A$, $p(A) = O$ or

$$ p(A) = (-1)^nA^n + c_{n-1}A^{n-1} + \cdots + c_1A + c_0I_n = O $$

The $n\times n$ matrix on the right-hand side of the equation is the $n\times n$ zero matrix, and $I_n$ is the $n\times n$ identity matrix.

1.

Verify the Cayley-Hamilton Theorem for the matrix

$$ \begin{bmatrix}\ \ 2\ & -2\ \\ -2\ & -1\ \end{bmatrix} $$

2.

Verify the Cayley-Hamilton Theorem for the matrix

$$ \begin{bmatrix}\ \ 6\ &\ \ 0\ &\ \ 4\ \\ -2\ &\ \ 1\ &\ \ 3\ \\ \ \ 2\ &\ \ 0\ &\ \ 4\ \end{bmatrix} $$

Use the ten properties of determinants to easily compute your determinant.

3.

Verify the Cayley-Hamilton Theorem for a general $2\times 2$ matrix $A$.

$$ A = \begin{bmatrix}\ a\ &\ b\ \\ \ c\ &\ d\ \end{bmatrix} $$

4.

For a nonsingular $n\times n$ matrix $A$, show that

$$ A^{-1} = \frac{1}{c_0}\left((-1)^{n+1}A^{n-1} - c_{n-1}A^{n-2} - \cdots - c_2A - c_1I_n \right) $$

Use this result to compute the inverse of the following matrix using the characteristic polynomial, matrix $A$ and matrix $I_2$.

$$ A = \begin{bmatrix}\ 1\ &\ 2\ \\ \ 3\ &\ 5\ \end{bmatrix} $$

5.

The Cayley-Hamilton Theorem is useful for calculating the positive integer powers $A^n$ of a square matrix $A$. For example, the characteristic polynomial for the matrix

$$ A = \begin{bmatrix}\ 3\ & -1\ \\ \ 2\ & -1\ \end{bmatrix} $$

is

$$ \begin{align*} p(\lambda) &= \begin{vmatrix} 3-\lambda & -1 \\ 2 & -1-\lambda \end{vmatrix} = (3-\lambda)(-1-\lambda) + 2 \\ &= -3 + \lambda - 3\lambda + \lambda^2 + 2 \\ &= \lambda^2 -2\lambda - 1 \end{align*} $$

Using the Cayley-Hamilton Theorem,

$$ \begin{align*} A^2 - 2A - I_2 &= O \\ \\ A^2 &= 2A + I_2 \\ \\ A^2 &= 2\begin{bmatrix}\ 3\ & -1\ \\ \ 2 & -1\ \end{bmatrix} + \begin{bmatrix}\ 1\ &\ 0\ \\ \ 0\ &\ 1\ \end{bmatrix} \\ \\ &= \begin{bmatrix}\ 7\ & -2\ \\ \ 4\ & -1\ \end{bmatrix} \end{align*} $$

Multiplying both sides of equation $A^2 = 2A + I_2$ yields

$$ A^3 = 2A^2 + A = 2(2A + I) + A = 5A + 2I $$

(a) Write $A^4$ in terms of $A$ and $I$.

(b) Find $A^5$ for the following matrix using the characteristic polynomial, $I$, $A$, and $A^2$.

$$ A = \begin{bmatrix}\ \ 0\ &\ \ 0\ &\ \ 1\ \\ \ \ 2\ &\ \ 2\ & -1\ \\ \ \ 1\ &\ \ 0\ &\ \ 2\ \end{bmatrix} $$

Use the ten properties of determinants to compute your determinant easily.

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