Matrix Transformation and Eigen value

$\spadesuit$Transition Matrix $\spadesuit$

Given a linear mappling $T$, then how to choose the basis of ${\mathcal R}^{n}$ and ${\mathcal R}^{m}$, the matrix representation of $T$ changes. In this section, we study the relationship between the matrix called a transition matrix which maps from the basis of ${\mathcal R}^{n}$ to the basis of ${\mathcal R}^{m}$ and the matrix representation.

Example 3..5  

Let $T : {\mathcal R}^{2} \longrightarrow {\mathcal R}^{2}$ be given by $T\left(\begin{array}{c} x\\ y \end{array}\right) = \left(\begin{array}{c} 3x - 4y\\ x + 5y \end{array}\right)$. Find the transition matrix $P$ which maps from the usual basis of $\{{\bf e}_{1},{\bf e}_{2}\}$ of $R^{2}$ to the basis $\{{\bf w}_{1} = \left(\begin{array}{r} 1\\ 3 \end{array}\right), {\bf w}_{2} = \left(\begin{array}{r} 2\\ 5 \end{array}\right) \}$. For $A = [T]_{\bf e}$, find $P^{-1}AP$.

Answer First of all, we find the matrix so that $P{\bf e}_{1} = {\bf w}_{1}, P{\bf e}_{2} = {\bf w}_{2}$.

$${\bf w}_{1} = \left(\begin{array}{r} 1\\ 3 \end{array}\right) = {\bf e}_{1} + 3{\bf e}_{2},$$

$${\bf w}_{2} = \left(\begin{array}{r} 2\\ 5 \end{array}\right) = 2{\bf e}_{1} + 5{\bf e}_{2}$$

Thus the transition matrix $P$ is given by

$$P = \left(\begin{array}{rr} 1&2\\ 3&5 \end{array}\right).$$

Next we find $P^{-1}AP$. By Example 3.1, $A = [T]_{\bf e} = \left(\begin{array}{cc} 3 & -4\\ 1 & 5 \end{array}\right)$. Thus,

$$P^{-1}AP = \left(\begin{array}{rr} -5&2\\ 3&-1 \end{array}\right... ...n{array}{rr} 77&124\\ -43&-69 \end{array}\right). \ensuremath{ \blacksquare}$$

Look this example carefully. Then $P^{-1}[T]_{\bf e}P = [T]_{\bf w}$. Is this always true?

Theorem 3..5  

Let $A$ be a matrix representation of the linear transformation $T : {\mathcal R}^{n} \longrightarrow {\mathcal R}^{n}$ relative to the basis $\{{\bf v}_{i}\}$ and $B$ be a matrix representation relative to the basis $\{{\bf w}_{i}\}$. Then the transition matrix $P$ from the basis $\{{\bf v}_{i}\}$ to the basis $\{{\bf w}_{i}\}$ satisfies $B = P^{-1}AP$.

Proof By Exercise3.2, the transposition matrix $P$ from the basis $\{{\bf v}_{i}\}_{i=1}^{n}$ to the basis $\{{\bf w}_{i}\}_{i=1}^{n}$ is regular matrix of the order $n$ and $P = (p_{ij})$ is given by the following:

$${\bf w}_{j} = p_{1j}{\bf v}_{1} + \cdots + p_{nj}{\bf v}_{n}, (j = 1,2,\ldots, n)$$

. Suppose that $A = (a_{ij}), B = (b_{ij})$. Then

$$T({\bf v}_{i}) = a_{1i}{\bf v}_{1} + \cdots + a_{mi}{\bf v}_{m}, (i = 1,2,\ldots,n) ,$$

$$T({\bf w}_{j}) = b_{1j}{\bf w}_{1} + \cdots + b_{mj}{\bf w}_{m}, (j = 1,2,\ldots,n)$$

implies that

$$(T({\bf v}_{1}),\ldots,T({\bf v}_{n})) = ({\bf v}_{1},\ldots,{\bf v}_{n})A,$$

$$(T({\bf w}_{1}),\ldots,T({\bf w}_{n})) = ({\bf w}_{1},\ldots,{\bf w}_{n})B$$

Then

$$(T({\bf w}_{1}),\ldots,T({\bf w}_{n})) = (T({\bf v}_{1}),\ldots,T({\bf v}_{n}))P$$

$$= (({\bf v}_{1},\ldots,{\bf v}_{n})A)P$$

$$= ({\bf v}_{1},\ldots,{\bf v}_{n})AP$$

$$= (({\bf w}_{1},\ldots,{\bf w}_{n})P^{-1})(AP)$$

$$= ({\bf w}_{1},\ldots,{\bf w}_{n})(P^{-1}AP)$$

Thus we have

$$P^{-1}AP = B$$

. $\blacksquare$

Suppose that $A$ and $B$ are square matrices for which there exists an invertible matrix $P$ such that $P^{-1}AP = B$. Then $B$ is said to be similar and denoted by $A \sim B$.

In the rest of the chapter, we study how to find a manageable form of matrix by choosing the regular matrix $P$ so that $P^{-1}AP$ is canonical form.

$\spadesuit$Eigenvalues and Eigenvectors $\spadesuit$

In this chapter we investigate the theory of a single linear operator $T$ on a vector space $V$ of finite dimension. In particular, we find conditions under which $T$ is diagonalizable.

A set complex numbers is denoted by ${ C}$ and the set of $n$ complex numbers is denoted by ${ C}^{n} = \{(c_{1},c_{2},\ldots,c_{n}), c_{i} \in { C} \}$.

Let $A$ be a square matrix of the order $n$. Then for $\lambda \in { C}$,

$$A{\mathbf x} = \lambda {\mathbf x} ({\mathbf x} \in { C}^{n}, {\mathbf x} \neq {\bf0}).$$

Then we say $\lambda$ is a eigenvalue of $A$ and ${\mathbf x}$ is a eigenvector of $A$ corresponds to $\lambda$.

For example, consider the linear transformation $T$ which maps the line $y = x + 2$to the line $Y = X$. This is a translation in the $y$ direction. Thus the vector $\left(\begin{array}{c} 0\\ y \end{array}\right)$ is translated to $\lambda \left(\begin{array}{c} 0\\ y \end{array}\right)$. The scalar $\lambda$ is the eigenvalue. Now we study how to obtain an eigenvalue and eigenvector.

Rewrite the equation $A{\mathbf x} = \lambda {\mathbf x}$. Then we have

$$(A - \lambda I){\mathbf x} = {\bf0}$$

This is homogeneous system. The necessary and sufficient conditions for which the eigen vector ${\mathbf x}$ exists is the system of linear equation has nonzero solution. By the theorem 2.5,

$$\vert A - \lambda I\vert = 0$$

Thus, the eigenvalue $\lambda$ can be found by the solution of the following equation with $t$ unknown:

$$\vert A - tI\vert = 0$$

Conversely, The solution $\lambda$ is the eigenvalue of $A$. Now the polynomial in $t$

$$\Phi_{A}(t) = \det(A - tI) = t^{n} + c_{1}t^{n-1} + \cdots + c_{n}$$

is called an characteristic polynomial of $A$ and the equation in $t$

$$\Phi_{A}(t) = \det(A - tI) = 0$$

is called an characteristic equation of $A$. From this, to find an eigenvalue, it is enough to solve the characteristic equation.

Example 3..6  

Find all eigenvalues of $A = \left(\begin{array}{rrr} 3&0&0\\ 0&2&-5\\ 0&1&-2 \end{array}\right)$.

Answer $\Phi_{A}(t) = \left \vert\begin{array}{rrr} 3-t & 0 & 0\\ 0 & 2-t & -5\\ 0 & 1 & -2-t \end{array}\right \vert = (3-t)(t^{2} + 1) = 0$. Thus the eigen values of $A$ are $\lambda = 3, \pm i .$ $\blacksquare$

As you can see even though the entries of the matrix are all real number, the eigenvalues might be complex numbers.

Example 3..7  

Find all eigenvalues and corresponding eigenvectors of $A = \left(\begin{array}{rrr} 0&1&1\\ 1&0&1\\ 1&1&0 \end{array}\right)$.

Answer $\Phi_{A}(t) = \left \vert\begin{array}{rrr} -t & 1 & 1\\ 1 & -t & 1\\ 1 & 1 & -t \end{array}\right \vert = -(t+1)^{2}(t-2).$ Thus, the eigenvalues of $A$ are $\lambda = 2,-1$. Now we find the eigenvector corresponding $\lambda$ of $A$.

For $\lambda = 2$, the eigenvector satisfies $(A - 2I){\mathbf x} = {\bf0}$ and nonzero. Solve this equation. We have

$$A - 2I = \left(\begin{array}{rrr} -2&1 & 1\\ 1& -2 & 1\\ 1&1&-2 \end{arr... ...htarrow} \left(\begin{array}{rrr} 1&1&-2\\ 1&-2&1\\ -2&1&1 \end{array}\right)$$

$$\stackrel{\begin{array}{cc} {}^{-R_{1} + R_{2}}\\ {}^{2R_{1} + R_{3}} \end{array}}{\longrightarrow} \left(\begin{array}{rrr} 1&1&-2\\ 0&-3&3\\ 0 & 3&-3 \end{array}... ...rrow} \left(\begin{array}{rrr} 1&1&-2\\ 0&1&-1\\ 0 & 0&0 \end{array}\right) .$$

Then the degree of freedom is 1. So let $x_{3} = \alpha$. Then

$${\mathbf x} = \left(\begin{array}{c} x_{1}\\ x_{2}\\ x_{3} \end... ...alpha \left(\begin{array}{c} 1\\ 1\\ 1 \end{array}\right) (\alpha \neq 0).$$

now for $\lambda = -1$, the eigenvector satisfies $(A - 3I){\mathbf x} = {\bf0}$ and nonzero. Solve this equation. We have

$$A+ I = \left(\begin{array}{rrr} 1&1&1\\ 1&1&1\\ 1&1&1 \end{arra... ...htarrow} \left(\begin{array}{rrr} 1&1&1\\ 0&0&0\\ 0 & 0&0 \end{array}\right)$$

Thus the degree of freedom is 2. So we let $x_{2} = \beta \neq 0, x_{3} = \gamma \neq 0$. Then

$${\mathbf x} = \left(\begin{array}{c} -x_{2} - x_{3}\\ x_{2}\\ x... ...(\begin{array}{r} -1\\ 0\\ 1 \end{array}\right). \ensuremath{ \blacksquare}$$

$\spadesuit$Cayley-Hamilton Theorem $\spadesuit$

For $\Phi_{A}(t) = t^{n} + c_{1}t^{n-1} + \cdots + c_{n}$, define

$$\Phi_{A}(A) = A^{n} + c_{1}A^{n-1} + \cdots + c_{n}I$$

Then we have the following theorem. This theorem is called (Cayley-Hamilton theorem).

Theorem 3..6  

[Cayley-Hamilton Theorem] Every matrix is a zero its characteristic polynomial. In other words, $\Phi_{A}(A) = {\bf0}$.

Proof Let $A$ be an arbitrary $n$-square matrix and let $\Phi_{A}(t)$ be its characteristic polynonmial; say,

$$\Phi_{A}(t) = \det(A - tI) = t^{n} + c_{1}t^{n-1} + \cdots + c_{n}$$

Now let $B(t)$ denote the ajoint of the matrix $A - t I$. The elements of $B(t)$ are cofactors of the matrix $A - t I$ and hence are polynomials in $t$ of degree not exceeding $n-1$. Thus,

$$B(t) = B_{1}t^{n-1} + B_{2}t^{n-2} + \cdots + B_{n}$$

Here $B_{i} (i = 1,2,\ldots,n)$ is $n$-square matrix. By the fundamental property of the classical ajoint

$$(A - tI)B(t) = \Phi_{A}(t)I$$

or

$$(A - tI)(B_{1}t^{n-1} + B_{2}t^{n-2} + \cdots + B_{n}) = ( t^{n} + c_{1}t^{n-1} + \cdots + c_{n})I .$$

Removing parentheses and equating the coefficients of corresponding powers ot $t$,

$$\begin{array}{rrl} -B_{1} &=& I,\\ -B_{2} + AB_{1} &=& c_{1}I,\\ \vdots&&\vdots\\ AB_{n}&=&c_{n}I \end{array}$$

Multiplying the above matrix equations by $A^{n},A^{n-1},\ldots,I$ respectively,

$${\bf0} = A^{n} + c_{1}A^{n-1} + \cdots + c_{n}I = \Phi_{A}(A). \ensuremath{ \blacksquare}$$

$\spadesuit$Eigenspace $\spadesuit$

Our purpose here is to find the regular matrix $P$ so that the matrix $A$ can be tranposed to a simpler matrix. In other words, to find the regular matrix $P$ such that $B = P^{-1}AP$.

The set of vectors such that

$$V(\lambda) = \{{\mathbf x} : (A - \lambda I){\mathbf x} = {\bf0}\}$$

is called an eigenspace corresponds to $\lambda$. This vector space is the same as the solution space formed by solution vectors ${\mathbf x}$ so that

$$(A - \lambda I){\mathbf x} = {\bf0}$$

Thus by the theorem 2.3, we have

$$\dim V(\lambda) = n - {\rm rank}(A - \lambda I)$$

Exercise3-4

1. Find the transposed matrix $P$ which maps the basis $\{{\bf v}_{1} = \left(\begin{array}{r} 1\\ -1 \end{array}\right), {\bf v}_{2} = \left(\begin{array}{r} 1\\ 1 \end{array}\right) \}$ of ${\mathcal R}^{2}$ to the basis $\{{\bf w}_{1} = \left(\begin{array}{r} 3\\ 1 \end{array}\right), {\bf w}_{2} = \left(\begin{array}{r} -1\\ 2 \end{array}\right) \}$.

2. Show that the transposed matrix $P$ which maps the basis $\{{\bf v}_{i}\}$ to the basis $\{{\bf w}_{i}\}$ of $R^{n}$ is regular.

3. Find all eigenvalues and all eigenvetors of the following matrices.

(a) $\left(\begin{array}{rr} 3&-1\\ 1&1 \end{array}\right)$ (b) $\left(\begin{array}{rrr} 2&1&0\\ 0&1&-1\\ 0&2&4 \end{array}\right)$ (c) $\left(\begin{array}{rrr} 1&4&-4\\ -1&-3&2\\ 0&2&-1 \end{array}\right)$

4. Find the eigenvalue of the square matrix $A$ which satisfies $A^{2} = A$

5. Let the eigenvalues of $A$ be $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}$. Then show that the eigenvalues of $A^{m}$ are $\lambda_{1}^{m}, \lambda_{2}^{m}, . . , \lambda_{n}^{m}$.

6. Given $A = \left(\begin{array}{rr} 3&1\\ -1&1 \end{array}\right)$. Find $A^{4},A^{-1}$ using Cayley-Hamilton theorem.

7. Suppose $X$ is the matrix of order 2. Find all $X$ satisfying $X^{2} - 3X + 2I = 0$.