MATH 323 Lecture 26

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Theorem 6.3.1

If $\lambda _{1},\ldots ,\lambda _{k}$ are distinct eigenvalues of an $n\times n$ matrix $A$ with corresponding eigenvectors ${\vec {x}}_{1},\ldots ,{\vec {x}}_{k}$ , then ${\vec {x}}_{1},\ldots ,{\vec {x}}_{2}$ are linearly independent.

${\begin{aligned}V&=\mathrm {Span} \{{\vec {x}}_{1},\ldots ,{\vec {x}}_{k}\}\\\dim V&=r\end{aligned}}$

Diagonalization

$A$ is said to be diagonalizable if there is a nonsingular matrix $X$ with

X^{-1}AX=D={\begin{pmatrix}\lambda _{1}&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &\lambda _{n}\end{pmatrix}}

where $D$ is a diagonal matrix.

Therefore $A\sim D$ by $A=XDX^{-1}$ .

Theorem 6.3.2

An $n\times n$ matrix $A$ is diagonalizable iff $A$ has $n$ linearly independent eigenvectors.

${\begin{aligned}A:&{\vec {x}}_{1},\ldots ,{\vec {x}}_{n}\quad {\text{linearly independent eigenvectors}}\\&\lambda _{1},\ldots ,\lambda _{n}\quad {\text{eigenvalues (not necessarily distinct)}}\\X&=({\vec {x}}_{1},\ldots ,{\vec {x}}_{n})\\D&={\begin{bmatrix}\lambda _{1}&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &\lambda _{n}\end{bmatrix}}\end{aligned}}$

Corollary

If $A$ is diagonalizable and $A=XDX^{-1}$ , then diagonal entries of $D$ are eigenvalues of $A$ .
$X=({\vec {x}}_{1},\ldots ,{\vec {x}}_{n})$ , but $X$ is not unique.
If eigenvalues are distinct, then $A$ is diagonalizable
If they are not distinct, then $A$ may or may not be diagonalizable
If $A=XDX^{-1}$ , then $A^{k}=XD^{k}X^{-1}=X{\begin{pmatrix}\lambda _{1}^{k}&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &\lambda _{n}^{k}\end{pmatrix}}X^{-1}$ .

Example

${\begin{aligned}A&={\begin{bmatrix}2&-3\\2&-5\end{bmatrix}}\\\lambda _{1}&=1&\lambda _{2}&=-4\\{\vec {x}}_{1}&=\left\langle 3,1\right\rangle &{\vec {x}}_{2}&=\left\langle 1,2\right\rangle \\X&={\begin{pmatrix}3&1\\1&2\end{pmatrix}}\\X^{-1}AX=D&={\begin{pmatrix}1&0\\0&-4\end{pmatrix}}\\&={\frac {1}{5}}{\begin{bmatrix}2&-1\\-1&3\end{bmatrix}}\,{\begin{bmatrix}2&-3\\2&-5\end{bmatrix}}\,{\begin{bmatrix}3&1\\1&2\end{bmatrix}}\end{aligned}}$

Matrix Exponential

$\mathrm {e} ^{x}=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}$ is a very important function.

We similarily define $\mathrm {e} ^{A}$ for a matrix $A$ to be

\mathrm {e} ^{A}=\sum _{n=0}^{\infty }{\frac {1}{n!}}A^{n}

If A is diagonalizable, then $\mathrm {e} ^{A}=X\mathrm {e} ^{D}X^{-1}$ , where $\mathrm {e} ^{D}$ is given by

$\mathrm {e} ^{D}={\begin{bmatrix}\mathrm {e} ^{\lambda _{1}}&\dots &0\\\vdots &\ddots &\vdots \\0&\dots &\mathrm {e} ^{\lambda _{n}}\end{bmatrix}}$

Example

$A={\begin{bmatrix}-2&-6\\1&3\end{bmatrix}}={\begin{bmatrix}-2&-3\\1&1\end{bmatrix}}\,{\begin{bmatrix}1&0\\0&0\end{bmatrix}}\,{\begin{bmatrix}1&3\\-1&-2\end{bmatrix}}$

$\mathrm {e} ^{A}={\begin{bmatrix}-2&-3\\1&1\end{bmatrix}}\,{\begin{bmatrix}\mathrm {e} &0\\0&1\end{bmatrix}}\,{\begin{bmatrix}1&3\\-1&-2\end{bmatrix}}={\begin{bmatrix}3-2\mathrm {e} &6-6\mathrm {e} \\\mathrm {e} -1&3\mathrm {e} -2\end{bmatrix}}$