MATH 323 Lecture 26

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Theorem 6.3.1

If ${\lambda }_1,\dots ,{\lambda }_k$ are distinct eigenvalues of an $n\times n$ matrix $A$ with corresponding eigenvectors ${\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_k$, then ${\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_2$ are linearly independent.

$\begin{array}{rl}V& =\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{{\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_k\}\\ \dim V& =r\end{array}$

Diagonalization

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

$X^{-1}AX=D=\left(\begin{array}{ccc}{\lambda }_1& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \dots & {\lambda }_n\end{array}\right)$

where $D$ is a diagonal matrix.

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

Theorem 6.3.2

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

$\begin{array}{rl}A:& {\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_n\text{linearly independent eigenvectors}\\ & {\lambda }_1,\dots ,{\lambda }_n\text{eigenvalues (not necessarily distinct)}\\ X& =({\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_n)\\ D& =\left[\begin{array}{ccc}{\lambda }_1& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \dots & {\lambda }_n\end{array}\right]\end{array}$

Corollary

1. If $A$ is diagonalizable and $A=XD{X}^{-1}$, then diagonal entries of $D$ are eigenvalues of $A$.
2. $X=({\overrightarrow{x}}_1,\dots ,{\overrightarrow{x}}_n)$, but $X$ is not unique.
3. If eigenvalues are distinct, then $A$ is diagonalizable
4. If they are not distinct, then $A$ may or may not be diagonalizable
5. If $A=XD{X}^{-1}$, then $A^k=X{D}^k{X}^{-1}=X\left(\begin{array}{ccc}{\lambda }_1^k& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \dots & {\lambda }_n^k\end{array}\right)X^{-1}$.

Example

$\begin{array}{rl}A& =\left[\begin{array}{cc}2& -3\\ 2& -5\end{array}\right]\\ {\lambda }_1& =1& {\lambda }_2& =-4\\ {\overrightarrow{x}}_1& =⟨3,1⟩& {\overrightarrow{x}}_2& =⟨1,2⟩\\ X& =\left(\begin{array}{cc}3& 1\\ 1& 2\end{array}\right)\\ X^{-1}AX=D& =\left(\begin{array}{cc}1& 0\\ 0& -4\end{array}\right)\\ & =\frac{1}{5}\left[\begin{array}{cc}2& -1\\ -1& 3\end{array}\right]\left[\begin{array}{cc}2& -3\\ 2& -5\end{array}\right]\left[\begin{array}{cc}3& 1\\ 1& 2\end{array}\right]\end{array}$

Matrix Exponential

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

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

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

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

${\mathrm{e}}^D=\left[\begin{array}{ccc}{\mathrm{e}}^{{\lambda }_1}& \dots & 0\\ ⋮& \ddots & ⋮\\ 0& \dots & {\mathrm{e}}^{{\lambda }_n}\end{array}\right]$

Example

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

${\mathrm{e}}^A=\left[\begin{array}{cc}-2& -3\\ 1& 1\end{array}\right]\left[\begin{array}{cc}\mathrm{e}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ -1& -2\end{array}\right]=\left[\begin{array}{cc}3-2\mathrm{e}& 6-6\mathrm{e}\\ \mathrm{e}-1& 3\mathrm{e}-2\end{array}\right]$

Application: Differential Equation

Solution to differential equation $y^{\prime }=ay$ is $y={\mathrm{e}}^{at}{y}_0$.

Similarly, the system of differential equations ${\overrightarrow{Y}}^{\prime }=A\overrightarrow{Y}$ has the solution $\overrightarrow{Y}={\mathrm{e}}^{tA}{\overrightarrow{Y}}_0$.

Final Exam Review

5 problems and an 11-point bonus problem

• Linear transformations (Null Space, Range, Basis)
• Orthogonality
• Orthogonalization (Gram-Schmidt process, $QR$ factorization)
• Eigenvalues and Eigenvectors
• Least Squares Problems

Note on Factoring Cubics

For a cubic polynomial $p(\lambda )=a{\lambda }^3+b{\lambda }^2+c\lambda +d$, if $p(m)=0$ for $m\in \mathbb{\mathbb{Z}}$, $m$ divides $d$:

$\begin{array}{rl}p(\lambda )& =a{\lambda }^3+b{\lambda }^2+c\lambda +d\\ & =(\alpha {\lambda }^2+\beta \lambda +\gamma )(\lambda -m)\end{array}$