MATH 323 Lecture 25

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Eigenvalues and Eigenvectors

Example

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

$p(\lambda )=\left|\begin{array}{ccc}2-\lambda & -3& 1\\ 1& -2-\lambda & 1\\ 1& -3& 2-\lambda \end{array}\right|=-\lambda (\lambda -1{)}^2$

Therefore, the eigenvalues are ${\lambda }_1=0$, ${\lambda }_2={\lambda }_3=1$

To find eigenvectors, find basis for $N(A-\lambda I)$:

1. $N(A-0\cdot I)=N(A)=⟨\alpha ,\alpha ,\alpha ⟩\alpha \in \mathbb{ℝ}$
2. $N(A-1\cdot I)=⟨3\alpha -\beta ,\alpha ,\beta ⟩$

Therefore, $⟨1,1,1⟩$ is an eigenvector belonging to ${\lambda }_1=0$, and $⟨3,1,0⟩$ and $⟨-1,0,1⟩$ are linearly independent eigenvectors belonging to ${\lambda }_2={\lambda }_3=1$.

Complex Eigenvalues

For an arbitrary $n\times n$ matrix, the characteristic polynomial will be $(-1{)}^n{\lambda }^n+\mathrm{\Theta }({\lambda }^{n-1})$.

Polynomials of degree $n$ always have exactly $n$ complex roots:

$\begin{array}{rl}p(\lambda )& =(a_{11}-\lambda )\dots (a_{nn}-\lambda )+\dots \\ & =(-1{)}^n(\lambda -{\lambda }_1)\dots (\lambda -{\lambda }_n)+\mathrm{\Theta }({\lambda }^{n-1})& =({\lambda }_1-\lambda )\dots ({\lambda }_n-\lambda )+\mathrm{\Theta }({\lambda }^{n-1})\end{array}$

$p(0)=\prod _{i=1}^n{\lambda }_i$ is the constant term of $p(\lambda )$.

The coefficient of the remaining ${\lambda }^{n-1}$ is given by $(-1{)}^{n-1}\sum _{i=1}^n{a}_{ii}$. This is called the trace of $A$.

Example

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

$\begin{array}{rl}{\lambda }_1& =1+2i\\ {\lambda }_2& =1-2i\end{array}$

$(A-{\lambda }_{1}I)\overrightarrow{x}=0⟶\left[\begin{array}{cc}1& i\\ 0& 0\end{array}\right]$

Therefore, $⟨-i,1⟩$ is an eigenvector belonging to ${\lambda }_1=1+2i$. If we multiply this eigenvector by $i$, we get $⟨1,i⟩$, which is a much nicer form.

Theorem 6.1.1

Let $A$ and $B$ be $n\times n$ matrices.

If $B$ is similar to $A$, then $A$ and $B$ have the same characteristic polynomials and the same eigenvalues (not eigenvectors).

Proof

Recall that $A\sim B⟺B=S^{-1}AS$

$p_B(\lambda )=|B-\lambda I|=|S^{-1}AS-\lambda I|=|S^{-1}(A-\lambda I)S|=|S^{-1}||A-\lambda I||S|=|A-\lambda I|=p_A(\lambda )$

Review: Complex Numbers

Every complex number $z=a+bi$ has a conjugate $\overline{z}=a-bi$.

A polynomial that has a root $z=a+bi$ will also have the conjugate as a root

corollary: if $\lambda =a+bi$ is an eigenvalue of a real matrix $A$, then $\overline{\lambda }=a-bi$ is also an eigenvalue.

if $z=a+bi$ is an eigenvector of complex eigenvalue ${\lambda }_1$, then $\overline{z}=a-bi$ is an eigenvector of the conjugate eigenvalue ${\overline{\lambda }}_1$.

multiplying a complex by its conjugate will always yield a real number

Conjugate of a Product of complex numbers is just the product of conjugates

Solving Systems of Linear Differential Equations

A linear differential equation is of the form

$\begin{array}{rl}{y_1}^{\prime }& =a_{11}{y}_1+\dots a_{1n}{y}_n\\ {y_2}^{\prime }& =a_{21}{y}_1+\dots a_{2n}{y}_n\\ \dots \\ {y_n}^{\prime }& =a_{n1}{y}_1+\dots a_{nn}{y}_n\end{array}$

Where $y_i$ are functions in $C^1[a,b]$.

Which can be written in the form ${\overrightarrow{Y}}^{\prime }=A\overrightarrow{Y}$.

Suppose $\overrightarrow{Y}(t)$ is of the form $⟨x_1{\mathrm{e}}^{\lambda t},x_2{\mathrm{e}}^{\lambda t},\dots ,x_n{\mathrm{e}}^{\lambda t}⟩$.

Then $\overrightarrow{Y}(t)={\mathrm{e}}^{\lambda t}\overrightarrow{x}$ and ${\overrightarrow{Y}}^{\prime }(t)=\lambda \overrightarrow{Y}(t)=\lambda {\mathrm{e}}^{\lambda t}\overrightarrow{x}$.

If $\overrightarrow{x}$ is an eigenvector of $A$, then $A\overrightarrow{Y}=A{\mathrm{e}}^{\lambda t}\overrightarrow{x}={\mathrm{e}}^{\lambda t}A\overrightarrow{x}={\mathrm{e}}^{\lambda t}\lambda x$

The set of solutions to ${\overrightarrow{Y}}^{\prime }=A\overrightarrow{Y}$ is an $n$-dimensional subspace, and if solutions ${\overrightarrow{Y}}_1,\dots ,{\overrightarrow{Y}}_n$ are linearly independent solutions, then any linear combination of ${\overrightarrow{Y}}_i$'s are also solutions.

Example

$\begin{array}{rl}{y_1}^{\prime }& =3y_1+4y_2\\ {y_2}^{\prime }& =3y_1+2y_2\end{array}$

$A=\left[\begin{array}{cc}3& 4\\ 3& 2\end{array}\right]$

${\lambda }_1=6,{\lambda }_2=-1$

$\begin{array}{rl}{\overrightarrow{x}}_1& =⟨4,3⟩\\ {\overrightarrow{x}}_2& =⟨1,-1⟩\\ {\overrightarrow{Y}}_1(t)& ={\mathrm{e}}^{6t}⟨4,3⟩\\ {\overrightarrow{Y}}_2(t)& ={\mathrm{e}}^{-t}⟨1,-1⟩\\ \overrightarrow{Y}& =c_1{\overrightarrow{Y}}_1(t)+c_2{\overrightarrow{Y}}_2(t)\\ & =\left[\begin{array}{c}4c_1{\mathrm{e}}^{6t}+c_2{\mathrm{e}}^{-t}\\ 3c_1{\mathrm{e}}^{6t}-c_2{\mathrm{e}}^{-t}\end{array}\right]\end{array}$

For initial value problem $\overrightarrow{Y}(0)=⟨6,1⟩$, $c_1=1$ and $c_2=2$.