MATH 323 Lecture 25

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Eigenvalues and Eigenvectors

Example

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

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

Therefore, the eigenvalues are $\lambda _{1}=0$, $\lambda _{2}=\lambda _{3}=1$

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

1. $N(A-0\cdot I)=N(A)=\left\langle \alpha ,\alpha ,\alpha \right\rangle \quad \alpha \in \mathbb {R}$
2. $N(A-1\cdot I)=\left\langle 3\alpha -\beta ,\alpha ,\beta \right\rangle$

Therefore, $\left\langle 1,1,1\right\rangle$ is an eigenvector belonging to $\lambda _{1}=0$, and $\left\langle 3,1,0\right\rangle$ and $\left\langle -1,0,1\right\rangle$ 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 $(-1)^{n}\lambda ^{n}+\Theta (\lambda ^{n-1})$.

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

${\begin{aligned}p(\lambda )&=(a_{11}-\lambda )\dots (a_{nn}-\lambda )+\dots \\&=(-1)^{n}(\lambda -\lambda _{1})\dots (\lambda -\lambda _{n})+\Theta (\lambda ^{n-1})&=(\lambda _{1}-\lambda )\dots (\lambda _{n}-\lambda )+\Theta (\lambda ^{n-1})\end{aligned}}$

$p(0)=\prod _{i=1}^{n}\lambda _{i}$ 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

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

${\begin{aligned}\lambda _{1}&=1+2i\\\lambda _{2}&=1-2i\end{aligned}}$

$(A-\lambda _{1}\,I){\vec {x}}=0\longrightarrow {\begin{bmatrix}1&i\\0&0\end{bmatrix}}$

Therefore, $\left\langle -i,1\right\rangle$ is an eigenvector belonging to $\lambda _{1}=1+2i$. If we multiply this eigenvector by $i$, we get $\langle 1,i\rangle$, 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 $A\sim B\iff B=S^{-1}AS$

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

Review: Complex Numbers

Every complex number $z=a+b\,i$ has a conjugate ${\bar {z}}=a-b\,i$.

A polynomial that has a root $z=a+b\,i$ will also have the conjugate as a root

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

if $z=a+b\,i$ is an eigenvector of complex eigenvalue $\lambda _{1}$, then ${\bar {z}}=a-b\,i$ is an eigenvector of the conjugate eigenvalue ${\bar {\lambda }}_{1}$.

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

${\begin{aligned}y_{1}'&=a_{11}\,y_{1}+\dots a_{1n}\,y_{n}\\y_{2}'&=a_{21}\,y_{1}+\dots a_{2n}\,y_{n}\\\dots \\y_{n}'&=a_{n1}\,y_{1}+\dots a_{nn}\,y_{n}\end{aligned}}$

Where $y_{i}$ are functions in $C^{1}[a,b]$.

Which can be written in the form ${\vec {Y}}'=A{\vec {Y}}$.

Suppose ${\vec {Y}}(t)$ is of the form $\langle x_{1}\mathrm {e} ^{\lambda t},x_{2}\mathrm {e} ^{\lambda t},\ldots ,x_{n}\mathrm {e} ^{\lambda t}\rangle$.

Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}(t) = \mathrm{e}^{\lambda t} \vec{x}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}'(t) = \lambda \vec{Y}(t) = \lambda \mathrm{e}^{\lambda t} \vec{x}} .

If ${\vec {x}}$ is an eigenvector of $A$, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A \vec{Y} = A \mathrm{e}^{\lambda t} \vec{x} = \mathrm{e}^{\lambda t} A \vec{x} = \mathrm{e}^{\lambda t} \lambda x}

The set of solutions to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}' = A \vec{Y}} is an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} -dimensional subspace, and if solutions Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}_1, \ldots, \vec{Y}_n} are linearly independent solutions, then any linear combination of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}_i} 's are also solutions.

Example

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} y_1' &= 3y_1 + 4y_2 \\ y_2' &= 3y_1 + 2y_2 \end{align}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = \begin{bmatrix} 3 & 4 \\ 3 & 2 \end{bmatrix}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_1 = 6, \lambda_2 = -1}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \vec{x}_1 &= \langle 4, 3 \rangle \\ \vec{x}_2 &= \langle 1, -1 \rangle \\ \vec{Y}_1(t) &= \mathrm{e}^{6t} \langle 4, 3 \rangle \\ \vec{Y}_2(t) &= \mathrm{e}^{-t} \langle 1, -1 \rangle \\ \vec{Y} &= c_1 \, \vec{Y}_1(t) + c_2 \, \vec{Y}_2(t) \\ &= \begin{bmatrix} 4c_1 \, \mathrm{e}^{6t} + c_2 \, \mathrm{e}^{-t} \\ 3c_1 \, \mathrm{e}^{6t} - c_2 \, \mathrm{e}^{-t} \end{bmatrix} \end{align}}

For initial value problem Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{Y}(0) = \langle 6, 1 \rangle} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_1 = 1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_2 = 2} .