Eigenvalues

Definition

Find $v\in {\mathbb{R}}^n\setminus \{0\}$ with $Av=\lambda v$ $(\lambda ,v)$ is an eigenvalue-eigenvector pair

Useful fact

Eigenvalues of $\left(\begin{array}{cc}a& b\\ b& a\end{array}\right)$ are $a\pm b$

Example: Fibonacci numbers

$g_0=\left(\begin{array}{c}1\\ 0\end{array}\right)$ $g_n=M^n{g}_0=\left(\begin{array}{c}{F}_{n+1}\\ F_n\end{array}\right)=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]\left(\begin{array}{c}{F}_n\\ F_{n-1}\end{array}\right)$ We find the eigenvalues of $M$ and the corresponding eigenvectors

$\left[\begin{array}{c}M-\lambda I\end{array}\right]$ has a not trivial null space $⟹[M-\lambda I]$ not invertible $⟹\det (M-\lambda I)=0$ $\det \left[\begin{array}{cc}1-\lambda & 1\\ 1& -\lambda \end{array}\right]={\lambda }^2-\lambda -1=0$ ${\lambda }_1=\frac{1+\sqrt{5}}{2},{\lambda }_2=\frac{1-\sqrt{5}}{2}$ (Golden ratio) For ${\lambda }_1$: $\left(\begin{array}{c}0\\ 0\end{array}\right)=\left[\begin{array}{cc}1-{\lambda }_1& 1\\ 1& -{\lambda }_1\end{array}\right]\left(\begin{array}{c}(v_1{)}_1\\ (v_1{)}_2\end{array}\right)$ $v_1=\left(\begin{array}{c}\frac{1+\sqrt{5}}{2}\\ 1\end{array}\right)$ $v_2=\left(\begin{array}{c}\frac{1-\sqrt{5}}{2}\\ 1\end{array}\right)$ $⟹span(v_1,v_2)={\mathbb{R}}^2$ $g_0=\left(\begin{array}{c}1\\ 0\end{array}\right)={\alpha }_1{v}_1+{\alpha }_2{v}_1$ $⟹{\alpha }_1=\frac{1}{\sqrt{5}},{\alpha }_2=-\frac{1}{\sqrt{5}}$ $g_n=M^n{g}_0=M^n\left(\frac{1}{\sqrt{5}}{v}_1-\frac{1}{\sqrt{5}}{v}_2\right)$ $=\frac{1}{\sqrt{5}}{M}^n{v}_1-\frac{1}{\sqrt{5}}{M}^n{v}_2$ $=\frac{1}{\sqrt{5}}{\lambda }_1^n{v}_1=\frac{1}{\sqrt{5}}{\lambda }_2^n{v}_2$

$$F_n=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^n-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$$

Example: complex Eigenvalue

$A=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$ $\det (A-\lambda I)=0$ ${\lambda }^2+1=0$ ${\lambda }_1=i,{\lambda }_2=-i$

Find $v\in {\mathbb{C}}^n$ with $Av=iv$ $(v\ne 0)$

$$Av=iv⟺v=\left(\begin{array}{c}{v}_1\in \mathbb{C}\\ v_2\in \mathbb{C}\end{array}\right)\text{ gives }\left(\begin{array}{c}-v_2\\ v_1\end{array}\right)=i\left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)$$

$v_1=a+ib,v_2=x+iy$ $⟹-x-iy=-b+ia⟹x=b,y=-a$ $⟹v=\left(\begin{array}{c}a+ib\\ b-ia\end{array}\right)=(a+ib)\left(\begin{array}{c}1\\ -i\end{array}\right)$

$⟹\left(\begin{array}{c}1\\ -i\end{array}\right)$ is an eigenvector of ${\lambda }_1=i$

Proposition 8.3.1

Lemma 8.2.8

If $(\lambda ,v)$ is an eigenvalue-eigenvector pair, then $(\bar{\lambda },\bar{v})$ is an eigenvalue-eigenvector pair.

Lemma 8.3.6

Let $A\in {\mathbb{R}}^{n\times n}$ and ${\lambda }_1,\dots ,{\lambda }_n$ its eigenvalues. ($\lambda$ can be a complex number)

• $\det (A)={\prod }_{i=1}^n{\lambda }_i$
• $Tr(A)={\sum }_{i=1}^n{\lambda }_i$

Proof $\det (A-\lambda I)=P(\lambda )=0$ (the characteristic polynomial) $\det (A-\lambda I)={\sum }_{\sigma \in {\prod }_n}sgn(\sigma ){\prod }_{i=1}^n(A-\lambda I{)}_{i,\sigma (i)}$ $=(a_{11}-\lambda )(a_{22}-\lambda )\dots (a_{nn}-\lambda )+\underset{\text{exponent }\le n-2}{\underbrace{a_{14}(a_{22}-\lambda )(a_{33}-\lambda )a_{41}}}\cdots +\dots .$ (If we do not use the identity $\sigma$, then we must have at least two elements not on the diagonal) $P_A(\lambda )=c_n{\lambda }^n+c_{n-1}{\lambda }^{n-1}+\cdots +c_0$ $=(-1{)}^n{\lambda }^n+(-1{)}^{n-1}\underset{\mathrm{Tr}(A)}{\underbrace{(a_{11}+a_{22}+\cdots +a_{nn})}}{\lambda }^{n-1}+\cdots +\underset{\det (A)}{\underbrace{c_0}}$ (*) Note that $P_A(0)=c_0=\det (A-0I)=\det (A)$ Now consider $P_A$ as product of roots $P_A=c_n(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_n)$ $P_A=(-1{)}^n(\lambda -{\lambda }_1)(\lambda -{\lambda }_2)\dots (\lambda -{\lambda }_n)$ (from *)) expands: $P_A=(-1{)}^n{\lambda }^n+(-1{)}^n{\lambda }^{n-1}(-{\lambda }_1-{\lambda }_2-\cdots -\lambda )+\cdots +(-1{)}^n(-1{)}^n({\lambda }_1{\lambda }_2\dots {\lambda }_n)$ $P_A=(-1{)}^n{\lambda }^n+(-1{)}^n\left(-{\sum }_{i=1}^n{\lambda }_i\right){\lambda }^{n-1}+\cdots +{\prod }_{i=1}^n{\lambda }_i$ $⟹c_0={\prod }_{i=1}^n{\lambda }_i=\det (A)$ $⟹(-1)\cdot -{\sum }_{i=1}^n{\lambda }^{n-1}=\mathrm{Tr}(A)⟹{\sum }_{i=1}^n{\lambda }_i=\mathrm{Tr}(A)$

math-tools

Method to prove that two matrices are similar

We set the characteristic equation equal

Example 1

Prove the eigenvalues of $A$ and $A^{\top }$ are the same $P_{A^{\top }}(\lambda )=\det (A^{\top }-\lambda I)=\det ((A-\lambda I{)}^{\top })=\det (A-\lambda I)=P_A(\lambda )$

Example 2

Prove $B$ similar to $A$, with $B=S^{-1}AS$ $P_B(\lambda )=\det (B-\lambda I)=\det (S^{-1}AS-\lambda S^{-1}S)=\det (S^{-1}(A-\lambda I)S)$ $=\det (S^{-1})\det (A-\lambda I)\det (S)=\det (A-\lambda I)=P_A(\lambda )$