# Eigenvectors and Eigenvalues

Consider a matrix

$A=\begin{pmatrix}0&0&1\\0&1&1\\-1&0&0\end{pmatrix}$

a) Find its eigenvalues and eigenvectors.

b) Construct a system of three normalized orthogonal eigenvectors and verify the completness condition.

c) Using eigenvectors of $A$ construct the transformation matrix $U$ and show that it diagonalizes $A$

d) Show that $e^{xA} =\cosh (x)+A*\sinh(x)$ (Use series expansion for the argument)

a)

To find eigenvalues $λ$  we set the equation $det ⁡(A-λ*I)=0$

$\begin{vmatrix} 0-λ&0&-1\\0&1-λ&0\\-1&0&0-λ)\end{vmatrix}=0$

or $λ^2 (1-λ)-(1-λ)=0$   or $(1-λ)(λ^2-1)=0$

$λ_{1,2}=1$ and $λ_3=-1$ so $+1$ is double degenerate

To find eigenvectors we set the equation $(A-λI)X=0$ for each value of $λ$

$λ=1$:    $\begin{pmatrix}-1&0&-1\\0&0&0\\-1&0&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0$  or $\left\{\begin{matrix}–x-z=0\\0=0\\-x-z=0\end{matrix}\right.$  so that $x=α$,  $y=β$,  $z=-α$
eigenvector is $X=\begin{pmatrix}α\\β\\-α\end{pmatrix}=α\begin{pmatrix}1\\0\\-1\end{pmatrix}+β\begin{pmatrix}0\\1\\0\end{pmatrix}$   where $α$ and $β$ can have all possible values

$λ=-1$: $\begin{pmatrix}1&0&-1\\0&2&0\\-1&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0$  or $\left\{\begin{matrix}x-z=0\\2y=0\\-x+z=0\end{matrix}\right.$  so that $x=α$,  $y=0$,  $z=α$

eigenvector is $X=\begin{pmatrix}α\\0\\α\end{pmatrix}=α\begin{pmatrix}1\\0\\1\end{pmatrix}$

b)

Construct 3 ortogonal eigevectors

$X_1=\begin{pmatrix}1\\0\\1\end{pmatrix}$,  $X_2=\begin{pmatrix}1\\1\\-1\end{pmatrix}$, $X_3=\begin{pmatrix}0\\1\\0\end{pmatrix}$

Testing for completeness: The initial matrix $A$ is 3×3 and there are 3 vectors above. They are linearly independent since the matrix having these vectors as columns has determinant nonzero.

$U=\begin{vmatrix}1&1&0\\0&1&1\\1&-1&0)\end{vmatrix}=0+0+1-0+1-0=2$

The transformation matrix to diagonalize A is the matrix having as columns a complete set of eigenvectors (the matrix above U). Since

$U^{-1}=(1/2)*\begin{pmatrix}1&0&1\\0&1&-1\\-1&2&1\end{pmatrix}$

One has

$U^{-1} AU=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&1\end{pmatrix}$

d)

$e^{xA}=∑ )1/n!)*(xA)^n=∑ (x^n/n!)*A^n$    but $A^{2n}=I$  and $A^{2n+1}=A$
$e^{xA}=∑ [x^{2n}/((2n)!)]*I+∑ [x^{2n+1}/(2n+1)!]*A=\cosh⁡ x+A*\sinh⁡ x$