$\begingroup$How do you find the eigenpairs of this?
Given two matrices $A=\begin{bmatrix} a& b\\b&-a\end{bmatrix}$ and $B=\begin{bmatrix} a& -b\\b&a\end{bmatrix}$ where $b \neq 0$, find all eigenpairs of $A$ and $B$, and comment if the matrices are diagonalizable or not.
I plugged in the characteristic equation $\det(A-\lambda I)=0$ and is unable to solve for $\lambda $. I got $\lambda ^2-a^2-b^2=0$ and it won't let me factorize. Any idea?
$\endgroup$02 Answers
$\begingroup$The characteristic polynomial of \begin{bmatrix} a & b \\ b & -a \end{bmatrix} is indeed $\lambda^2-a^2-b^2$. The eigenvalues are $\sqrt{a^2+b^2}$ and $-\sqrt{a^2+b^2}$.
Since $$A-\sqrt{a^2+b^2}\,I= \begin{bmatrix} a-\sqrt{a^2+b^2} & b \\ b & -a-\sqrt{a^2+b^2} \end{bmatrix} $$ has rank one, an eigenvector $\begin{bmatrix} x \\ y\end{bmatrix}$ satisfies $$ \begin{bmatrix} a-\sqrt{a^2+b^2} & b \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = 0 $$ so we can take $$ \begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix} b \\ \sqrt{a^2+b^2}-a\end{bmatrix} $$ and this is nonzero, so it is a basis for the eigenspace. The other eigenvectors are the non zero scalar multiples of this one.
Similarly for the other eigenvalue.
For the matrix \begin{bmatrix} a & -b \\ b & a \end{bmatrix} the characteristic polynomial is $\lambda^2-2a\lambda+(a^2+b^2)$, whose roots are $a+ib$ and $a-ib$.
Assuming $a$ and $b$ real, the first matrix is diagonalizable, because it has distinct real eigenvalues. The second matrix is diagonalizable over the complex numbers, but not over the reals.
$\endgroup$$\begingroup$If you have $\lambda^2 = a^2 + b^2$, then the eigenvalue is $$\lambda = \pm \sqrt{a^2+b^2}.$$
From there you can find the eigenvectors.
Edit:
If we want to find a particular $(x_1, x_2)$ such that $\begin{bmatrix} z_1 & z_2 \\ 0 & 0\end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix}=0$, we can just pick $x_1=-z_2, x_2=z_1.$
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