If $b\ne 0$, orthogonally diagonalize \begin{bmatrix}a&0&b\\0&a&0\\b&0&a\end{bmatrix} I'm trying to calculate the eigenvalues as the first step of this problem, but it's leading me down this rabbit hole of countless computations to find all the eigenvalues. I'm confident I'm supposed to solve this a different way that is less tedious. Anyone able to give me a hand?
 A: Let $A= \begin{bmatrix}a&0&b\\0&a&0\\b&0&a\end{bmatrix}\implies |A-xI_3|= \begin{vmatrix}a-x&0&b\\0&a-x&0\\b&0&a-x\end{vmatrix}=(a-x)^3-b^2(a-x).$
Solving $(a-x)^3-b^2(a-x)=0,$ we have, $x=a$ or $(a-x)^2-b^2=0\implies x\in\{a-b,a+b\}$.
Hence, the eigenvalues are $a,a-b$ and $a+b$.
A: Let $M$ be your matrix. Then $M-aI$ has a row of zeroes so one eigenvalues is $a$. The other eigenvalues must then add up to $2a$ (since the trace of the $3×3$ matrix is $3a$) and have determinant $a^2-b^2$ (taking minors of the second row shows the $3×3$ determinant is $a(a^2-b^2)$). Thus the eigenvalues end up $\{a,a\pm b\}$.
A: the eigenvalues of
$$
\left(
\begin{array}{cc}
0&1 \\
1&0 \\
\end{array}
\right)
$$
are $1,-1$  with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}& \frac{-1}{\sqrt 2}\\
\frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
0&0&1 \\
0&0&0 \\
1&0&0 \\
\end{array}
\right)
$$
are $1,0,-1$  with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
0&0&b \\
0&0&0 \\
b&0&0 \\
\end{array}
\right)
$$
are $b,0,-b$  with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
a&0&b \\
0&a&0 \\
b&0&a \\
\end{array}
\right)
$$
are $a+b,a,a-b$  with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
