I have the following matrix
$$A=\begin{equation}
\begin{pmatrix}
0 & a & -b\\
-a & 0 & c\\
b & -c & 0
\end{pmatrix}
\end{equation}$$
I have to prove that $M=A^2+I$ is idempotent knowing that $a^2+b^2+c^2=1$.
I can calculate M using brute force as
$$M=
\left( \begin{array}{ccc}
0 & a & -b \\
-a & 0 & c \\
b &-c & 0
\end{array} \right)
%
\left( \begin{array}{ccc}
0 & a & -b \\
-a & 0 & c \\
b &-c & 0
\end{array} \right)
+
%
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array} \right)
$$
I obtain $$ M=\left( \begin{array}{ccc}
c^2 & bc & ac \\
bc & b^2 & ab \\
ac & ab & a^2
\end{array} \right)$$
I have tried to solve this by brute force, but I cannot prove that $M=M^2$, that is what I need to show that $M$ is idempotent since my result by brute force contains too many terms to simplify.
Can someone explain to me how to do it?
Best Answer
Giulio hint already solves the problem, but here's an alternative solution:
The characteristic polynomial of $A$ is $x(x^2+1)$ meaning it has eigenvalues $0,-i,i$ and it is diagonalizable. Then $M = A^2+I$ has eigenvalues $1,0,0$ and it is still diagonalizable. This concludes that $M$ is idempotent