How do you find a symmetric matrix $A ∈\Bbb R^{3×3}$, knowing that:
- $A^n$ $= A$ for some $n > 1$
- $A$ has three distinct eigenvalues
- and we are given two (orthogonal) eigenvectors $[1, 2, 2]^T$ and $[2, 1, −2]^T$.
The problem specifies that there are several possible solutions and I need to compute only one of them. However, I don't see any possible solution. I'm stuck here:
If $A^n$ $= A$, then $D^n$ $=D$, then its eigenvalues to the power of n are equal to themselves. I can think of only $0$ and $1$ to satisfy this property, but the problem asks for three distinct eigenvalues. What is another possible distinct eigenvalue?
I have also derived an expression for the third eigenvector, which is $[y+4z, y, z]^T$, but this doesn't quite help. Especially as we can choose any real values for $y$ and $z$, which basically means that there are infinitely many solutions for the matrix $A$.
Thank you for your help.
There are in fact only $6$ such matrices, corresponding to the $6$ permutations of $\{0,1,-1\}$. Each such matrix $A$ would then have $A^n=A$ for all odd $n$.
The third eigenvector $(x, y, z)$ must be orthogonal to the other two. So, $x+2y+2z=0$ and $2x+y-2z=0$. Solving this, one gets $y=-x$ and $z=\frac{1}{2}x$, so the third eigenvector must be of the form $(x, -x, \frac{1}{2}x)$. Any nonzero scalar multiple of an eigenvector is again an eigenvector (corresponding to the same eigenvalue), so $(1, -1, \frac{1}{2})$ is an eigenvector.
The given eigenvectors, however, are not orthonormal unit vectors. Each eigenvector may be divided by its length to get a unit vector. One has $||(1, 2, 2)||=||(2, 1, -2)||=3$ and $||(1, -1, \frac{1}{2})||=\frac{3}{2}$.
So, the resulting unit vectors are $(\frac{1}{3}, \frac{2}{3}, \frac{2}{3})$, $(\frac{2}{3}, \frac{1}{3}, -\frac{2}{3})$, and $(\frac{2}{3}, -\frac{2}{3}, \frac{1}{3})$.
It thus follows that the orthogonal diagonalization of $A$ is given by $\begin{pmatrix}\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\end{pmatrix}D\begin{pmatrix}\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\\frac{2}{3}&\frac{1}{3}&-\frac{2}{3}\\\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\end{pmatrix}^T$, where $D$ is a diagonal matrix with the eigenvalues $0$ and $\pm{1}$ on the diagonal.