Do eigenvectors of a Toeplitz matrix form an orthogonal set?

Question

It is true for a $2 \times 2$ Toeplitz matrix (put values $a$ and $b$ in the first row and $b$ and $a$ in the second and work out), but when I tried it for a $3 \times 3$, it turns out to be a bit difficult.

Any help or reference to my question is greatly appreciated. Thanks.

Answer 1

Let $A=\pmatrix{0&1&0\\ 0&0&1\\ 8&0&0}$. Then $\det(xI-A)=x^3-8$ and $A$ has three distinct, simple eigenvalues. That is, $A$ has an eigenbasis and all its eigenspaces are one-dimensional. However, $A^TA-AA^T=\pmatrix{63&0&0\\ 0&0&0\\ 0&0&-63}\ne0$. Therefore $A$ is not normal, i.e. it has two different eigenspaces (corresponding to two different eigenvalues) that are not orthogonal to each other.