I have a math problem I am struggling with:
If a linear transform $A: \mathbb{R}^n\to\mathbb{R}^n$ and we have a basis of $\mathbb{R}^n$ of eigenvectors of $A$, can't we just orthonormalize them and get a matrix $P$ such that $P^{-1}=P^\mathrm{T}$ and thus $P^\mathrm{T}AP$ is diagonal?
Solution (so far):
Let $B$ be the matrix of eigenvectors of $A$.
Now, I know that if $B$ is orthogonalizable, then the rest of the problem is true, and I also know that B is diagonalizable, because it is a basis of $\mathbb{R}^n$ and is therefore linearly independent. I also know (from prodding the professor) that the conjecture is false (i.e. we can't just orthonormalize $B$). I can't see how the process of normalizing the vectors in a matrix would cause a problem, so it must come down to whether the [linearly independent] vectors of $B$ are orthogonalizable.
Question:
Which matrix of linearly independent vectors is not orthogonalizable? (and why?) Thanks