# Is every set of independent eigenvectors of an orthogonally diagonalizable matrix orthogonal?

Let $f : V \rightarrow V$ be a linear operator, where $V$ is a finite dimensional inner product space over $\mathbb{C}$ of dimension $n$. Suppose we are given that $f$ has $n$ orthogonal eigenvectors $u_1\ldots,u_n$, corresponding to the eigenvalues $\lambda_1,\ldots,\lambda_n$. Now suppose $v_1,\ldots,v_n$ is another set of linearly independent eigenvectors of $f$ for the same eigenvalues $\lambda_1,\ldots,\lambda_n$ respectively.

Question: Are these "new" eigenvectors also orthogonal?

If yes, please provide a proof. If not, a counterexample.

To avoid the trivial case, assume that at $\lambda_1,\ldots,\lambda_n$ are NOT pairwise distinct.

• This reads like a verbatim homework assignment. Did you try anything at all? – Umberto P. Sep 6 '18 at 11:55

Consider the $2 \times 2$ identity matrix, and the standard basis. Then look at any other basis.