Determine if 2 eigenvectors are orthogonal.

Question

Let $$ T:V \to V $$ Be a linear operator.

And:

$$ 0 \neq u,v \in V $$

Such that:

$$ T(v) = \lambda_1v, \quad T(u) = \lambda_2u $$

NOTE: It may be that $\lambda_1 = \lambda_2$

The question asks what are the conditions in which it holds that: $$ (u,v) = 0 $$

Now i have the following options:

1. If $T$ is normal
2. Only if $T$ is normal
3. Only if $T$ is normal and $\lambda_1 \neq \lambda_2$
4. All the answers are incorrect

What i think:

If $T$ is normal, eigenvectors of different eigenvalues are orthogonal.

Therefore, it must hold that if $3$ holds, so $(u,v) = 0$

Yet, maybe can i say more? Am i missing something?

Answer 1

Normal only $(\lambda_1=\lambda_2)$ does not garantee $(u,v)=0$, for exemple suppose as given $\lambda_1=\lambda_2=\lambda$, $$ T(v) = \lambda v, \quad T(u) = \lambda u$$ and $(u,v)=0$ take $w=u+v$ then also $T(v) = \lambda v, \quad T(w) = \lambda w$ but $(v,w)\neq 0$, to have a certain orthogonality $T$ normal and $\lambda_1\neq \lambda_2$.

(For normal matrices there exist a set of orthonormal eigenvectors)

Now the answer here could be four as one can easily construct non normal matrices of dimension $n>2$ having only two orthogonal eigenvectors