# Eigenvectors associated to distinct eigenvalues are orthogonal (with T a normal operator)

Let be V a inner product space over $$\mathbb{C}$$, $$T$$ a normal operator in $$V$$ and $$u,v \in V$$ two eigenvectors of T corresponding to different eigenvalues. Prove that $$u$$ and $$v$$ are ortogonal.

I was trying to prove this fact, until I found this proof in a Lineal Algebra Book (Friedberg):

Proof: \begin{align*} \lambda_1 \left \langle u,v \right \rangle &=\left \langle \lambda _1 u,v \right \rangle\\ &=\left \langle Tu,v \right \rangle\\ &=\left \langle u,T^{*}v \right \rangle\\ &=\left \langle u,\overline{\lambda}_{2} v \right \rangle\\ &=\lambda_{2}\left \langle u,v \right \rangle \end{align*} and since, $$\lambda_{1}\neq \lambda_{2}$$ (both eigenvalues) $$\Rightarrow \left \langle u,v \right \rangle=0$$

Nerverthless I still have a doubt in one step. I don't understand why there appears $$\lambda_{2}$$. I think that the correct step should be $$\left \langle u,T^{*}v \right \rangle=\left \langle u,\overline{\lambda}_{1}v \right \rangle=\lambda_{1}\left \langle u,v \right \rangle$$. I'm conscious that if I'm correct then the proof is incorrect, but I still don't understand why is $$\lambda_{2}$$. Can you help me to understand this please?

Extra note: I also know that with $$V$$ a inner product space over $$\mathbb{C}$$ and T a normal operator it satisfies:

• If $$\lambda$$ is an eigenvalue of $$T$$ $$\Rightarrow$$ $$\overline{\lambda}$$ is a eigenvalue of $$T^{*}$$.

And I know $$\lambda \neq \overline{\lambda}$$, but this (unless I'm not understanding well) supports what I've exposed. I hope I have made myself understood well.

• Hint: If $T$ is normal, then $\|(T-\lambda I)x\|^2=\|(T^*-\overline{\lambda}I)x\|^2$ holds for all $x$ and $\lambda$, which means that $T^*x=\overline{\lambda}x$ iff $Tx=\lambda x$. Sep 11, 2020 at 23:42

The answer came to my mind! The reason is because $$u,v$$ are distinct eigenvectors linked to a eigenvalue respectively. So,

\begin{align*} \left \langle \lambda_{1} u,v \right \rangle &= \left \langle Tu,v \right \rangle \end{align*}

Then by properties of $$T^{*}$$ we have:

\begin{align*} \left \langle Tu,v \right \rangle &= \left \langle u, T^{*}v \right \rangle \end{align*}

And here is the key step! $$T^{*}$$ is applied in the vector $$v$$ (and no in the vector $$u$$, so this was what I wasn't understanding well in my original question). And as $$v$$ and $$u$$ are two vectors with two different eigenvalues, in this step necessarily has to be another eigenvalue:

\begin{align*} \left \langle u, T^{*}v \right \rangle &=\left \langle u ,\overline{\lambda}_{2} v\right \rangle \\ &=\lambda_{2}\left \langle u,v \right \rangle \end{align*}

And, to finish, since $$\lambda_1 \neq \lambda_{2}$$ $$\Rightarrow$$ $$\left \langle u,v \right \rangle=0$$

\begin{align*} \therefore u \text{ and } v \text{ are orthogonal} \end{align*}

• Seems right. Perhaps you would have saved some time if the problem was phrased as '... and $u, v \in V$ two eigenvectors of $T$ associated to different eigenvalues $\lambda_{1}$ and $\lambda_{2}$...' Sep 11, 2020 at 4:43