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.

  • $\begingroup$ 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 $Tx=\lambda x$. $\endgroup$ Sep 11, 2020 at 23:42

1 Answer 1


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*}

  • 1
    $\begingroup$ 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 $\lambda_{2}$...' $\endgroup$ Sep 11, 2020 at 4:43

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