# The Unapologetic Mathematician

## Eigenvalues and Eigenvectors of Normal Transformations

Let’s say we have a normal transformation $N$. It turns out we can say some interesting things about its eigenvalues and eigenvectors.

First off, it turns out that the eigenvalues of $N^*$ are exactly the complex conjugates of those of $N$ (the same, if we’re working over $\mathbb{R}$. Actually, this isn’t even special to normal operators. Indeed, if $T-\lambda I_V$ has a nontrivial kernel, then we can take the adjoint to find that $T^*-\bar{\lambda}I_V$ must have a nontrivial kernel as well. But if our transformation is normal, it turns out that not only do we have conjugate eigenvalues, they correspond to the same eigenvectors as well!

To see this, we do almost the same thing as before. But we get more than just a nontrivial kernel this time. Given an eigenvector $v$ we know that $\left(N-\lambda I_V\right)v=0$, and so it must have length zero. But if $N$ is normal then so is $N-\lambda I_V$:

\displaystyle\begin{aligned}\left(N-\lambda I_V\right)\left(N-\lambda I_V\right)^*&=\left(N-\lambda I_V\right)\left(N^*-\bar{\lambda}I_V\right)\\&=NN^*-\lambda I_VN^*-\bar{\lambda}NI_V+\lambda\bar{\lambda}I_VI_V\\&=N^*N-\lambda N^*I_V-\bar{\lambda}I_VN+\lambda\bar{\lambda}I_VI_V\\&=\left(N^*-\bar{\lambda}I_V\right)\left(N-\lambda I_V\right)\\&=\left(N-\lambda I_V\right)^*\left(N-\lambda I_V\right)\end{aligned}

and so acting by $\left(N-\lambda I_V\right)^*$ gives the same length as acting by $\left(N-\lambda I_V\right)$. That is:

$\displaystyle0=\lVert\left(N-\lambda I_V\right)v\rVert=\lVert\left(N-\lambda I_V\right)^*v\rVert=\lVert N^*v-\bar{\lambda}v\rVert$

thus by the definiteness of length, we know that $N^*v-\bar{\lambda}v$. That is, $v$ is also an eigenvector of $N^*$, with eigenvalue $\bar{\lambda}$.

Then as a corollary we can find that not only are the eigenvectors corresponding to distinct eigenvalues linearly independent, they are actually orthogonal! Indeed, if $v$ and $w$ are eigenvectors of $N$ with distinct eigenvalues $\lambda$ and $\mu$, respectively, then we find

\displaystyle\begin{aligned}(\lambda-\mu)\langle v,w\rangle&=\langle\bar{\lambda}v,w\rangle-\langle v,\mu w\rangle\\&=\langle N^*v,w\rangle-\langle v,Nw\rangle\\&=\langle v,Nw\rangle-\langle v,Nw\rangle=0\end{aligned}

Since $\lambda-\mu\neq0$ we must conclude that $\langle v,w\rangle=0$, and that the two eigenvectors are orthogonal.

August 6, 2009