Let $A$ be a normal matrix. Prove that if $|\lambda| = 1$ for all eigenvalues $\lambda$ of $A$ then $A$ is unitary.

Question

Let $A$ be a normal matrix. Prove that if $|\lambda| = 1$ for all eigenvalues $\lambda$ of $A$ then $A$ is unitary.

Normal is defined as : $A^*A = AA^*$.

I am unable to find a theorem that makes $|\lambda| =1$ useful in anyway. I have been trying to use Schur's theorem to make $U^*AU = T$ where the diagonal of $T$ will be eigenvalues of $A$.

Can someone please help out with this proof?

Answer 1

Let $A x_i = \lambda x_i$, where $\lvert \lambda_i \rvert = 1$ and $\langle x_i, x_j \rangle = \delta_{ij}$ with $1 \leq i, j \leq n$. Write every vector $x \in \mathbb{C}^n$ as $x = \sum_{i = 1}^n \alpha_i x_i$, then it follows that

$$A^\ast A x = A^\ast \left ( \sum_{i = 1}^n \alpha_i A x_i \right ) = A^\ast \left (\sum_{i =1}^n \alpha_i \lambda_i x_i \right).$$

Since $A^\ast x_i = \bar{\lambda}x_i$ holds for all $i = 1, \dots, n$, this equation becomes

$$A^\ast Ax = \sum_{i = 1}^n \alpha_i \lambda_i A^\ast x_i = \sum_{i = 1}^n \alpha_i \lvert \lambda_i \rvert^2 x_i = \sum_{i = 1}^n \alpha_i x_i = x.$$

Thus $A^\ast Ax = x$ for every $x \in \mathbb{C}^n$ and hence $A^\ast A = I$.

Answer 2

One answer just using the definition of adjoint is this: Let $v$ be an unit eigenvector with eigenvalue $\lambda$, that is $\lvert v \rvert^2 = 1$. Then $$1 = \lvert\lambda\rvert^2 \lvert v\rvert^2 = \rvert\lambda v\rvert^2 = \lvert Av \rvert^2 = \langle Av, Av\rangle = \langle v, A^{*}Av\rangle = \langle v, AA^{*}v\rangle = \langle A^{*}v, A^{*} v\rangle$$ Also, $$ |A^{*}Av\rvert ^2 =\langle A^{*}Av, A^*Av\rangle = \langle A v, AA^*Av\rangle = \langle Av, A^{*}AAv\rangle = \langle AAv, AAv\rangle= \lvert AA v\rvert^2 = 1$$ Therefore $$ \lvert v - A^{*}Av\rvert^2 = \langle v - A^{*}A v, v-A^{*}Av \rangle = \langle v, v \rangle - \langle v, A^{*}Av\rangle - \langle A^{*}A v, v\rangle + \langle A^{*}A v, A^{*}Av\rangle = 0$$ This means $v=A^{*}Av$ for all unit eigenvectors. Since we can take a basis of unit eigenvectors and the transform is the identity in that basis, then it is the identity everywhere.