Verify my proof that eigenvalues of a unitary matrix have absolute value 1

Question

Show that eigenvalues of a unitary matrix have absolute value 1.

Proof below. Please verify, critique, or improve. Note: Many proofs are available; this question is to verify this proof.

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Notation: $M^H$ is the conjugate transpose of $M$.

Let $Q^HQ = I, Qx = \lambda x, x \neq 0$. Then $$\begin{align} \lambda \bar \lambda x^Hx &= (\lambda x)^H(\lambda x) \quad \text{(by linearity)} \\ &= (Qx)^H(Qx) \\ &= x^HQ^HQx \\ & = x^Hx. \end{align}$$

Since $x \neq 0$, then $x^Hx \neq 0$, and $\lambda \bar \lambda = 1$, so $|\lambda| = 1$.

Answer 1

Yes, the argument you are using is standard. Minor details:

• you are missing the absolute value bars at the start, it should be $|\lambda|^2$;

• I wouldn't say that the first equality is due to "linearity", rather you are using the compatibility of the product of matrices with that of numbers, and that $(\lambda x)^H=\overline\lambda\,x^H$;

• instead of assuming that $x\ne0$ you can assume that $x^Hx=1$ (just replace $x$ with $x/\|x\|$) and the argument is marginally simpler.