Let $A$ be a normal matrix. Prove $\left \| (A+I)v \right \| = \left \| (A^{*}+I)v \right \|$

Question

Need some help with the following:
Let $A$ be a normal $N \times N$ matrix over $\mathbb{C}$. Prove $\left \| (A+I)v \right \| = \left \| (A^{*}+I)v \right \|$ for any $v\in \mathbb{C}^{n}$.

Is it some how related to A eigenvalues?

Answer 1

For any normal matrix $N$ we have

$$\|Nx\|^2 = \langle Nx, Nx\rangle = \langle N^*Nx, x\rangle = \langle NN^*x,x\rangle = \langle N^*x,N^*x\rangle = \|N^*x\|^2, \forall x\in\mathbb{C}^n$$

Now, if $A$ is normal then $A + I$ is also normal:

\begin{align} (A + I)(A+I)^* &= (A+I)(A^*+I) \\ &= AA^* + A + A^* + I \\ &= A^*A + A + A^* + I \\ &= (A^*+I)(A+I)\\ &= (A+I)^*(A+I) \end{align}

So we conclude $$\|(A+I)v\| = \|(A+I)^*v\| = \|(A^*+I)v\|$$

for all $v \in \mathbb{C}^n$.

Answer 2

Hint: Prove it if $A$ is diagonal, and then diagonalize.