Norm of a normal matrix equals the norm of its conjugate transpose Let $N$ be an $n\times n$ complex matrix. Show that if $N$ is normal, then $\left\Vert Nx \right\Vert=\left\Vert N^\ast x\right\Vert$ for all $x\in\mathbb{C}^n$ where $N^\ast$ is the conjugate transpose of $N$.
Does it make sense if the norm isn't specified?
Assuming it is an inner-product space I've done the following but this wasn't specified:
$$\begin{align}\left\Vert Nx\right\Vert &=\sqrt{\langle Nx,Nx\rangle}\\
&=\sqrt{\langle x,N^\ast Nx\rangle}\\
&=\sqrt{\langle x,NN^\ast x\rangle}\\
&=\sqrt{\langle N^\ast x,N^\ast x\rangle}\\
&=\left\Vert N^\ast x\right\Vert\end{align}$$
A pointer to the right direction would be greatly appreciated. ✌️
 A: As the question is stated, I would presume that the norm being used was intended to be the standard norm of $\ \mathbb C^n\ $: $\ \left\Vert x\right\Vert = \sqrt{x^*x}\ $, in which case the OP's answer is perfectly fine.  In fact, it's not necessarily true that $\ \left\Vert Nx \right\Vert=\left\Vert N^\ast x\right\Vert$ for all $x\in\mathbb{C}^n\ $ for any other norm of $\ \mathbb C^n\ $, even if the norm comes from an inner product, since the adjoint $\ N^\dagger\ $ of $\ N\ $ with respect to that inner product will not necessarily be its conjugate transpose, and will not necessarily commute with it just because its conjugate transpose does.   For concreteness, here's a counterexample.
Let $\ N=\begin{pmatrix}1&1&0\\ 0&1&1\\ 1&0&1\end{pmatrix}\  $ (lifted from Wikipedia), which is easily shown to be normal. Let $\ D= \begin{pmatrix}5&-2&3\\ -2&5&-2\\ 3&-2&5\end{pmatrix}\ $ (constructed to be positive definite), $\ \left\Vert x\right\Vert_D = \sqrt{x^* D x}\ $ for $\ x\in \mathbb C^3\ $, and $\ y=\begin{pmatrix}1\\ 0\\ 2\end{pmatrix}\ $. Then $\ Ny = \begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\ $, $\ \left\Vert Ny\right\Vert_D = \sqrt{56}\ $, $\ N^*y = \begin{pmatrix}3\\ 1\\ 2\end{pmatrix}\ $, and $\ \left\Vert N^*y\right\Vert_D = \sqrt{86}\ne \left\Vert Ny\right\Vert_D\ $.
