# 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. ✌️

• Not all normed vector spaces are inner product spaces, so you cannot assume an inner product exists. – Adam Francey Feb 25 at 0:11
• I don’t see why using the inner product is a problem. The OP was asked to prove the result for $x \in \mathbb{C}^n$ which is an inner product space. – Jordan Green Feb 25 at 0:46
• @JordanGreen No, $\mathbb{C}^n$ isn't an inner product space all on its own. It also needs an inner product. The question tells us the norm exists, so we know that the normed vector space ${\displaystyle (\mathbb{C}^n,\|\cdot \|)}$ exists. This does not allow us to assume that an inner product $\langle \cdot ,\cdot \rangle$ and inner product space $(\mathbb{C}^n \langle \cdot ,\cdot \rangle)$ inducing that norm exist. – Adam Francey Feb 25 at 0:59

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\$$.