Prove or disprove:
- There exists A such that $\|Av\|=\|v\|$ for any $v \in R^n$, but $A$ is not a normal matrix
- There exists A such that $\|Av\|=\|v\|$ for any $v \in R^n$, but $A$ is not a unitary matrix
I can't find an example for A, but also can't prove it doesn't exists. I also understand that $1 \implies 2$.
Would the answer be different if the space was $C^n$?
In the below, $\{e_1,\dots,e_n\}$ denote the standard basis vectors.
It helps to write out the norm using dot products. In this case, we have
$$ \|Av\|^2 = (Av)^*(Av) = v^*A^*A v \equiv v^*v = \|v\|^2 $$
In fact, we can rewrite this as $v^*(A^*A - I)v = 0$, as I will leave to you to verify.
So the question is now if $B = A^*A - I$ satisfies $v^*Bv = 0$ for every $v$ in $\mathbb{R}^n$ (or $\mathbb{C}^n$), what do we know about $A$?
For the case of $\mathbb{C}^n$, I will leave to you to verify that this is enough to guarantee that $B = 0$, which is to say that $A$ is unitary (hint: plug in $v = e_j$, then $v = e_j + e_k$ and $v = e_j + ie_k$).
For $\mathbb{R}^n$, it ends up that all we know is that $B$ is skew-Hermitian, i.e. that $B^* = -B$. In order to verify that $B$ must indeed be skew-Hermitian, note that if $B$ has entries $(b_{jk})$, then we can plug in $v = e_j$, then $v = e_j + e_k$ to find $$ e_j^* B e_j = b_{jj} = 0\\ (e_j + e_k)^*B(e_j + e_k) = b_{jj} + b_{jk} + b_{kj} + b_{kk} = 0 \implies\\ b_{jk} = -b_{kj} $$ You could prove that any skew-Hermitian matrix $B$ satisfies $v^*Bv = 0$ for every single $v \in \mathbb{R}^n$, though that's not strictly necessary here.
So, we now know that it is enough to have $A^*A - I$ be a skew-Hermitian matrix. However, $A^*A - I$ must always be Hermitian, i.e. $(A^*A - I)^* = A^*A - I$ (verify that this is the case). Since $A^*A - I$ is both Hermitian and skew-Hermitian, it must be the zero matrix. So, $A^*A = I$, which is to say that $A$ is unitary.
So, the answer to both questions, as it ends up, is false, whether you're in $\mathbb{R}$ or $\mathbb{C}$.