Unitary in the Orthocomplement of a Matrix

Question

Given a matrix $A \in M_n(\Bbb{C})$, does there exist a unitary $U \in M_n(\Bbb{C})$ such that $tr(AU)=0$?

Geometrically it seems true: thinking of $A$ as a vector in the plane, and unitaries as points on the unit-circle, then I should be able to rotate the line defined by $A$ through a $90$ deg. angle and intersect the circle at some point $U$, which means $A$ and $U$ are orthogonal. I just can't figure out how to put this geometric intuition into a proof...But perhaps this intuition is misguided

The condition $tr(AU)=0$ basically says that $A$ and $U$ are orthogonal. In fact, I'm really trying to show that the unitary group intersects the orthocomplement of $A$. With this in mind, I tried arguing by contradiction using a Hahn-Banach separation theorem, but I couldn't see how to finish the proof...

Another idea was to take the linearly independent columns of $A$, then use the Gram-Schmidt process to form an orthonormal basis for $\Bbb{C}^n$, and then use the vectors as the column of $U$. But I'm not sure this would work...In fact, it won't. If $A \in M_2(\Bbb{C})$ is the matrix whose first column is $(1,0)^T$ and second column is $(-1,1)^T$, the orthonormal basis formed from these is $(1,0)^T$ and $(0,1)^T$, meaning that the identity is the associated unitary. But clearly it doesn't satisfy the condition $Tr(AU)=0$.

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Here is my attempt using Hahn-Banach. Since $\{A\}^\perp$ is closed and $U(n)$, the $n \times n$ unitary group, is compact, there exists a linear functional $\phi : M_n(\Bbb{C}) \to \Bbb{C}$ and $s,t \in \Bbb{R}$ such that

$$Re \phi (U) < t < s < Re \phi (X)$$

for all $U \in U(n)$ and $X \in \{A\}^\perp$. Alternatively, it says $Re \phi(U(n)) \subseteq (-\infty, t)$ and $Re \phi (\{A\}^\perp) \subseteq (s, \infty)$. Since $-U(n) \subseteq U(n)$, this implies $Re \phi(-U(n)) \subseteq (-\infty, t)$ or $Re \phi (U(n)) \subseteq (-t, \infty)$. Also $iU(n) \subseteq U(n)$, so $Re \phi(iU(n)) \subseteq (-\infty, t)$ or $Im \phi (U(n)) \subseteq (-t, \infty)$. Similarly, $-iU(n) \subseteq U(n)$ shows that $Im(\phi(U(n)) \subseteq (-\infty, t)$. Hence

$$\phi U(n) \subseteq (-t,t) \times (-t,t) \subseteq \Bbb{C}$$

Doing the same for $\{A\}^\perp$, we get

$$\phi (\{A\}^\perp) \subseteq (-s,s) \times (-s,s) \subseteq \Bbb{C}$$

By Riesz Representation theorem, there is some $C \in M_n(\Bbb{C})$ such that $\phi( \cdot) = Tr(C \cdot )$...

This is basically as far as I can get. I can't figure out what the contradiction is going to look like. I'll continue to think about it, however.

Answer 1

Suppose $A$ has the SVD $A = U_0\Sigma V_0^{\mathrm{H}}$, then $\mathrm{tr}(AU) = 0$ is equivalent to $\mathrm{tr}(\Sigma V_0^{\mathrm{H}}UU_0) = 0$. Let $P$ be the circular shift matrix defined as $$ P_{ij} = \begin{cases} 1,& \text{if}~j = (i~\text{mod}~n) + 1,\\ 0,& \text{otherwise}, \end{cases} $$ then $P$ is unitary and $\text{tr}(\Sigma P) = 0$.

Hence, we have $\mathrm{tr}(AU) = 0$ if we choose $U = V_0PU_0^{\mathrm{H}}$.

Answer 2

Yes, this is true. Because every matrix can be written as a linear combination of unitary matrices. So if $tr\left(AU\right)=0$ for all unitary $U\in M_{n}\left(\mathbb{C}\right)$, then $tr\left(AB\right)=0$ for all $B\in M_{n}\left(\mathbb{C}\right)$. Thus $tr\left(A^{*}A\right)=0$, which holds iff $A=0$.