# Is every square traceless matrix unitarily similar to a zero-diagonal matrix?

This question asks for the symmetric case, but after consideration I believe that any complex square matrix with zero trace is unitarily similar to a matrix with zero diagonal. This answer to another related question has a demonstration of the not necessarily unitary affirmative.

Is the unitary case known true or false already?

For reference, this is what makes me think it is true:

Consider the set of values from the diagonal one pair at a time, say $d_0$ and $d_1$. We have the principal submatrix $$\pmatrix{d_0 & x \\ y & d_1}$$ The general unitary transform (for any $c$ and $s$ such that $cc^* + ss^* = 1$) is \begin{align} & \pmatrix{c & s \\ -s^* & c^*}\pmatrix{d_0 & x \\ y & d_1}\pmatrix{c^* & -s \\ s^* & c} \\ = & \pmatrix{cd_0 + sy& cx+sd_1 \\ -d_0s^* + c^*y & -s^*x+c^*d_1}\pmatrix{c^* & -s \\ s^* & c} \\ = & \pmatrix{\vert c \vert^2d_0 +\vert s\vert^2d_1 + cs^*x + c^*sy & -csd_0 - s^2y + c^2x + csd_1\\ -c^*s^*d_0 +(c^*)^2y - (s^*)^2x + csd_1& \vert s \vert^2d_0 +\vert c\vert^2d_1 - c^*sy - cs^*x} \\ \end{align} The question at this point is if for some $c$ and $s$ can we have zero in the bottom right: $$\vert c \vert^2d_0 +\vert s\vert^2d_1 = (cs^*)x + (c^*s)y$$

From this point I visualize on the complex plane.

The left side is in terms of only magnitudes. Parameterizing the magnitude ratio of $c$ and $s$ gives the value on a line between the points $d_0$ and $d_1$.

The RHS (right hand side) is arbitrary in terms of complex angle. If $x$ and $y$ are large enough, then some angle for $c^*s$ (and opposite angle for $cs^*$) gives equality. The endpoints of the LHS line where $c=0$ or $s=0$ coincide with right hand side zero. At the middle points on the path between $d_0$ and $d_1$, the circle of angle possibilities for the right hand side grows, thus (if $x$ and $y$ are large enough) the possibility of equality exists with appropriate choice of angle for $c$ and $s$.

For smaller values of $x$ and $y$, then a point closer to zero is attainable. For $x=0$ and $y=0$ a midpoint between $d_0$ and $d_1$ is closer to zero because the pair may be chosen as such due to the zero trace. Thus an iterative method converging to zero for all points is possible.

As this argument is not terribly rigorous, I am wondering if the result is already known? Or would it be worth my time to formalize the argument?

## 3 Answers

The answer is affirmative. In the real case, I have already given a constructive proof in my answer to another question you cited. For the general case, if $A$ is traceless, then $0$ is the sum as well as the mean of the eigenvalues of $A$. In other words, $0$ lies inside the convex hull of the eigenvalues of $A$. However, the field of values (a.k.a. numerical range) $F(M)$ of any complex square matrix $M$ is a convex set that includes the spectrum of $M$ as a subset. Therefore $0\in F(A)$ as well. Hence there exists a unit vector $x$ such that $x^\ast Ax=0$. Extend this vector to a unitary matrix $U$, the $(1,1)$-th entry of $U^\ast AU$ becomes zero. Perform the same operation recursively, $A$ is unitarily similar to a zero-diagonal matrix.

The above proof, however, is merely existential and it has employed a more advanced result (the convexity of field of values) in matrix theory. I would be glad to see a neat, elementary and constructive proof. If you could devise one, please don't hesitate to post it.

• I am happy with just "there exists a unit vector $x$ such that $x^*Ax=0$. That does sum things up quite nicely! Though I guess it is not as nice for constructive purposes. When my current work gets less active, I may go back to finishing the constructive algorithm. Commented Dec 29, 2012 at 19:24
• I have completed the unitary $2 \times 2$ case, which you might be interested to see here. Commented Feb 26, 2013 at 3:50

The proof can be simplified a bit if we take the following consideration at first.

Any complex matrix $$M$$ can be written as $$M = A + iB$$ where $$A,B$$ are both Hermitian. We just set $$A = (M+M^{*})/2$$ and $$B = -i(M-M^{*})/2$$. Since $$M$$ is traceless then $${\rm Tr}(A)={\rm Tr}(B)=0$$.

At first, we use a unitary that diagonalizes $$B$$. Then we use the discrete Fourier transform, which maps traceless diagonal matrices to matrices with $$0$$ diagonal. The result will be $$M' = A' + iB'$$, where $$B'$$ has $$0$$ diagonal and $${\rm Tr}(A')=0$$.

We then just need a way to transform matrices $$\pmatrix{d_0 & x \\ y & d_1}$$, where $$d_0,d_1$$ are real and $$d_0\le 0 \le d_1$$ to the form $$\pmatrix{0 & * \\ * & d_0+d_1}$$. This is a simpler problem, I believe, because we need to solve $$\vert c \vert^2d_0 +\vert s\vert^2d_1 = (cs^*)x + (c^*s)y$$ only for real $$d_0, d_1$$.

We then use those transformations recursively to $$M'$$ $$-$$ by touching only two indices at a time with a two-level unitary.

The idea is taken from this answer.

the answer is found in Horn-Johnson: Matrix Analysis.
(Chapter 2, Section 2, Unitary equivalence, Problem 3.)
It has an iterative character - first proving it for 2x2 matrices. Best regards, L. L.

• I’m looking at the book now (2nd version). It doesn’t seem to be problem 3 of chapter 2, section 2. It doesn’t seem to be in any of the problems of that section, though the section is about Unitary Similarity. Commented Dec 9, 2021 at 15:57