Let $A$ be a traceless $2\times 2$ complex matrix.
Its SVD reads $A=UDV^\dagger$, or in dyadic notation,
$$A=s_1 u_1 v_1^\dagger+s_2 u_2 v_2^\dagger,$$
with $\langle u_i,u_j\rangle=\langle v_i,v_j\rangle=\delta_{ij}$ and $s_i\ge0$.
The left (right) singular vectors of $A$ are $(u_1,u_2)$ and $(v_1,v_2)$, and its singular values are $s_1,s_2$.
The trace condition $\operatorname{Tr}(A)=0$ translates, in terms of its SVD, into $$s_1\langle v_1,u_1\rangle+s_2\langle v_2,u_2\rangle=0.$$
However, numerically, I find that the stronger condition $\langle u_1,v_1\rangle=\langle u_2,v_2\rangle=0$ holds. In words, the left and right singular vectors corresponding to the same singular values are always orthogonal.
You can use the following Mathematica snippet to verify it directly:
With[{mat = # - Tr[#]/2 IdentityMatrix@2 & @ RandomComplex[{-1 - I, 1 + I}, {2, 2}]},
SingularValueDecomposition@mat // Dot[ConjugateTranspose@#[[1]], #[[3]]] & // Chop // MatrixForm
]
This snippet generates random complex matrices $A$ by sampling the components from the uniform distribution in $[0,1]$, and then removing $\operatorname{Tr}(A) I/2$ to get a traceless matrix. The output is the inner product between the different left and right singular vectors, and the zeros on the diagonal correspond to the orthogonality that is the subject of this question.
At the same time, this is clearly false for many matrices, in particular normal ones, for which $u_i=v_i$.
Still, is there a way to see why left and right singular vectors are "often" orthogonal for traceless matrices?
Best Answer
If $A=USV^\ast$ is a singular value decomposition of a non-normal traceless $2\times2$ matrix $A$, then $V^\ast U$ must possess a zero diagonal.
Write $-\det(A)$ in polar form as $de^{i\theta}$. By dividing $A$ by $e^{i\theta/2}$ and by a change of orthonormal basis, we may assume that $-\det(A)=d\ge0$ and $V=I$. We want to show that $U$ has a zero diagonal.
Since $A$ has a zero trace, $A^2=dI$. Therefore $USUS=dI$.
If $A$ is singular, then $SUS=0$. Since $A$ is not normal, $S=\operatorname{diag}(s,0)$ for some $s>0$. The equality $SUS=0$ thus implies that $u_{11}=0$. As $U$ is unitary, $u_{22}$ must also be zero. Hence $U$ has a zero diagonal.
If $A$ is nonsingular, then $d>0$. From $USUS=dI$, we get $(USU^\ast)U^2 = \left(dS^{-1}\right)(I)$. By the uniqueness of polar decompositions of nonsingular matrices, we have $U^2=I$. As $U\ne\pm I$ (otherwise $A=\pm S$ is normal), the spectrum of $U$ must be equal to $\{1,-1\}$. Hence the trace of $U$ is zero. If the diagonal of $U$ is nonzero, since $A=US$ also has zero trace, $S$ must be a scalar matrix and $A=US$ is normal, which is a contradiction. Therefore $U$ has a zero diagonal.