TITLE: Simultaneous "orthonormalization" in $\mathbb{C}^4$
QUESTION [20 upvotes]: Let $A$ be a positive, invertible $4 \times 4$ hermitian complex matrix. 
So we have a positive sesquilinear form $\langle Av,w\rangle$. Say that a pair $(v,w)$ of vectors in $\mathbb{C}^4$ is good for $A$ if they are orthogonal and have the same norm relative to the hermitian form given by $A$, i.e., $\langle Av,w\rangle = 0$ and $\langle Av,v\rangle = \langle Aw,w\rangle$.

Let $A$, $B$, and $C$ be positive, invertible $4 \times 4$ complex matrices. 
  Can we always find a pair of nonzero vectors which is simultaneously good for $A$, $B$, and $C$?


It was suggested that the answer should be yes for generic $A$, $B$, and $C$. Note that this implies it is true for all $A$, $B$, and $C$, as the set of triples for which it holds is closed. (If $(A_n, B_n, C_n) \to (A,B,C)$ then let $(v_n,w_n)$ be a good pair for $A_n$, $B_n$, and $C_n$ with $\langle A_nv_n,\rangle = \langle A_nw_n,w_n\rangle = 1$ and let $(v,w)$ be a cluster point of $(v_n,w_n)$.)

REPLY [5 votes]: I can now prove the existence of a good pair if, after rescaling so $A = I_4$, some nonzero Hermitian matrix in the span of $B$ and $C$ has a repeated eigenvalue. (But as I learned from Robert Bryant here, that generally will not be the case.)
But in this special case, since $(v,w)$ is good for $A$, $B$, and $C$ iff it is good for every Hermitian matrix in their span, wlog we can assume $B$ has an (at least) double eigenvalue. Subtracting a scalar multiple of $A = I_4$, we can assume this double eigenvalue is $0$.
If $0$ is a triple eigenvalue, then it is easy to find a pair of vectors in this three-dimensional eigenspace which is good for $I_4$ and $C$, and that solves the problem. The solution is also easy if the two nonzero eigenvalues of $B$ have opposite sign: say $B = {\rm diag}(a,-b,0,0)$ with $a,b > 0$, wlog with $\frac{1}{a^2} + \frac{1}{b^2} = 1$. Then let $W = \left[\matrix{\frac{1}{a}&0&0\cr\frac{1}{b}&0&0\cr 0&1&0\cr 0&0&1}\right]$, so
that $W^*BW = 0$. Then find a pair of vectors $v_0,w_0 \in \mathbb{C}^3$ which is good for $I_3$ and $W^*CW$ (easy) and set $v = Wv_0$, $w = Ww_0$.
The hard case is the one where both nonzero eigenvalues have the same sign. The case where they are equal is the one I treated in an earlier answer, which I'm retaining below. If they are not equal, wlog say $B = {\rm diag}(1,a,0,0)$ with $a > 1$. Similarly to the case where $a = 1$ presented below, it will suffice to find $0 \leq \lambda \leq 1$ and a $2\times 2$ unitary $U$ such that $$sC_1s + cUC_2^*s + sC_2U^*c + cUC_3U^*c$$ is a scalar multiple of $I_2$, where $C = \left[\matrix{C_1&C_2\cr C_2^*&C_3}\right]$ and $s = {\rm diag}(\sqrt{\lambda},\sqrt{\lambda/a})$, $c = {\rm diag}(\sqrt{1-\lambda},\sqrt{1 - \lambda/a})$. This is done as in Robert Bryant's solution when $a = 1$ where again, when $\lambda = 0$ the expression is just $UC_3U^*$, which becomes the Hopf map when you pass to the $S^3$-$S^2$ picture (which we can do if there is no solution to the problem). But it's a little harder here because the $\lambda = 1$ extreme no longer reduces to a constant map. However, not so much harder because some computation shows that the image of $S^3$ under the $\lambda=1$ map misses a point on $S^2$, and is therefore null homotopic, leading to the same contradiction.

Previous answer:
It follows from Robert Bryant's brilliant answer to this question that the answer is yes in an important special case, when $A = I_4$ (as pointed out in the comments, there is no essential loss of generality in assuming this) and $B$ is a rank $2$ projection.
Namely, work in an orthonormal basis that diagonalizes $B$ and write $C = \left[\matrix{C_1&C_2\cr C_2^*&C_3}\right]$ where $C_1$, $C_2$, and $C_3$ are $2\times 2$ matrices and $C_1$ and $C_3$ are Hermitian. We seek a rank $2$ projection $P$ with the property that $PBP$ and $PCP$ are both scalar multiples of $P$; if so, then any orthonormal vectors $v$ and $w$ in the range of $P$ will be good for $A$, $B$, and $C$.
If $C_3$ is a scalar multiple of $I_2$ then $P = I_4 - B$ is the desired projection. Otherwise, according to the answer cited above we can find $a \geq 0$ and a $2\times 2$ unitary $U$ such that $$C_1 + a(C_2U + U^*C_2^*) + a^2U^*C_3U$$ is a scalar multiple of $I_2$. A short computation then shows that $P = \frac{1}{1 + a^2}\left[\matrix{I_2& aU^*\cr aU&a^2I_2}\right]$ has the desired properties.