# Simultaneous “orthonormalization” in $\mathbb{C}^4$

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 $(v_n,w_n)$.)

• Where does this question arise? – Noam D. Elkies Dec 12 '17 at 4:07
• Quantum error correction. You have some self-adjoint matrices $A_i$ and a "code" is an orthogonal projection $P$ such that $PA_iP$ is a scalar multiple of $P$ for all $i$. – Nik Weaver Dec 12 '17 at 4:24
• @NikWeaver: I thought so. In that case, because the conditions are linear in $A$, $B$, and $C$, this is only a question about the linear span of the three Hermitian matrices, so your condition can be relaxed to simply asking that some linear combination of $A$, $B$, and $C$ be positive definite, which you might as well take to be $A$, and then you might as well take $A$ to be $I_4$ and $B$ to be diagonal with maximally distinct eigenvalues that sum to zero, since, by a linear isomorphism of $\mathbb{C}^4$ and basis change, you can reduce to this case. Maybe this will simplify your problem. – Robert Bryant Dec 12 '17 at 18:28
• Quotienting out the symmetries $(v,w) \mapsto (re^{i\alpha} v, re^{i\beta} w)$, I count 13 real degrees of freedom and 9 constraints, so generically one would expect a lot of solutions. But to prove this rigorously may require a computationally intensive amount of real algebraic geometry. – Terry Tao Dec 12 '17 at 20:56
• @NikWeaver your question is tagged "linear algebra" for a good reason, and in linear algebra "positive matrix" might very well mean entrywise positive. – Dima Pasechnik Dec 19 '17 at 14:59

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.

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.

• I should say, work in an orthonormal basis which makes $B = \left[\matrix{I_2&0\cr 0&0}\right]$. – Nik Weaver Dec 28 '17 at 2:30
• Note that the answer is also yes if $B$ is a rank $1$ or $3$ projection. For then we can compress $B$ to a scalar multiple of $I_3$ on a $3$ dimensional subspace of $\mathbb{C}^4$, and then we just have to compress $C$ to a scalar multiple of $I_2$ on a $2$ dimensional subspace of that, which is easy. – Nik Weaver Dec 28 '17 at 4:06
• I haven't been following the previous answers and discussion, so I may have missed something: but in your original post, isn't B supposed to be invertible? – Yemon Choi Dec 28 '17 at 4:06
• @YemonChoi: oh, you're right. But this was an inessential restriction, as adding a scalar multiple of $A = I_4$ to $B$ doesn't change the problem. – Nik Weaver Dec 28 '17 at 4:08
• This partial solution now appears in a paper: arxiv.org/abs/1802.07394 – Nik Weaver Feb 22 '18 at 21:00

Here is an argument that will give the existence of such good pairs, provided certain connectedness property (even a weaker property) holds, which seems reasonable, but I haven't checked the details.

Let me first try to rephrase the "goodness" condition in terms of double ratios, where I use hermitian forms instead of matrices for simplicity, i.e. write $A(v,w)$ instead of $\langle Av, w\rangle$.

Suppose a pair $(v,w)$ is good for both $A$ and $B$. Then $A(v,v)=A(w,w)$ and $B(v,v)=B(w,w)$ imply that the double ratio

$$R(A,B;v,w):= \frac{A(v,v)}{A(w,w)} : \frac{B(v,v)}{B(w,w)} = 1.$$

Vice versa, suppose $R(A,B;v,w)=1$. Then I can always scale one of the vectors, say $v$, to achieve $A(v,v)=A(w,w)$, which together with the double ratio relation will simultaneously imply $B(v,v)=B(w,w)$. Thus, up to scaling, the "goodness" of $(v,w)$ for both $A$ and $B$ is equivalent to their double ratio being equal to $1$, together with the orthogonality $A(v,w)=B(v,w)=0$. In view of this equivalence, we shall in sequel always mean "good up to scaling (of one of the vectors)" without explicitly mentioning it.

Now consider any pair $(v,w)$ for which $R(A,B;v,w)\ne 1$. Note that $R$ is always positive. Then switching $v$ and $w$ leads to the inverse of their double ratio: $$R(A,B;w,v) = R(A,B;v,w)^{-1}.$$ Hence one of these ratios is $<1$ and the other is $>1$.

Next, let us bring up the orthogonality and consider the (real-algebraic) set $O(A,B)$ of all $(v,w)$ with $A(v,w)=B(v,w)=0$. I can arbitrary choose $v$ and then $w$ in the intersection of its both orthogonal complements with respect to $A$ and $B$. In particular, $O(A,B)$ is connected. Furthermore, for any pair $(v,w)\in O(A,B)$, either it is already good for $A,B$ or any path in $O(A,B)$ connecting $(v,w)$ with $(w,v)$ must have pairs with double ratio on both sides of $1$, hence the path must contain at least one good pair.

Fixing $v$, the set of all $w$ for which $(v,w)$ is $(A,B)$-good and $C$-orthogonal, if nonempty, is generically a 1-torus (generating a complex line bundle via rescaling), when $A,B,C$ are also generic. Here we rely on the property that generically the $(A,B,C)$-orthogonal complement of $v$ is 1-dimensional. The whole real-alebraic variety $G(A,B;C)$ of all pairs $(v,w)$ good for $A,B$ and $C$-orthogonal is therefore a semialgebraic torus bundle (with possible singular fibers corresponding to degenerations of the orthogonal complements) over a semialgebraic subset of codimension $1$ in $\mathbb C^4$.

Furthermore, I think that $G(A,B;C)$ should be connected, due to the above strong property that it must meet every path connecting pairs $(v,w)$ and $(w,v)$. I haven't checked the details that may require some use of topology.

Now, assuming $G(A,B;C)$ is connected, we can repeat the above path argument with the double ratios $R(A,C;v,w)$ to achieve the same conclusion, i.e. every path in $G(A,B;C)$ connecting $(v,w)$ with $(w,v)$ must contain a pair good for all $A,B,C$.

Thus, to complete the arguments, it would suffice to find a single pair $(v,w)\in G(A,B;C)$ which is in the same connected component as the flip $(w,v)$ (which is weaker than the connectedness of $G(A,B;C)$).

• It seems like after the first few paragraphs, "good" means "good up to scaling one of $v$ and $w$", right? – Nik Weaver Dec 20 '17 at 21:23
• I'm confused by the claim that "fixing $v$, the set of all $w$ for which $(v,w)$ is $(A,B)$-good and $C$-orthogonal is generically a 1-torus". Fixing $v$, the set of $w$ which are orthogonal to $Av$, $Bv$, and $Cv$ is generically a one (complex) dimensional subspace of $\mathbb{C}^4$, and all $w$ in this subspace will have the same double ratio $R(A,B;v,w)$. So I would think that for generic $v$ this set is empty. – Nik Weaver Dec 20 '17 at 21:26
• So I'm thinking that for generic $v$ there is one dimension of $w$ satisfying $\langle Av,w\rangle = \langle Bv,w\rangle = \langle Cv,w\rangle = 0$, and single values of $R(A,B;v,w)$ and $R(A,C;v,w)$ for such $w$. We know that if you switch $v$ and $w$ then these values both invert, and we want to find $v$ for which both are simultaneously $1$. It does seems like a topological problem at this point (though the fact that there are singular $v$ where $w$ is underdetermined bothers me). – Nik Weaver Dec 20 '17 at 21:43
• @NikWeaver You are right, I have made the changes. "Good" can be assumed up to scaling (of one of the vectors). "Fixing $v$" was meant to be when the fiber is nonempty, and generic refers to the top-dimensional semialgebraic strata. – Dmitri Zaitsev Dec 21 '17 at 2:04
• @NikWeaver The set of "admissible" $v$ for which there is (at least one) $w$ with $(v,w)\in G(A,B;C)$ is semialgebraic of codimension 1 in $\mathbb C^4$, and is also closed. So a singular $v$ always has some $w$ in the fiber. What may only happen, is that the $(A,B,C)$-orthogonal complement jumps in dimension for a singular $v$. You can roughly think of $G(A,B; C)$ as a real-algebraic cone projecting to a semialgebraic cone of codimension 1 in $\mathbb C^4$. – Dmitri Zaitsev Dec 21 '17 at 2:22

As R. Bryant pointed out, we may choose A=Id and focus on B,C. Since B and C are positive and invertible, there exist matrices R,S with B=R*R and C=S*S (by the Kelly-Vaught theorem) which are invertible. Assume that an orthonormal pair {v,w} which is good for both B and C exists generically, so that (v+iw,v-iw)=0. We have the conditions (Rv,Rv)=(Rw,Rw) & (Rv,Rw)=0 and similarly for S; where here (.,.) denotes the Hermitian inner product (please forgive my poor notation!). These in turn imply that (R(v+iw),R(v-iw))=(S(v+iw),S(v-iw))=0. Hence, the six vectors {R(v+iw),R(v-iw),S(v+iw),S(v-iw),v+iw, v-iw} have linear relations among them (as we are working in four dimensions). This is certainly not true for generic R and S (unless they commute and under other special conditions).

• I don't quite follow the last part ... if $x = v +iw$ and $y = v - iw$ then for any $x$ the vectors $x$, $Bx$, and $Cx$ span an at most 3 (complex) dimensional subspace of $\mathbb{C}$, so there is always a $y$ orthogonal to all three of them. – Nik Weaver Dec 18 '17 at 17:37
• *subspace of $\mathbb{C}^4$ – Nik Weaver Dec 18 '17 at 19:56
• Indeed there is no obvious benefit to working in terms of $v \pm iw$ rather than $v$ and $w$, as they will satisfy the same conditions. – Nik Weaver Dec 18 '17 at 21:09