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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 $(v_n,w_n)$.)

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    $\begingroup$ Where does this question arise? $\endgroup$ – Noam D. Elkies Dec 12 '17 at 4:07
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    $\begingroup$ 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 $i$. $\endgroup$ – Nik Weaver Dec 12 '17 at 4:24
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    $\begingroup$ @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. $\endgroup$ – Robert Bryant Dec 12 '17 at 18:28
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    $\begingroup$ 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. $\endgroup$ – Terry Tao Dec 12 '17 at 20:56
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    $\begingroup$ @NikWeaver your question is tagged "linear algebra" for a good reason, and in linear algebra "positive matrix" might very well mean entrywise positive. $\endgroup$ – Dima Pasechnik Dec 19 '17 at 14:59
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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.

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  • $\begingroup$ I should say, work in an orthonormal basis which makes $B = \left[\matrix{I_2&0\cr 0&0}\right]$. $\endgroup$ – Nik Weaver Dec 28 '17 at 2:30
  • $\begingroup$ 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. $\endgroup$ – Nik Weaver Dec 28 '17 at 4:06
  • $\begingroup$ 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? $\endgroup$ – Yemon Choi Dec 28 '17 at 4:06
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    $\begingroup$ @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. $\endgroup$ – Nik Weaver Dec 28 '17 at 4:08
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    $\begingroup$ This partial solution now appears in a paper: arxiv.org/abs/1802.07394 $\endgroup$ – Nik Weaver Feb 22 '18 at 21:00
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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 $G(A,B;C)$).

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  • $\begingroup$ It seems like after the first few paragraphs, "good" means "good up to scaling one of $v$ and $w$", right? $\endgroup$ – Nik Weaver Dec 20 '17 at 21:23
  • $\begingroup$ 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. $\endgroup$ – Nik Weaver Dec 20 '17 at 21:26
  • $\begingroup$ 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). $\endgroup$ – Nik Weaver Dec 20 '17 at 21:43
  • $\begingroup$ @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. $\endgroup$ – Dmitri Zaitsev Dec 21 '17 at 2:04
  • $\begingroup$ @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 $\mathbb C^4$. $\endgroup$ – Dmitri Zaitsev Dec 21 '17 at 2:22
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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).

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  • $\begingroup$ 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. $\endgroup$ – Nik Weaver Dec 18 '17 at 17:37
  • $\begingroup$ *subspace of $\mathbb{C}^4$ $\endgroup$ – Nik Weaver Dec 18 '17 at 19:56
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    $\begingroup$ 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. $\endgroup$ – Nik Weaver Dec 18 '17 at 21:09

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