21
$\begingroup$

Edit: since we seem a bit deadlocked at this point, let me weaken the question. It's fairly easy to see that the set of 8-tuples of reals which can be the eigenvalues of a matrix of the desired form is closed. We know from jjcale and Caleb Eckhardt that its complement is nonempty. Is its complement dense? That is, would a generic 8-tuple not be the eigenvalues of such a matrix?


First, here is a baby version of the question, that I already know the answer to. Consider complex Hermitian $4\times 4$ matrices of the form $$\left[\begin{matrix}a I_2&A\cr A^*&b I_2\end{matrix}\right]$$ where $A \in M_2(\mathbb{C})$ and $a,b \in \mathbb{R}$ are arbitrary. Can any four real numbers $\lambda_1 \leq \lambda_2 \leq \lambda_3\leq \lambda_4$ be the eigenvalues of such a matrix, or is there some restriction? Answer: there is a restriction, we must have $\lambda_1 + \lambda_4 = \lambda_2 + \lambda_3$.

The real question is: what are the possible eigenvalues of Hermitian $8\times 8$ matrices of the form $$\left[\begin{array}{c|c}aI_4&A\cr \hline A^*&\begin{matrix}bI_2& B\cr B^*&cI_2\end{matrix}\end{array}\right]$$ with $a,b,c\in\mathbb{R}$, $A \in M_4(\mathbb{C})$, and $B \in M_2(\mathbb{C})$? Can any eight real numbers be the eigenvalues of such a matrix? (I suspect not. If they could, that would tell you that any Hermitian $8\times 8$ matrix is unitarily equivalent to one of this form.)

$\endgroup$
  • $\begingroup$ Perhaps useful: if you let $M$ be your $8\times8$ matrix, then $$ \det(M-xI_4)=\det((b-x)(c-x)I_2-B^2)\det\left((a-x)I_4-A\left(\begin{matrix}(b-x)I_2& B\cr B^*&(c-x)I_2\end{matrix}\right)^{-1}A^*\right) $$ $\endgroup$ – AccidentalFourierTransform Mar 5 '18 at 16:24
  • $\begingroup$ @AccidentalFourierTransform: wait, so any eigenvalues of the bottom right $4\times 4$ are eigenvalues of the whole matrix? $\endgroup$ – Nik Weaver Mar 5 '18 at 16:57
  • $\begingroup$ It appears so. Also, note that the eigenvalues of $B^2$ are of the form $(b-\lambda_M)(c-\lambda_M)$, with $\lambda_M$ an eigenvalue of $M$. Beware: I may have messed up the algebra in the formula above. I'm just using the well-known identity $$ \det\left( \begin{array}{c|c} A & B\\ \hline C & D \end{array} \right)=\det(D)\det(A-BD^{-1}C)$$ but you should double check I didn't make any algebraic mistake before concluding anything from it. Cheers! $\endgroup$ – AccidentalFourierTransform Mar 5 '18 at 17:12
  • 3
    $\begingroup$ I don't know the answer to the general question but I think there are examples to show that not any 8 numbers will work. For example you can not get a rank one projection unitarily equivalent to a matrix of your form. If $a\neq0$, then the rank must be at least 4--likewise if $b\neq0$ or $c\neq0$ then the rank must be at least 2. But $a=b=c=0$ forces the trace to be 0. I think this should rule out the case of seven 0s and one 1. $\endgroup$ – Caleb Eckhardt Mar 6 '18 at 20:38
  • 1
    $\begingroup$ @CalebEckhardt: beautiful! Can you write this as an answer? $\endgroup$ – Nik Weaver Mar 6 '18 at 22:35
4
+300
$\begingroup$

Let $V$ be the real vector space of the $8\times 8$ matrices of the form given in the question.

Where is an open set in $\mathbb{R}^8$ of possible $8$-tuples of eigenvalues of matrices in $V$.

Proof :

Choose $M_1 \in V$ such that $M_1$ has no degenerated eigenvalues.

Let $v_1,...,v_8$ be an orthonormal base of eigenvectors of $M_1$ .

Then first order perturbation theory tells us that it suffices to show that the map

$V \rightarrow \mathbb{R}^8$, $M \mapsto (v_1^* M v_1,...,v_8^* M v_8)$ has rank $8$.

So I need $7$ additional matrices $M_2,...,M_8$ in $V$ such that the matrix $X = (v_i^* M_j v_i)_{ij}$ is nonsingular .

Let $f(a,b,c,A,B)$ be the corresponding matrix.

Choose $$ A_1 = \left[\begin{matrix}4&2&3&4\cr5&6&7&8\cr9&10&11&12\cr13&14&15&16\end{matrix}\right] $$ .

Then choose

$M_1 = f(2.7,1,-1,A_1,diag(2,1))$,

$M_2 = f(1,0,0,0,0)$,

$M_3 = f(0,1,0,0,0)$,

$M_4 = f(0,0,1,0,0)$,

$M_5 = f(0,0,0,diag(1,1,1,2),0)$,

$M_6 = f(0,0,0,diag(1,0,0,0),diag(0,1))$,

$M_7 = f(0,0,0,diag(0,1,0,0),diag(1,2))$,

$M_8 = f(0,0,0,diag(0,0,1,0),diag(3,2))$.

This gives $det X = 21.661...$ .

$\endgroup$
  • $\begingroup$ So the set of possible 8-tuples is closed and somewhere dense, but not all of $\mathbb{R}^8$. Of course that rules out it being defined by some linear relation among the eigenvalues. So this tells me a lot. $\endgroup$ – Nik Weaver Mar 11 '18 at 17:52
  • $\begingroup$ Just one question: why 2.7? $\endgroup$ – Nik Weaver Mar 11 '18 at 17:52
  • $\begingroup$ My first choice of $M_1$ was not general enough, so I changed both a and A . $\endgroup$ – jjcale Mar 11 '18 at 18:20
  • $\begingroup$ Well, thank you. This is really helpful. $\endgroup$ – Nik Weaver Mar 11 '18 at 19:07
16
$\begingroup$

There is a general framework for answering questions like this although I don't know the answer in this case. You are asking for which coadjoint orbits of $U(8)$ the moment polytope for the action of the subgroup $SU(4) \times SU(2) \times SU(2)$ contains the origin. There is a method for computing these polytopes in Berenstein, Arkady; Sjamaar, Reyer, Coadjoint orbits, moment polytopes, and the Hilbert-Mumford criterion, J. Am. Math. Soc. 13, No.2, 433-466 (2000). ZBL0979.53092..

There is a way of packaging all the coadjoint orbits together, by saying that each coadjoint orbit is a symplectic quotient of the cotangent bundle of the group. This rephrases the problem as that of computing the moment polytope for the $U(8)$ action on $T^*(U(8)/SU(4) \times SU(2) \times SU(2))$. This is going to be a convex polyhedral cone.

General theory says that you can compute the affine space spanned by the polytope from the generic stabilizer (the subgroup of elements fixing a generic point). In your case, the generic stabilizer is the same as the generic stabilizer of $SU(4) \times SU(2) \times SU(2)$ on the quotient of Lie algebras $\mathfrak{u}(8)/{\mathfrak{su}}(4) \oplus \mathfrak{su}(2) \oplus \mathfrak{su}(2)$.

In the 4 by 4 case you mentioned, to find the space perpendicular to the moment polytope you want to compute the generic stabilizer of $G = SU(2) \times SU(2)$ on $V = \mathfrak{u}(4)/\mathfrak{su}(2) \oplus \mathfrak{su}(2)$. Except for the multiples of the 2 by 2 identity this quotient can be identified with $2 \times 2$ matrices. The generic stabilizer is only defined up to conjugacy, and it's a bit tricky to find the equation for a moment polytope as opposed to one of its Weyl conjugates. The stabilizer $G_B$ at a matrix $B$ is the subgroup of $(A_1,A_2)$ so that $A_1 B A_2^{-1} = B$. Any $B$ is diagonal up to left and right multiplication, and so any diagonal $B$ with generic eigenvalues gives the generic stabilizer. But taking $B$ diagonal the resulting hyperplane doesn't meet the positive Weyl chamber; this is related to the fact that the answer that you want is going to depend on how the eigenvalues are ordered, so its better to take $B$ antidiagonal. (Once one accepts that the generic stabilizer is abelian, any full rank $B$ is ok.) Take $$B = \left[ \begin{array}{cc} 0 & b_{12} \\ b_{21} & 0 \end{array} \right] , \ |b_{12}| \neq |b_{21}|, \ \ A_2 = \left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] .$$ Then we want $$A_1 = B A_2 B^{-1}$$ to be special unitary. Since $$ B A_2 B^{-1} = \left[ \begin{array}{ll} a_{22} & (b_{12}/b_{21}) a_{21} \\ (b_{21}/ b_{12}) a_{12} & a_{11} \end{array} \right] $$ is unitary only if $a_{12} = a_{21} =0$, we have $$A_2 = \operatorname{diag}(t^{-1},t), \ \ t= a_{22}, \ A_1 = \operatorname{diag}(t,t^{-1}) .$$ Hence the generic stabilizer is the set of matrices $$diag(t,t^{-1},t^{-1},t)$$ for some complex $t$ with norm one. The Lie algeba of the stabilizer is the span of $(1,-1,-1,1)$, and the perpendicular is the space of $(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$ satisfying

$$\lambda_1 - \lambda_2 - \lambda_3 + \lambda_4 = 0 .$$

In the 8 by 8 case you want to understand, the quotient $\mathfrak{u}(8)/\mathfrak{su}(4) \oplus \mathfrak{su}(2) \oplus \mathfrak{su}(2)$ is (after forgetting about diagonal matrices) identified with the space of pairs $(A,B)$ where $A$ is 4 by 4 and $B$ is 2 by 2. Generically such matrices are full rank, and take $B$ to be a generic antidiagonal matrix implies that the the matrices in $SU(2) \times SU(2)$ must be of the form diag$(t,t^{-1},t^{-1},t)$. But then we want $$ A\,\text{diag}(t,t^{-1},t^{-1},t) A^{-1} $$
to be special unitary which is not the case for generic $A$. So the generic stabilizer has trivial Lie algebra, which means that the moment polytope is full rank (that is, no linear equations are satisfied).

In a previous version of this answer, I accidentally took $A$ to be generic "unitary" and got the wrong answer that the cone has codimension one. However, my previous wrong answer does suggest something about the facets of the cone. General theory says that the hyperplanes at the boundary of the cone are perpendicular to one-dimensional stabilizers. If one takes $A$ to be a permutation matrix then one gets an element with one-dimensional stabilizer. So I wonder whether the cone you are looking for is the cone whose facets are among those defined by equalities $$ \lambda_{\sigma(1)} - \lambda_{\sigma(2)} - \lambda_{\sigma(3)} + \lambda_{\sigma(4)} + \lambda_{\sigma(5)} - \lambda_{\sigma(6)} -\lambda_{\sigma(7)} + \lambda_{\sigma(8)} = 0 $$
where $\sigma$ ranges over elements of the eighth symmetric group. The Berenstein-Sjamaar paper would answer this with enough work. (It is a Schubert calculus computation.)

$\endgroup$
  • 1
    $\begingroup$ Thank you --- this is going to take some work for me to digest. $\endgroup$ – Nik Weaver Mar 7 '18 at 20:44
  • 2
    $\begingroup$ Counterexample : a = b = 1, c = -1, A=diag(2,1,1,1), B = 0 . $\endgroup$ – jjcale Mar 8 '18 at 20:55
  • 1
    $\begingroup$ I'm also not sure about the specific linear relation mentioned in the post here -- I tried generating random matrices of the form in the original post, and they didn't satisfy any linear relationship between the eigenvalues with all coefficients in $\{-3,-2,-1,0,1,2,3\}$. $\endgroup$ – Kevin P. Costello Mar 8 '18 at 21:42
  • 1
    $\begingroup$ Agreed - my "identity matrix" was not sufficiently generic. Once one takes a truly generic 4x 4 matrix, there are no solutions, which implies that the convex polyhedral cone is maximal rank. $\endgroup$ – Chris Woodward Mar 8 '18 at 22:02
  • 2
    $\begingroup$ Being a convex polyhedral cone of maximal rank, it could be much smaller than the full Weyl chamber. Because the cone is maximal rank, the facets of the cone you are looking for (relative to the Weyl chamber) are a subset of the hyperplanes corresponding to one-dimensional stabilizers. Taking A to be the identity gives a one-dimensional stabilizer and so one particular hyperplane, while taking A to be a different permutation matrix would give others. I wonder if the hyperplanes at the boundary of the cone are of the form (sum of four eigenvalues) = (sum of four other eigenvalues). $\endgroup$ – Chris Woodward Mar 9 '18 at 16:34
8
$\begingroup$

This is a partial answer that shows an obstruction to certain eigenvalue sequences. First, I claim that if $M$ is rank one then it isn't similar to something of the stated form.

Take a matrix $M$ of the given form and suppose the rank is $0$ or $1.$ If $a\neq0$, then the rank of $M$ is at least 4. If $b\neq0$ or $c\neq0$ then the rank of $M$ is at least 2. Since $M$ is rank 0 or 1 we must have $a=b=c=0.$ Since we have a 0 diagonal, if $A\neq0$, then the rank of $M$ would have to be at least 2 so we must have $A=0$. Similarly, the 0 diagonal and $B\neq0$ forces the rank to be at least 2 so we must have $B=0$, so $M$ is rank $0.$

That eliminates some eigenvalue sequences. Now notice that if the rank of $M$ is $\leq 3$ and $M$ is positive semidefinite this forces $a=0$ and $A=0$ so you are back in the case that you know how to deal with. This will eliminate some other eigenvalue sequences.

$\endgroup$
  • $\begingroup$ Nice. The set of tuples $\vec{\lambda} \in \mathbb{R}^8$ which can be the eigenvectors of such a matrix is closed, and you have shown its complement is nonempty ... but is it nowhere dense? $\endgroup$ – Nik Weaver Mar 7 '18 at 12:48
  • 1
    $\begingroup$ The case rank $M \le 3$ and positive semidefinite is a special case of my answer if one considers $ \Vert M \Vert I - M $ . $\endgroup$ – jjcale Mar 7 '18 at 19:15
7
$\begingroup$

Consider the case where the $8\times 8$ matrix is positive semidefinite and assume that the 5 largest eigenvalues are all equal. Then by the argument of Federico Poloni they equal a. Then it follows $A = 0$ and therefore the four smallest eigenvalues are restricted like in the $4\times 4$ case .

$\endgroup$
  • $\begingroup$ Okay, I get this now. Yes, this shows a restriction on the possible eigenvalues. Thank you! $\endgroup$ – Nik Weaver Mar 7 '18 at 20:36
2
$\begingroup$

I treat the real case. The eigenvalues of your matrix $M$ are the critical points of the restriction of the mapping $x \mapsto \langle Mx,x \rangle$ to the unit sphere. You observe that $$ \langle Mx,x\rangle = a (x_1^2+x_2^2+x_3^2+x_4^2) + b (x_5^2+x_6^2+x_7^2+x_8^2) + 2 \langle A (x_5, x_6, x_7,x_8)^*, (x_1, x_2,x_3,x_4)^*\rangle + 2 \langle B (x_7,x_8)^*,(x_5,x_6)^*\rangle. $$ So if $x$ in the unit sphere is a critical point there is a real number $\lambda$ such that $$ A \left( \begin{matrix}x_5\\x_6\\x_7\\x_8 \end{matrix} \right) =\left( \frac{\lambda-2a}{2}\right) \left( \begin{matrix}x_1\\ x_2\\ x_3\\ x_4\end{matrix}\right) $$ and $$ A^* \left( \begin{matrix} x_1\\x_2\\x_3\\x_4\end{matrix} \right) + 2 \left( \begin{matrix} B \left( \begin{matrix} x_7 \\ x_8\end{matrix} \right) \\ B^* \left( \begin{matrix} x_5 \\ x_6\end{matrix} \right)\end{matrix} \right) = \left( \frac{\lambda-2b}{2}\right) \left( \begin{matrix}x_5\\ x_6\\ x_7\\ x_8\end{matrix}\right). $$ Then multiplying the first equality by $A^*$ yields $$ A^*A \left( \begin{matrix}x_5\\x_6\\x_7\\x_8 \end{matrix} \right) =\left( \frac{\lambda-2a}{2}\right) \left[ \left( \frac{\lambda-2b}{2}\right) \left( \begin{matrix}x_5\\ x_6\\ x_7\\ x_8\end{matrix}\right) - 2 \left( \begin{matrix} B \left( \begin{matrix} x_7 \\ x_8\end{matrix} \right) \\ B^* \left( \begin{matrix} x_5 \\ x_6\end{matrix} \right)\end{matrix} \right) \right] = \left( \frac{\lambda-2a}{2}\right) \left[ \left( \frac{\lambda-2b}{2}\right) I_4 - 2 \left( \begin{matrix} 0 & B \\ B^* & 0 \end{matrix} \right) \right] \left( \begin{matrix}x_5\\ x_6\\ x_7\\ x_8\end{matrix}\right). $$ Let $a,b$ and $A,B$ be fixed. The set of eigenvalues $\mu_1, \mu_2, \mu_3, \mu_4$ of the matrix $$ \left( \begin{matrix} 0 & B \\ B^* & 0 \end{matrix} \right) $$ is contained in the set $\{\pm \lambda_1, \pm \lambda_2\}$ where $\lambda_1, \lambda _2$ are the two non-negative eigenvalues of $B^*B$.

$\endgroup$
1
$\begingroup$

It should follow from Cauchy's interlacing theorem (with compression $\begin{bmatrix}I_2\\0\end{bmatrix}$) that your first example has two eigenvalues greater than $a$ and two smaller. The same holds for $b$, considering $\begin{bmatrix}0\\I_2 \end{bmatrix}$ as compression; so in practice you have 2 eigenvalues larger than $\max(a,b)$ and two smaller than $\min(a,b)$.

Similarly, in your second example you have four eigenvalues greater than $a$ and four smaller. This argument doesn't tell all the story (for instance, in your first example it does not reveal the constraint on the sums that you have found), but at least it shows in a more general setting that there are some restrictions on the eigenvalues.

$\endgroup$
  • 3
    $\begingroup$ Interesting, but since $a$ and $b$ are arbitrary this doesn't rule out that any eight real numbers could be the eigenvalues of such a matrix ... $\endgroup$ – Nik Weaver Mar 5 '18 at 16:54
  • $\begingroup$ @NikWeaver Hmm, good point, you are correct. $\endgroup$ – Federico Poloni Mar 5 '18 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.