Lets say $A, B$ are rank 1 hermitian real matrices. It follows immediately that $A\otimes A$ and $B\otimes B$ individually are rank 1. My question is how does the largest eigenvalue of $A+B$ relate to the largest eigenvalue of $A\otimes A+B\otimes B$. I have phrased the question as if A,B are 2 by 2 matrices, but I would like to know the intuition in general for arbitrary dimensions. Any references would also be appreciated!


As $A$ is rank-1 Hermitian, its only nonzero eigenvalue is $a=\operatorname{tr}(A)$. Define $b$ similarly for $B$.

Weyl's inequality states that if $X$ and $Y$ are Hermitian matrices of the same sizes, and $\lambda_k(\cdot)$ denotes the $k$-th smallest eigenvalue of a matrix, then $$ \lambda_k(X)+\lambda_\min(Y)\le\lambda_k(X+Y)\le\lambda_k(X)+\lambda_\max(Y).\tag{1} $$ It follows that $$ \max\left\{\lambda_\max(X)+\lambda_\min(Y), \ \lambda_\max(Y)+\lambda_\min(X)\right\}\le\lambda_\max(X+Y)\le\lambda_\max(X)+\lambda_\max(Y).\tag{2} $$ Now, as $a$ is the only nonzero eigenvalue of $A$, we have $\lambda_\max(A\otimes A)=a^2$ (even when $a$ is negative) and $\lambda_\min(A\otimes A)=0$ and similarly for $B\otimes B$. So, $(2)$ gives $$ \max\{a^2,b^2\}\le\lambda_\max(A\otimes A+B\otimes B)\le a^2+b^2\tag{3} $$ and both bounds are sharp: the lower bound is attained when $A=\operatorname{diag}(a,0,0,\ldots,0)$ and $B=\operatorname{diag}(0,b,0,\ldots,0)$, while the upper bound is attained if take $B=\operatorname{diag}(b,0,0,\ldots,0)$ instead.

Note that the eigenvalues of $A\otimes A+B\otimes B$ are not solely determined by $a$ and $b$. More specifically, by spectral decomposition, we can always write $A=auu^\ast$ and $B=bww^\ast$ for some unit vectors $u,w$. Let $w=cu+sv$ for some unit vector $v\perp u$. By picking some appropriate phase angles for $w$ and $v$, we may assume that $c$ and $s$ are real nonnegative and hence $s=\sqrt{1-c^2}$. So, by a change of basis, we may assume that $$ A=\pmatrix{a&0\\ 0&0}\oplus0,\quad B=b\pmatrix{c^2&sc\\ sc&s^2}\oplus0. $$ It follows that $c = \sqrt{\dfrac{\operatorname{tr}(ABA)}{aba}}$. Therefore, the eigenvalues of $A+B$ and $A\otimes A+B\otimes B$ are completely determined by $a,b$ and $c$.

  • $\begingroup$ Thanks for your detailed answer. Just to ensure we are on the same page, suppose a=b=1, then the eigen values of A+B are (1-c),(1+c)? (because the trace of A+B is 2 and the determinant is (1-c^2)). ALso could you give a little bit more information about the last step on how you could write A and B in terms of the direct sum of matrices? I sort of lost you there. $\endgroup$ – Annonymous Apr 15 '16 at 9:34
  • $\begingroup$ @Annonymous I don't get your comment. Why is the determinant of A+B equal to $1-c^2$? $\endgroup$ – user1551 Apr 15 '16 at 10:13
  • $\begingroup$ Sry, that isnt right, determinant of A+B is ofcourse 0 cause it involves a direct sum with the 0 submatrix. The top left 2 x 2 submatrix has trace 2 and determinant $1-c^2$ $\endgroup$ – Annonymous Apr 15 '16 at 12:43
  • $\begingroup$ @Annonymous I see. Yes, the only two possibly nonzero eigenvalues of $A+B$ are the two roots of $x^2-(a+b)x+ab(1-c^2)$. In the particular case where $a=b=1$, they become $1\pm c$. So, you see, they are not determined solely by $a$ and $b$. You also need $c$. The case for $A\otimes A+B\otimes B$ are similar. In an appropriate basis, it is the direct sum of a zero block and a $4\times4$ matrix, whose entries are determined by $a,b$ and $c$. So you also need $c$ to calculate its eigenvalues. $\endgroup$ – user1551 Apr 15 '16 at 14:16
  • $\begingroup$ And you cannot infer its eigenvalues from the eigenvalues of $A+B$ because $A+B$ has only two nonzero eigenvalues, but you need three parameters a,b,c to determine the eigenvalues of $A\otimes A+B\otimes B$. In this sense, getting an inequality is the best we can do. $\endgroup$ – user1551 Apr 15 '16 at 14:18

This is not a full answer, but I wish it can help.

Let $\lambda_{max}$ be the largest eigenvalue of $A+B$,

and $\Lambda_{max}$ be the largest eigenvalue of $C=A \otimes A + B \otimes B$.

I have done extensive simulations (see Matlab program and accompanying figure: stars for $\Lambda_{max}$, circles for $\lambda_{max}$): something general and interesting can be conjectured about a certain threshold situated here around $s=1.75$, yielding broadly two cases:

  • either $s\leq\lambda_{max}\leq\Lambda_{max}$ or the reverse situation:

  • $\Lambda_{max}\leq\lambda_{max}\leq s$.

  • But this classification suffers exceptions that should be analysed in a more thorough manner.

Did you have a similar conjecture ?

The part of analysis I have done is as follows (but I have not enough time to devote to it).

The term hermitian can be replaced here by symmetric $n \times n$.

$A$ and $B$ being rank one, their resp. spectra are $\{\lambda_A,0,0,\cdots,0\}$ and $\{\lambda_B,0,0,\cdots,0\}$ ($n-1$ zeros) with $\lambda_A,\lambda_B$ real because they are symmetric.

$A+B$, in general is thus a symmetric rank 2 matrix, meaning that it has two non zero real eigenvalues.

The spectra of $A \otimes A$ and $B \otimes B$ are thus resp. $\{\lambda_A^2,0,0,\cdots,0\}$ and $\{\lambda_B^2,0,0,\cdots,0\}$ (with $n^2-1$ zeros), meaning that they are both rank 1 matrices. Their sum $C=A \otimes A + B \otimes B$ is thus a rank 2 matrix in general.

Here is the Matlab program and (below) the graphical display

clear all;close all;hold on i=complex(0,1); for k=1:50 U=rand(3,1); A=U*U'; KA=kron(A,A); V=rand(3,1); B=V*V'; KB=kron(B,B); eKAB=eig(KA+KB); I=find(abs(eKAB)>0.01); LKAB=eKAB(I)'; eAB=eig(A+B); I=find(abs(eAB)>0.0001); LAB=eAB(I)'; plot(k*i+max(LAB),'or'); plot(k*i+max(LKAB),'*b'); plot(k*i+[max(LAB),max(LKAB)]) end;

  • $\begingroup$ Thanks for your answer. I am not quite sure I follow your 3 observations earlier on. From where did you get this 1.75, doesnt this constant seem too specific for a rather generic question of mine? I had conjectures similar to what you mention, but I wanted something more specific actually like the answer earlier by @user1551. $\endgroup$ – Annonymous Apr 15 '16 at 9:36
  • $\begingroup$ I agree for the 1.75 which indeed is very specific to the range of values of the entries of matrices $A$ and $B$ (it is the meaning of the word "here"). If I have confirmed a conjecture that you have done, at least, this has not been done for nothing... $\endgroup$ – Jean Marie Apr 15 '16 at 17:14

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.