4
$\begingroup$

Question: Suppose that $\Lambda$ is an $N × N$ real-valued diagonal matrix and $Q$ is real symmetric. Suppose that $tr\ \Lambda \neq 0$ and $tr\ Q \neq 0$. Prove there is a vector $v \in \mathbb{R}^N$ such that $$v^T\Lambda v = tr\ \Lambda, v^TQv = tr\ Q $$

My thought:

Decompose the $\Lambda$ as $\Lambda = U^T\Lambda U, \ \forall\ U^TU = I $ and $Q = U^T\Lambda_1 U$.

So $$v^T\Lambda v = (Uv)^T\Lambda (Uv)= tr\ \Lambda$$ and $$v^TQ v = (Uv)^T\Lambda_1 (Uv) = tr \ Q = tr\ \Lambda_1$$ Intutively,the structure is same so they could preserve the same property. But how could I prove the $v$ in two formulas is the same one?

Could anyone help me out? Thank in advance!

$\endgroup$
5
  • $\begingroup$ From $Q = U_q^T\Lambda_1 U_q$, (fix such $U_q$), and use it on $(U_qv)^T\Lambda (U_qv) = tr \Lambda$. $\endgroup$ – induction601 Dec 10 '17 at 0:58
  • $\begingroup$ @induction601 Thanks for your comment! But I could not get your point. How do you know that the $v$ in two formulas is the same one? $\endgroup$ – stander Qiu Dec 10 '17 at 1:10
  • $\begingroup$ I guess $U^TU = I$ not necessarily implies $\Lambda = U^T\Lambda U$. $\endgroup$ – Alex Ravsky Dec 12 '17 at 2:45
  • $\begingroup$ @AlexRavsky Yep, you are right. Then the problem becomes exist $U_q$ such that $U_q^T\Lambda U_q = \Lambda$ and $Q= U_q^T\Lambda_1U_q$. Well, it seems more complexed. $\endgroup$ – stander Qiu Dec 12 '17 at 3:27
  • 1
    $\begingroup$ I thing this is a wrong problem. I guess it is about a simultaneous diagonalization of two (symmetric) matrices and I recall that it (maybe ?) is possible iff these matrices commute. Whereas I expect that to answer the original question we need some simple calculations. $\endgroup$ – Alex Ravsky Dec 12 '17 at 3:53
3
+50
$\begingroup$

The assumption that $\Lambda$ is diagonal is a premature simplification. We can just assume that $\Lambda$ is real symmetric. Also, by scaling $\Lambda$ and $Q$ if necessary, we may assume that both matrices have traces $1$. Thus we want to solve $v^T\Lambda v=v^TQv=1$ when $\Lambda$ and $Q$ are real symmetric matrices of traces $1$.

Let $D=\Lambda-Q$. By a change in orthonormal basis, we may assume that $D$ is a traceless diagonal matrix. Let $S\subset\mathbb R^N$ be the set of all vectors whose entries belong to $\{-1,1\}$. Then $\frac1{2^N}\sum_{v\in S}v^TQv=\operatorname{tr}Q>0$ and hence $v^TQv>0$ for some $v\in S$. Since $D$ is a traceless diagonal matrix, we also have $v^TDv=\operatorname{tr}D=0$.

Therefore $v^T\Lambda v=v^T(Q+D)v=v^TQv>0$. Now we can scale $v$ to make $v^T\Lambda v=v^TQv=1$.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for your answer. How do you know that $\frac1{2^N}\sum_{v\in S}v^TQv=\operatorname{tr}Q$ ? $\endgroup$ – stander Qiu Dec 13 '17 at 0:55
  • 1
    $\begingroup$ @standerQiu This is true in general for every square matrix $Q$, not only the symmetric ones. Just count the sums. Off-diagonal sums are zero because contributions from each pair of $v$s that differ by exactly one entry will cancel out each other. $\endgroup$ – user1551 Dec 13 '17 at 7:07

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.