MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Provided two diagonal real matrix which has positive entries, $V$ and $U$.

Find a real matrix $A$, satisfying $A^TA=a^2I$ for some scalar $a$, to minimise


where the matrix norm could be an induced one, or in form of $|M|^2_{F}=\mathrm{tr}(M^TM)$.

I believe the problem is quite useful, however I am not sure where I can find the related materials. A numerical approach is also welcome.

I found some related works , I think I can program the general framework for non-linear optimisation problem with unitary constraints. But since $(*)$ is only a quadratic form. I wonder if there are some improvements.

remark: there are two trivial cases, namely $V=U$ or $U=I$.

share|cite|improve this question
Maybe this comment is misguided, but bobye only stipulated $U$ and $V$ be diagonal. $A$ is apparently a scalar multiple of an orthogonal matrix, since $A^{T}A/a^{2} =I$. – drbobmeister Apr 20 '11 at 8:41
Oops, i read $A$ to be a diagonal matrix, not $A^TA$ as such! – Suvrit Apr 20 '11 at 9:56
The trace of $M^TM$ is the square of the Frobenius (=Schur, Hilbert-Schmidt) norm, but this norm is not an induced one. For instance because the norm of $I_n$ is $\sqrt n$ instead of $1$. – Denis Serre Apr 20 '11 at 10:55
Sorry, I got it. The L_2 induced norm is given as the largest singular value of M. – bobye Apr 20 '11 at 11:00
... the largest singular value of $M^TM$. – bobye Apr 20 '11 at 11:01
up vote 2 down vote accepted

The set of matrices $A$ is a cone, smooth away from the origin. With the Frobenius norm $\sqrt{{\rm Tr}(M^TM)}$, you can use differential calculus. The minimum is achieved at some $A$. If $A\ne0_n$, that is $a\ne0$, the admissible variations are $\delta A=\rho A+AB$ with $\rho\in\mathbb R$ and $B$ skew-symmetric. When writing $$\delta|A^TVA-U|^2=0,$$ you obtain on the one hand that $V^TAV$ commuttes with $U$, and on the other hand that the trace of $(A^TVA-U)A^TVA$ vanishes. Let me assume for the sake of simplicity that $U$ has pairwise distinct diagonal entries $u_i$ (=eigenvalues). Then $A^TVA$ must be diagonal; its diagonal entries $w_i$ are equal to $a^2v_{\sigma(i)}$ for some permutation $\sigma$ (consider the eigenvalues). The second requirement gives us the value of $a$: $$a^2=\frac{\vec u\cdot\vec v_\sigma}{|\vec v|^2}.$$

The value of $|A^TVA-U|^2$ is then $|\vec u|^2-(u\cdot\vec v_\sigma)^2|\vec v|^{-2}$. To minimize it, we must choose the permutation $\sigma$ that puts $v_\sigma$ in the same ordering as $u$. Say that if $u_1\ge\cdots\ge u_n$ then we reorder $\vec v$ in descreasing form as well.

share|cite|improve this answer
As you said, a permutation with a scale transform will minimise the Frobenius norm. Thank you anyway. I tested the Frobenius norm by a general proposed optimisation program. The result is also what you proved. I think Frobenius norm is not what I need in application. I may consider induced norms. – bobye Apr 20 '11 at 13:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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