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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

$\left|A^TVA-U\right|\quad\quad(*)$

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 $U=I$.

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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
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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
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1 Answer 1

up vote 1 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.

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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
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