7
$\begingroup$

$M,N$ are $2\times2$ real matrices, and $MN=NM$. Then, for any three real numbers $x,y,z$, we have

$$4xz\det(xM^2+yMN+zN^2)\geq(4xz-y^2)\big(x\det(M)-z\det(N)\big)^2 $$

some thought:

1). calculate directly, we got $$ \det(A-xB)=x^2\det(B)-\big(\operatorname{Tr}(A)\operatorname{Tr}(B)-\operatorname{Tr}(AB)\big)x+\det(A) $$ (where $A,B$ are $2\times2$ matrices).

2). $ 4xz\cdot \det(xM^2+yMN+zN^2)= 4x^3z\cdot \det(M-mN)\det(M-nN) $

But I don't know how to go ahead. Thanks a lot!

$\endgroup$
  • $\begingroup$ Could you explain how you got the identity $\operatorname{det}(A-xB) = x^2\operatorname{det}(B)-\big(\operatorname{Tr}(A) \operatorname{Tr}(B) - \operatorname{Tr}(AB)\big)x + \operatorname{det}(A)$? $\endgroup$ – Mussé Redi Apr 4 '14 at 8:13
  • $\begingroup$ @MusséRedi calculate directly $\endgroup$ – ziang chen Apr 4 '14 at 15:03
  • $\begingroup$ Here's one way to proceed. Observe: (a) The set of diagonalizable matrices is dense in the set of all matrices. Thus, it's sufficient to prove the inequality for diagonalizable matrices. (b) Commuting matrices have the same eignvectors (same eigenspaces). Combining this with (a) shows it's sufficient to prove the inequality for diagonal matrices $M$ and $N$. That seems like a big simplification, but I haven't worked through the details, so maybe it's still hard to prove even for diagonal $M$ and $N$. $\endgroup$ – Will Nelson Apr 5 '14 at 6:19
3
$\begingroup$

Had the OP revealed the origin of the question, I might be able to devise a nicer solution. For the time being, I can only solve the problem by a brute-force approach. As $M$ and $N$ commute, there are only three possibilities:

  1. At least one of the two matrices is nonsingular. When $M$ is nonsingular, the inequality is equivalent to $$4xz \det(xI + yNM^{-1} + z(NM^{-1})^2) \ge (4xz-y^2) (x-z \det(NM^{-1}))^2.$$ So, we may assume that $M=I$. The inequality then becomes $$4xz\det(xI + yN + zN^2) \ge (4xz-y^2) (x-z \det(N))^2.$$ One can verify that the difference between the two sides is equal to $\left[(x+z\det(N))y + 2xz\operatorname{trace}(N)\right]^2$. Hence the inequality holds.
  2. $M,N$ are singular and diagonalisable. We may assume that $M=\operatorname{diag}(m,0)$ and $N=\operatorname{diag}(0,n)$ or $\operatorname{diag}(n,0)$ for some $m,n\in\mathbb R$. When $N=\operatorname{diag}(0,n)$, the inequality reduces to $4x^2z^2m^2n^2\ge0$; when $N=\operatorname{diag}(n,0)$, the inequality reduces to $0\ge0$.
  3. $M$ and $N$ are nilpotent matrices. We may assume that they are both scalar multiples of $\pmatrix{0&1\\ 0&0}$. The inequality then reduces to $0\ge0$.

I must stress that this answer is rather unsatisfactory, as it does not really explain how the inequality arises. I hope someone can come up with a more revealing answer.

$\endgroup$
  • $\begingroup$ It's really sufficient to show for non-singular $M$. If the inequality failed for some singular $M$, then it would have to fail for all matrices $M'$ that are sufficiently close to $M$, and some of those matrices are non-singular. In my opinion, that that makes the proof nicer. $\endgroup$ – Will Nelson Apr 7 '14 at 1:43
  • $\begingroup$ @WillNelson If we use a continuity argument, we need to explain why, in any small neighbourhood of any pair of commuting singular $M,N$, there is always a pair of commuting $M',N'$ such that at least one of them is nonsingular. Since the matrices are only 2x2, explicit construction of $M',N'$ should not be too hard. But if I do so, I believe the answer would become longer than its present form. So I finally opted to consider the singular case separately, without using a continuity argument. $\endgroup$ – user1551 Apr 7 '14 at 5:21
  • 1
    $\begingroup$ I don't know the origin of the question. I found it when surfing on the Internet $\endgroup$ – ziang chen Apr 7 '14 at 8:27
  • $\begingroup$ Thanks! How did you verify $$4xz\det(xI + yN + zN^2) - (4xz-y^2) (x-z \det(N))^2 = \left[(x+z\det(N))y + 2xz\operatorname{trace}(N)\right]^2 $$ $\endgroup$ – ziang chen Apr 7 '14 at 8:28
  • $\begingroup$ @ziangchen Just calculate both sides in terms of the entries of $N$. $\endgroup$ – user1551 Apr 7 '14 at 9:48

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.