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$A$ is hermitian iff $A=M-N$ for some $M,N$ positive semidefinite matrix

Suppose $A=M-N$

to show: $A^*=A$

then as $M,N$ are psd, there exists $U,W$ unitary matrix and $D_1,D_2$ diagonal matrices consisting of eigen values ($\ge 0)$ of $M,N$ respectively such that $M=U^*D_1U,N=W^*D_2W$

$A^*=M^*-N^*=U^*D_1^*U-W^*D_2^*W=U^*D_1U-W^*D_2W =M-N=A$

Could anyone tell me how to prove the otherway?

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  • $\begingroup$ You do not need the eigenvalue decomposition. You are using the fact that $M,N$ are Hermitian (otherwise, $D_1,D_2$ would not be real diagonal), so you can just say that $A^* = M^* - N^* = M - N = A$. $\endgroup$ – Vedran Šego Oct 12 '14 at 11:47
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Since $A$ is hermitian, there exist an unitary matrix $U$ a real diagonal matrix $D$ such that $A=UDU^{*}$.

Now $D$ has some positive diagonal entries, some negative diagonal and some zeros on the diagonals (not strictly like this, but you get the idea). Let $D_+$ be the matrix obtained from $D$ by replacing the negative entries by $0$ and let $D_{\bf-}$ be the matrix obtained from $D$ by replacing the positive entries by $0$. One has $D=D_++D_\bf -$. Also $D_+$ is positive semidefinite and so is $-D_\bf -$.

One gets $A=UD_+U^{*}-UD_{\bf-}U^{*}$ with $UD_+U^{*}$ and $UD_-{\bf }U^{*}$ both positive semidefinite, because similarity preserves eigenvalues and hence 'definiteness' too.

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    $\begingroup$ It is clear from the proof that the above statement could also state that $M,N$ commute (they need not always, but they can always be chosen as such). $\endgroup$ – Vedran Šego Oct 12 '14 at 11:45
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    $\begingroup$ Btw, don't you need the unitarity of $U$ as well? Without it, $UD_+U^{-1}$ and $UD_-U^{-1}$ need not be Hermitian, so not positive definite either (unless you prove that they are, but it's easier to just go for the unitarity of $U$). $\endgroup$ – Vedran Šego Oct 12 '14 at 12:01
  • $\begingroup$ @VedranŠego You're right, I didn't check that. $\endgroup$ – Git Gud Oct 12 '14 at 12:02
  • $\begingroup$ @Une Femme Douce Please note Vedran's comment above. I fixed my answer. $\endgroup$ – Git Gud Oct 12 '14 at 12:03
  • $\begingroup$ @VedranŠego Thanks, Gitgud, Thanks $\endgroup$ – Marso Oct 12 '14 at 12:05

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