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

$A$ 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?

• 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 $A^* = M^* - N^* = M - N = A$. – Vedran Šego Oct 12 '14 at 11:47

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.
• 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). – Vedran Šego Oct 12 '14 at 11:45
• 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 $U$). – Vedran Šego Oct 12 '14 at 12:01