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I have two positive semidefinite matrices $A$ and $B$. Is it necessarily true that $$ rank(A+B) = rank(A^2+A+B) $$ ?

It is easy to see that $rank(A+B) \le rank(A^2+A+B)$, but for any example I try, I end up with the ranks on both sides always being equal.

user75267
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1 Answers1

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Yes, the two ranks are equal. For any positive semidefinite matrix $P$, we have $Px=0$ if and only if $x^\ast Px=0$. It follows that $$\ker(A+B)=\ker(A)\cap\ker(B)=\ker(A^2+A)\cap\ker(B)=\ker(A^2+A+B).$$

user1551
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  • Great, thats a clear and simple answer. Another related question: Is it necessarily true that

    $$ rank(XA+B) = rank((X+AY)A + B) $$ ?

    Here $X$ is a diagonal matrix with all positive diagonal entries, while $A$, $B$ and $Y$ are arbitrary positive semidefinite matrices. (This simplifies to the original question when $A = B = I$.)

     – user75267 Apr 30 '13 at 23:21
  • Nevermind, got the answer using a similar proof. Thanks! – user75267 Apr 30 '13 at 23:41
  • @user75267 Are you sure? Let $X=\pmatrix{100&0\ 0&2}$, $A=\pmatrix{8&-4\ -4&4}$, $Y=\pmatrix{4&2\ 2&77/58}$ and $B=0$. Then $XA+B$ is nonsingular but $XA+AYA+B$ is singular. – user1551 May 01 '13 at 00:58
  • That seems like a valid example, not sure what I'm doing wrong. Here's my thinking right now:

    $$ \ker(XA+AYA+B)= \ker(XA)\cap \ker(AYA) \ker(B) = \ker(A) \cap \ker(AYA)\ \cap \ker(B)=\ker(A) \cap \ker(B)=\ker(XA) \cap \ker(B)=\ker(XA+B) $$ since $\ker(XA)=\ker(A)$ and $\ker(A) \subseteq \ker(AYA)$. Any clues on where I'm going wrong?

     – user75267 May 01 '13 at 16:19
  • @user75267 For simplicity, let's consider the special case where $B=0$. Then you are arguing that $\ker(XA+AYA) = \ker(A) \cap \ker(AYA) = \ker(A) = \ker(XA)$. Among the three equalities here, the first one is false. – user1551 May 01 '13 at 17:04
  • Why so? Since $XA$ and $AYA$ are both positive semidefinite, shouldn't the equality be valid by your original argument? – user75267 May 01 '13 at 18:21
  • @user75267 No. That $X$ and $A$ are positive semidefinite doesn't make their product $XA$ positive semidefinite ($XA$ is not necessarily Hermitian in the first place). – user1551 May 01 '13 at 21:29
  • Ok, got it. I was blinded by the fact that $XA$ is positive semidefinite in my problems of interest. Thanks! – user75267 May 01 '13 at 23:26