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Let $A$ and $B$ be two symmetric positive semidefinite matrices.

Is it true that $\operatorname{Null}(A+B) = \operatorname{Null}(A) \cap \operatorname{Null}(B)$?

I think this is wrong, but cannot find a simple counterexample.

Lalit Tolani
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Guillaume
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1 Answers1

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Okay, so the answer is yes, it wasn't that hard in the end. Here is a detailed proof:

If $x \in Null{A} \cap Null{B}$ then it is trivial to see that $x \in Null{A+B}$. If $x \in Null{A+B}$, then \begin{equation} 0 = \langle (A+B)x,x \rangle = \langle Ax,x \rangle + \langle Bx,x \rangle, \end{equation}

where by positive semi-definiteness we have $\langle Ax,x \rangle \geq 0$ and $\langle Bx,x \rangle \geq 0$. The sum of nonnegative numbers being nonegative, we deduce that \begin{equation} \langle Ax,x \rangle=\langle Bx,x \rangle=0. \end{equation}

Since $\langle Ax,x \rangle=0$, we deduce from the fact that $A$ is symmetric positive semi-definite that $Ax=0$. Similarly, $Bx=0$, which concludes the proof.

Guillaume
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  • Sorry about my unuseful suggestion above, but I have a problem with your claim "Since ⟨Ax,x⟩=0, we deduce from the fact that A is symmetric that Ax=0". See, for example if $A=\begin{pmatrix}1&0\0&-1\end{pmatrix}$ and $x=\begin{pmatrix}1\1\end{pmatrix}$ this is not true. However, for symmetric positive semi-definite $A$ it's true, but I don't see anything less powerful than a spectral decomposition of $A$ to prove it. –  Jun 02 '21 at 17:54
  • You are correct, I was focused on the spectral decomposition (hence the symmetry) but the semi-definiteness is essential as well. – Guillaume Jun 24 '21 at 13:18
  • This was asked later... https://math.stackexchange.com/questions/4173715/singular-matrix-and-positive-semidefiniteness –  Jun 24 '21 at 14:11