Let $A$ be an $n \times n$ symmetric positive definite matrix, and let $M$ be an $n \times m$ matrix.

Show that the matrix $M^tAM$ is symmetric positive definite (spd).

I am trying to use the definition of an spd matrix, showing that the inner product is $\geq 0$. So, $\langle x^tM^tAM,x\rangle \geq 0$ and since $A$ is spd, it is also diagonazable $A=V^tLV$. Am I going in the right direction?


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    $\begingroup$ What happens when $M$ is the zero-matrix? $\endgroup$ – Moritz Mar 4 '15 at 20:03
  • $\begingroup$ @Moritz right, but $A$ positive semi definite implies that $M^tAM$ is positive semi definite. $\endgroup$ – Surb Mar 4 '15 at 20:04
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    $\begingroup$ I only see "symmetric positive definite" with no "semi". Sorry, just kidding, I know what is meant. $\endgroup$ – Moritz Mar 4 '15 at 20:06

For every $x$ we have $$\langle M^tAMx,x\rangle = \langle AMx,Mx\rangle = \langle Ay,y\rangle\geq 0$$ since $\langle Ay,y\rangle \geq 0$ for every $y$, and $$(M^tAM)^t = M^tA^t(M^t)^t = M^tA^tM=M^tAM$$ since $A$ is symmetric.

As noted by @Moritz and @Mnifldz this shows only that $M^tAM$ is positive semi-definite. If you want to guarantee that $M^tAM$ is positive definite, then you'll need to ask for $\ker(M)=\{0\}$. Then $y\neq 0$ in the proof above and $A$ positive definite implies $\langle Ay,y\rangle>0$. Otherwise, if $\ker(M)\neq \{0\}$, pick any $x\in \ker(M)$ with $x \neq 0$ and note that $\langle M^TAMx,x\rangle = \langle 0,x\rangle = 0$.

  • $\begingroup$ What about the case where $m>n$ and we are guaranteed that there is some nontrivial $x \in \ker M$? I think your proof only shows positive semi-definiteness. $\endgroup$ – Mnifldz Mar 4 '15 at 20:11
  • $\begingroup$ @Mnifldz You are right, my proof only shows positive semi-definiteness. This was already noticed by Moritz in the OP comments. To get positive-definiteness, we need $\ker(M)=\{0\}$ (as you just said). $\endgroup$ – Surb Mar 4 '15 at 20:17

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