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If $A$ is a symmetric and positive definite matrix and matrix $B$ has linearly independent columns , is it true that $B^T A B$ is symmetric and positive definite?

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up vote 3 down vote accepted

Only hints, because it is a homework question:

Symmetric just means $M^T=M$. Given $(AB)^T=B^TA^T$, the symmetry should be easy to check for you.

If $B$ has linearly independent columns, for $v\ne 0$, you'll also have $Bv\ne 0$ (why?). So given $w=Bv$, what is $w^TAw$?

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Thank you for the hints! It took a while for me to understand and form a proof. But eventually, it helped. – Ashwin Sep 20 '12 at 10:05

If the matrices are real yes: take $x\in\Bbb C^d$. Then $Bx\in \Bbb C^d$ hence $x^tB^tABx=(Bx)^tA(Bx)\geq 0$ and if $x\neq 0$, as $B$ is invertible $Bx\neq 0$. $A$ being positive definite we have $x^tB^tABx>0$.

But if the matrices are real it's not true: take $A=I_2$, $B:=\pmatrix{1&0\\0&i}$, then $B^tAB=\pmatrix{1&0\\0&-1}$ which is not positive definite. But it's true if you replace the transpose by the adjoint (entrywise conjugate of the transpose).

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