Proof: Symmetric and Positive Definite 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?

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).
