Let $B$ be some matrix in $\Bbb{R}^{k*n}$ and define $A=B^TB\hspace{0.5cm}(A\in\Bbb{R}^{n*n}$)
How can I show that $A$ is symmetric and positive-semidefinite and that A is strictly positive-definite if and only if $ker(B)=\{0\}$?
I've tried this: $x^TAx=x^TB^TBx=(Bx)^T(Bx)\geq0$, hence it is positive-semidefinite. Then, how can I show that it is symmetric?
Best Answer
Keep going.
Showing that $A$ is symmetric is trivial: $$ A^t=(B^tB)^t=B^t(B^t)^t=B^tB=... $$
Notice that $y^ty=\|y\|^2\ge 0$ for any vector $y$; particularly for $y=Bx$. This, with your attempt together, gives you the positive semidefiniteness of $A$.
Suppose $A$ is strictly positive definite. Then your attempt shows that $Bx=0$ implies $x=0$. On the other hand, if $Bx=0$, then your attempt also shows that $x^tAx=0$ and thus $x=0$.