How can I show these properties of $B^TB$?

Question

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?

Does this answer your question? Is $AA^T$ a positive-definite symmetric matrix? — Jean Marie, Nov 14 '21 at 20:57

score 1 · Accepted Answer · 2021-11-14T20:35:05.827

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 $x=0$.

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