# A and B are nxn matrices, $A =B^TB$. Prove that if rank(B)=n, A is pos def, and if rank(B)<b, A is pos semi-def.

$A$ and $B$ are $n\times n$ matrices, A = $B^{T}B$. Prove that if $\operatorname{rank}(B)=n$, then $A$ is positive definite, and if $\operatorname{rank}(B)<n$, $A$ is positive semi-definite.

My current understanding is that if $\operatorname{rank}(B)=n$, then $\operatorname{rank}(B^{T}B)=n$ then $\operatorname{rank}(A)=n$, making $A$ positive definite.

Similarly, if $\operatorname{rank}(B)\le n$, then $\operatorname{rank}(B^{T}B)\le n$ then $\operatorname{rank}(A)\le n$, making $A$ positive semi-definite.

This doesn’t feel like a complete proof. Is there anything I am missing?

#### Solutions Collecting From Web of "A and B are nxn matrices, $A =B^TB$. Prove that if rank(B)=n, A is pos def, and if rank(B)<b, A is pos semi-def."

You have to show that for every column vector $v$, $v^TAv$ is positive.

So, $v^TAv=v^TB^TBv=(Bv)^T(Bv)=\sum_{i=1}^n(Bv)_i^2\ge0$, and if rank(B)=n and $v\neq0$, then $Bv\neq0$ and hence $\sum_{i=1}^n(Bv)_i^2>0$