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?

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