The definition of positive definteness that I'm working with is:
"A real n by n matrix $\mathbf{A}$ is positive definite if for all conformable, non-zero vectors $\mathbf{x}$ (n by 1), the following is satisfied: $\mathbf{x^T}\mathbf{A}\mathbf{x}>0$"
So I have the following theorem that a symmetric matrix $\mathbf{A}$ (n by n) is positive definite iff it can be decomposed as $\mathbf{A}=\mathbf{C}\mathbf{C^T}$ where $\mathbf{C}$ is a square matrix (n by n) of full rank.
However I also have the following statement: Consider the matrix $\mathbf{S} = \mathbf{B}\mathbf{B^T}$ where $\mathbf{S}$ is n by n and $\mathbf{B}$ is n by m, then the matrix $\mathbf{S}$ is positive definite if $\mathbf{B}$ is of full row rank.
Is this statement true or false? I thought I could just prove it from the theorem, but because the theorem only applies to square matrices and since $\mathbf{B}$ doesn't have to be square, I can't apply the theorem.
True. That's a multiple duplicate but I can't find them...
Hint 1: every matrix of the form $BB^T$ is positive semidefinite.
Hint 2: for such matrices, $\ker BB^T=\ker B^T$.
Hint 3: $B$ has full row rank if and only if $B^T$ is injective.