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Let $ A \in \mathbb{R}^{n \times m}, $ with $1<m<n$, where $ \text{rank}(A)=m $ (i.e., $ A $ is a tall matrix of full column rank). Let $ B \in \mathbb{R}^{n \times n} $ be a positive definite matrix. Can we say anything about the definiteness of the product $ A^{\text{T}}BA $?

My gut feeling is since $A$ is of full column rank and $m<n$ that $ A^{\text{T}}BA $ is positive definite; however, I cannot find a way to show this. I have seen many posts about the definiteness of matrix product where all matrices are either positive definite or positive semi definite; however, I cannot seem to find anything quite like the question I have presented.

A little background, I am working on some research where I need $ A^{\text{T}}BA $ to be invertible. If invertibility cannot be guaranteed then I need to take another approach.

AdamsK
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    By assumptions $\ker A={0},$ hence $A^TBA$ is positive definite. – Ryszard Szwarc Jun 01 '23 at 15:10
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    @AnneBauval There is no symmetry assumption for $B.$ So representation $B=C^TC$ may be not available. – Ryszard Szwarc Jun 01 '23 at 15:22
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    @AnneBauval In the complex case the self adjointness follows. But in this particular real problem symmetry is not needed. If $\langle BAx,Ax\rangle= 0,$ then $Ax=0,$ since $B$ is positive definite. Then $x=0,$ since the columns of $A$ are linearly independent. – Ryszard Szwarc Jun 01 '23 at 15:31

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