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I want to show:

Given $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times p}$ with $rank(B)=p$ and $C(B)\subset C(A)$. Prove $B'AB$ is positive definite, where $C(B)$ denotes the column space of matrix $B$.

The steps below are what I have done so far.

Since $B'AB \in \mathbb{R}^{p \times p}$ , intuitively, I consider to prove its definiteness from the definition; i.e., for arbitrary nonzero (column) vector $x\in\mathbb R^p$, it suffices to show that $x'B'ABx >0$.

By $rank(B)=p$ we have $B$ is column full rank, then it follows $B'B$ is positive definite ($x'B'Bx >0$). Then how to use the condition $C(B)\subset C(A)$ to get $x'B'ABx >0$?

John Stone
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    Some condition is missing. You could have $A = [-1,0;0,1]$ and $B = [-1,0]^T$ and the matrix $B^TAB = [-1]$ and is not positive definite. – Balaji sb Aug 09 '23 at 02:57
  • @Balajisb Yes, your check is right and thanks! However, I've no idea what conditions should I add? – John Stone Aug 09 '23 at 03:16
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    Clearly what you want is true if $A$ is positive definite independent of what $B$ is. – Balaji sb Aug 09 '23 at 05:39
  • @Balajisb Yes, you are right. However, the condition $A$ is positive definite independent of $B$ is too strong; i.e., from it we can directly get $B'AB$ is positive definite without $C(B)\subset C(A)$ (see here) ! If we change the condition $C(B)\subset C(A)$ to $C(B)\supset C(A)$, is it true? – John Stone Aug 09 '23 at 13:11
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    Instead of just saying $C(B) \subset C(A)$, if we say more specifically: $C(B) \subseteq span(v: Av = \lambda v, \lambda > 0)$ then it will work. I dont think you can relax this any further. – Balaji sb Aug 09 '23 at 14:35
  • @Balajisb Great, I will try to prove it based on your specific condition. Tks! – John Stone Aug 11 '23 at 07:38

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