prirodnogbk

2022-07-03

Consider the $m×m$-matrix $B$, which is symmetric and positive definite (full rank). Now this matrix is transformed using another matrix, say $A$, in the following manner: $AB{A}^{T}$. The matrix A is $n×m$ with $n. Furthermore the constraint $rank\left(A\right) is imposed.
My intuition tells me that $AB{A}^{T}$ must be symmetric and positive semi-definite, but what is the mathematical proof for this? (why exactly does the transformation preserve symmetry and why is it that possibly negative eigenvalues in $A$ still result in the transformation to be PSD? Or is my intuition wrong)?

Kiana Cantu

For symmetry: note that in general, we have $\left(AB{\right)}^{T}={B}^{T}{A}^{T}$, hence $\left(ABC{\right)}^{T}={C}^{T}{B}^{T}{A}^{T}$. With that, we see that
$\left(AB{A}^{T}{\right)}^{T}={A}^{TT}{B}^{T}{A}^{T}=AB{A}^{T}.$
For positive semidefiniteness: an $n×n$ symmetric matrix $M$ is positive semidefinite if (and only if) ${x}^{T}Mx\ge 0$ whenever $x\in {\mathbb{R}}^{n}$. We note that
${x}^{T}\left(AB{A}^{T}\right)x=\left({x}^{T}A\right)B\left({A}^{T}x\right)=\left({A}^{T}x{\right)}^{T}B\left({A}^{T}x\right).$
Because $B$ is positive definite (and hence positive semidefinite), we must have ${y}^{T}By\ge 0$ for $y={A}^{T}x$. Thus, ${x}^{T}\left(AB{A}^{T}\right)x\ge 0$, so that $AB{A}^{T}$ is indeed positive semidefinite.

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