If a real $n\times n$ matrix $U$ satisfies $U+U^T\geq 0$ (i.e., positive semi-definite), does its unitarily similar counterpart $V = W U W^T$, with $WW^T = W^T W = I$, also satisfy $V+V^T \geq 0$?
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Shouldn't the transformation be $WUW^\dagger$ for unitary similarity? – NDewolf Aug 21 '20 at 09:48
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1By $A \geq 0$, do you mean that $A$ is positive semidefinite or that $A$ has positive entries? – Ben Grossmann Aug 21 '20 at 09:48
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1@BenGrossmann positive semidefinite - it is stated in the question =) – TZakrevskiy Aug 21 '20 at 09:49
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@BenGrossmann By $\geq 0$, I mean positive semidefinite. Also, the matrices $U,V,W$ are not symmetric. – unobservable_node Aug 21 '20 at 10:01
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By the way, $W$ is not required to be unitary for this to hold. It also works for an arbitrary linear operators (such that $W U W^*$ makes sense) -- see the answer of Fred below. – Rammus Aug 21 '20 at 10:05
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@NDewolf By $W^\dagger$, you mean pseudo inverse? I am following the definition of unitary similarity as in Chapter 2 of Matrix Analysis by Horn and Johnson. – unobservable_node Aug 21 '20 at 10:07
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By $\dagger$ I meant the Hermitian adjoint, which is also often (as in the book) denoted by an asterisk $*$. – NDewolf Aug 21 '20 at 10:13
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1Oh ok. All matrices are real. That is, the unitary similarity is same as the unitary congruence in this case. – unobservable_node Aug 21 '20 at 10:37
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We have $V+V^{T}=W(U+U^{T})W^{T}.$ It follows that
$$((V+V^{T})x|x)=((U+U^{T})W^{T}x|W^{T}x) \ge 0.$$
Fred
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Actually, $V+V^T = WUW^T + W^T U^T W$. The matrices $U,V,W$ are not symmetric. – unobservable_node Aug 21 '20 at 09:59
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@unobservable_node $(ABC)^T = C^T B^T A^T$, you do not need the matrices to be symmetric. – Rammus Aug 21 '20 at 10:04
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Yeah, you are right :) ... I don't know what I was confused about. Thanks for the answer. – unobservable_node Aug 21 '20 at 10:10
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@Fred Does this also hold for general similar matrices, i.e., $V=WUW^{-1}$? – unobservable_node Aug 21 '20 at 12:43
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