Trace of symmetric matrix product
Since $B$ is rank one and positive-semi-definite (has to be p.s.d. and not p.d. since it is rank deficient) matrix, you have $B = u u^{T}$ for some $u \neq 0$. And so using $\mathrm{tr}(Auu^T) =\mathrm{tr}(u^TAu) = 0$, it follows that $u$ is an isotropic vector of $A$.
EDIT: Thanks to Loup Blanc for pointing out the mistake.
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Drew Brady
Updated on January 10, 2022Comments
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Drew Brady almost 2 years
I was thinking about the following linear algebra question, and I feel there should be a good answer as to necessary and sufficient conditions.
Let A be a symmetric, square real matrix with non-negative entries. For which rank 1 positive (non-negative) definite matrices B is it true that Tr(AB) = 0?
I've been thinking about this, and it seems tricky. I was thinking by analogy to the vector case where $c^T x = 0$ if $x$ and $c$ is orthogonal. Is there a similar notion for matrices?
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user1551 over 5 yearsBy the tracial property, if $B=vv^T$, then $\operatorname{tr}(AB)=v^TAv$.
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Confounded almost 2 years@user1551 Hi. Would you happen to know if there is also a result for $tr(ABAB)$ where $A$ is symmetric and $B = v v^T$? Many thanks
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user1551 almost 2 years@Confounded If $B=vv^T$, the trace of $ABAB$ is the square of the trace of $AB$. So, this is the same question.
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Confounded almost 2 years@user1551 Thank you for your reply, but I am not sure I fully understand it. The trace of $ABAB$ is not the square of the trace of $AB$, it is the trace of the square of $AB$, or are you saying that in this case the two things are equal? Thank you.
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user1551 almost 2 years@Confounded They are equal, because $B=vv^T$.
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Admin over 5 yearsYour sentence "it follows that..." is false.
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onetimething over 5 yearsI'm sorry, but I might be missing something. Why is it not true? $u^TAu$ is a scalar, and thus it's trace equals the value itself. Further, $u^T A u = \sum_{i,j} u_i u_j A_{ij}$. Since all entries of $A$ are non-negative, the only combination that works is when the condition holds.
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Admin over 5 yearsWhen $A$ is diagonalized, $u^TAu=\sum_i\lambda_i u_i^2$; note that the $(\lambda_i)$ and the $(u_i)$ may be $<0$. $u$ is an isotropic vector for $A$, that gives one quadratic relation. Here the hypothesis $A_{i,j}\geq 0$ is useless.
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onetimething over 5 yearsYes, you're right. Should I delete my answer?
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Admin over 5 yearsNo. Say that $u$ is an isotropic vector for $A$.