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Let $X$ be $v\times b$ matrix with $X1_{b}=r1_{v}$ and $X^T1_{v}=k1_{b}$ where $r,k$ are scalars. Then can we say that $I_v-\frac1{kr}XX^T$ is positive semi definite?

I was thinking in the lines of projections. If we prove that $B:=\frac1{kr}XX^T$ is a projection matrix. Then $I-B=B^\perp$ which is psd. However, I can not prove it. So, any help/suggestion are welcome.

Max
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1 Answers1

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First, observe that $\dfrac{1}{kr}XX^T$ is a stochastic (indeed, doubly stochastic) matrix, since $XX^T\mathbf{1}_v = kX\mathbf{1}_b = kr\mathbf{1}_v$.

We know that the largest eigenvalue of a stochastic matrix is $1$, which implies that every eigenvalue of $I_v - \dfrac{1}{kr}XX^T$ is non-negative. This proves that it is positive semidefinite.

M. Vinay
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  • This seems obvious to me, and should be to the OP or anyone else who's dealing with positive semidefinite matrices and such, but still, for anyone who can't see the implication: For any square matrix $A$, $\lambda$ is an eigenvalue of $A$ if and only if $1 - \lambda$ is an eigenvalue of $I - A$ (where $I$ is the identity matrix of the same order as $A$). – M. Vinay May 05 '16 at 15:33