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How to prove this matrix is positive semi-definite?

Mathematics Asked by Tree23 on November 9, 2021

Define $K:mathbb{R}^{2}timesmathbb{R}^{2} to mathbb{R}$ by

$$K(a,b)=Kleft((a_{1},a_{2}),(b_{1},b_{2})right)=a^{T}b+cosleft(frac{(a_{2}-b_{2})pi}{3}right)=a_{1}b_{1}+a_{2}b_{2}+cosleft(frac{(a_{2}-b_{2})pi}{3}right).$$

Let $a_{1},a_{2},cdots a_{n}inmathbb{R}^{2}$ are arbitrary vectors. And $A$ is $ntimes n$ matrix with $A_{ij}=K(a_{i},a_{j}).$

I want to prove that $A$ is positive semi-definite.But I don’t know how to. Any help will be thanked.

One Answer

An $ntimes n$ matrix whose $r,s$-th entry is $a_{r,s}=cos(t_r-t_s)$, for some constants $t_1,ldots,t_n$ is positive semidefinite, in the sense that if $zinmathbb C^n$, then $sum_{r,s}a_{r,s}z_rbar z_sge 0$. This can be seen by recognizing $cos t$ as the characteristic function of a random variable, namely $cos t=E[e^{itX}]$, where $X=pm 1$ with probabilities $1/2$ and $1/2$. (Which is just a complicated way of saying $cos t=(e^{it}+e^{-it})/2$.) One can check this by appealing to Bochner's theorem, or directly: $$sum_{r,s}cos(t_r-t_s)z_rbar z_s=sum_{r,s}E[e^{i(t_r-t_s)X}] z_rbar z_s\=sum_{r,s}E[e^{it_rX}z_re^{-it_sX}bar z_s]=Eleft[ left|sum_r e^{it_rX}z_rright|^2right] ge 0.$$

The OP's matrix is the sum of such a matrix and of a Gram matrix, and hence is psd as well.

Answered by kimchi lover on November 9, 2021

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