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The below quote is copy from "Problems and Theorems in Linear Algebra" Author is : V.Prasolov

Let $A=||a_{ij}||_{1}^{n}$ be an Hermitian matrix, if $A$ is positive definite, then the matrix $||a_{ij}||_{1}^{k}$ corresponds to the restriction of a positive definite Hermitian form $x^*Ax$ to a subspace and ,therefore, $|a_{ij}|_{1}^{k}>0$.


Question: Assume $A$ is the matrix of operator $\mathcal{A}$ on the space $V$ under the basis $\{\alpha_1, \cdots, \alpha_n\}$. Does this sentence

then the matrix $||a_{ij}||_{1}^{k}$ corresponds to the restriction of a positive definite Hermitian form $x^*Ax$ to a subspace

means there exists a subspace $M \subset V$,such that $\mathcal{A}$ restrict to it is $||a_{ij}||_{1}^{k}$(under some property basis of $M$)? How to find this subspace? Or the author means something else ?

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What the author is saying is that if you let $M=\mbox{span }\,\{\alpha_1,\ldots,\alpha_k\}$ and you write $A'$ for the "left upper corner of size $k$" of $A$, then for any nonzero $y\in M$ you have $$ y^*A'y=\begin{bmatrix}y\\0\end{bmatrix}^*\,A\,\begin{bmatrix}y\\0\end{bmatrix}>0 $$ (where the zeroes in the matrices above have size $(n-k)\times1$). So $A'$ is positive definite.

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