For $v_1,\dots ,v_n\in\mathbb{C}^n$ define $A=(\langle v_i,v_j\rangle)_{i,j=1}^n$. Prove that $A$ is a non negative linear operator on $\mathbb{C}^n$.

Question

Consider $\mathbb C^n$ as an inner product space with the standard inner product $\langle \cdot, \cdot \rangle$. For $v_1,\dots ,v_n\in \mathbb{C}^n$ define the $n\times n$ matrix $$A=(\langle v_i,v_j \rangle)_{i,j=1}^n.$$ Prove that $A$ is a non negative linear operator on $\mathbb C^n$. Does the converse hold? Prove or give a counterexample.

This link that Vincent pointed out in the comments seems to answer the non-negativity part. I am copying the necessary part of that answer here-

Let $x$ be the matrix $$ x=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix}. $$ Then $$ x^*x=\begin{bmatrix} v_1^*v_1&v_1^*v_2&\cdots&v_1^*v_n\\ v_2^*v_1&v_2^*v_2&\cdots&v_2^*v_n\\ \vdots & \vdots & \ddots & \vdots \\ v_n^*v_1&v_n^*v_2&\cdots&v_n^*v_n\\ \end{bmatrix} =\begin{bmatrix} \langle v_1, v_1 \rangle & \langle v_1, v_2\rangle & \cdots &\langle v_1, v_n \rangle \\ \langle v_2, v_1 \rangle & \langle v_2, v_2\rangle & \cdots &\langle v_2, v_n \rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle v_n, v_1 \rangle & \langle v_n, v_2\rangle & \cdots &\langle v_n, v_n \rangle \end{bmatrix}. $$ As $x^*x$ is positive-semidefinite, $\det x^*x\geq0$.

Also, the linearity of $A$ trivially follows from the linearity of matrix multiplications.

For the converse, I'm not sure how to proceed. I guess, the converse of the given statement is something like "if $A$ is a non-negative linear operator on $\mathbb C^n$, then it's entries are of the form $\langle v_i,v_j \rangle$ for $v_1,\dots ,v_n\in \mathbb{C}^n$". I feel like it's not true, but I can't produce any counterexamples. Please help me.

Also the information in this question (and its answer) might help: https://math.stackexchange.com/q/1202022/101420 — Vincent, Dec 10 '21 at 08:42
@Sayan No, the fact that the diagonal entries of $A$ are positive is not enough information to deduce that $A$ is a non-negative operator. — Ben Grossmann, Dec 10 '21 at 14:42
@Sayan The link that Vincent found doesn't only "help", it completely answers your question. If there's a part of the answer given there that you don't understand, please explain this in your question. — Ben Grossmann, Dec 10 '21 at 14:52
@BenGrossmann you wouldn't believe but actually I was just thinking of calling you again to answer this one (I know you from that question on $T$-conductors). Life can give such sweet coincidences sometimes :) — Sayan Dutta, Dec 10 '21 at 15:44
@BenGrossmann Okay, returning to my problem, I should have mentioned that properly when I added Vincent's link to my question and sorry for that. But, as far as I can understand, that link doesn't say anything about the converse. I guess, the converse of the given statement is something like "if $A$ is a non-negative linear operator on $\mathbb C^n$, then it's entries are of the form $\langle v_i,v_j \rangle$ for $v_1,\dots ,v_n\in \mathbb{C}^n$". The linked question doesn't seem to answer that. — Sayan Dutta, Dec 10 '21 at 15:44
@SayanDutta You're right, that's a good point. The important thing to know for the converse is that for every non-negative operator $A$, there exists an operator $X$ such that $A = X^X$. This is typically proved in two ways. The one that I suspect you may have seen is to note that there exists a positive operator $X$ that is a "square root" of $A$, i.e. $X^2 = A$; because $X$ is self-adjoint, we have $X^2 = X^X = A$. The other approach is use the Cholesky decomposition. — Ben Grossmann, Dec 10 '21 at 16:30
@BenGrossmann I haven't seen any of the proofs you mentioned. So, I just wanted to ask how do you arrive at a proof of the fact that such an $X$ (which satisfies $X^2=A$) exists? I can't see how to prove that. — Sayan Dutta, Dec 10 '21 at 17:47
@SayanDutta Are you aware of the spectral theorem? We know that $A$ is orthogonally diagonalizable with non-negative eigenvalues, so if we build the same matrix with the square roots of the eigenvalues we end up with the square root $X$. — Ben Grossmann, Dec 10 '21 at 17:52

For $v_1,\dots ,v_n\in\mathbb{C}^n$ define $A=(\langle v_i,v_j\rangle)_{i,j=1}^n$. Prove that $A$ is a non negative linear operator on $\mathbb{C}^n$.

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