# Proof about positive semi-definite matrix

I used the following proof as an answer on the following question about a positive semi-definite matrix, and the OP accepted this answer, however the wikipedia page about positive semi-definite matrices made me doubt, so any thoughts or even a counterexample would be welcome. I have overwritten the question in order to make this question self-sustaining.

$\textbf{The statement}$: Let $A$ be a $n \times n$ complex invertible matrix. Denote $B = A^{-1}$. Furthermore, we denote the $i$th row of $B$ by $B_{i\bullet}$. Let $\alpha \geq \|B_{i\bullet}\|_2^2$. I have tried to prove that $$\alpha A^HA - e_ie_i^T$$ is a positive semi-definite matrix. Here $e_i$ denotes the columnvector having $1$ in the $i$th entry and $0$ in the other entries (so the $i$th standard basis vector.

$\textbf{My attempt to proof}:$ What I thought is the following: let $z$ be a complex nonzero columnvector, then we want to show that $$z^H(\alpha A^HA - e_ie_i^T)z \geq 0.$$ This is equal with showing that $$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0.$$ We have that the $2$-norm $\|x\|_2^2 = \langle x, x \rangle$, where $\langle \cdot, \cdot \rangle$ denotes the inner product of vectors. Moreover, you can compute yourself that $z^He_ie_i^Tz = |z_i|^2$, where $z_i$ is the $i$th entry of the vector $z$.

Let us now focus on the term $\alpha(Az)^H(Az)$ for which we have the following inequality: $$\alpha(Az)^H(Az) \geq \|B_{i\bullet}\|_2^2 \|Az\|_2^2$$ because of the definition of $\alpha$. Let us now use the inequality of Cauchy-Schwarz, which states that for every $p$-norm we have that $|x^Hy| \leq \|x\|_p\|y\|_p$. Applying this for $p = 2$ and the fact that $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is a strictly increasing function on $\mathbb{R}^+$, we have that $$\|B_{i\bullet}^H\|_2^2 \|Az\|_2^2 \geq |B_{i\bullet} Az|^2$$ (we can write $\|B_{i\bullet}^H\|_2^2 = \|B_{i\bullet}\|_2^2$ for a vector).

If we now note that $Az$ is a linear combination of the columns of $A$ with coefficients $z_j$ (the entries of the vector $z$) and we also have that $B_{i\bullet}a_j = \delta_{i,j}$ where $a_i$ denotes the $i$th column of the matrix $A$, we have that $$|B_{i\bullet} Az|^2 = |B_{i\bullet}a_iz_i|^2 = |z_i|^2.$$

As a result, we find that $$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0 \geq |z_i|^2 - |z_i|^2 = 0,$$ which concludes this proof.

This is what I thought should be a proof. However, I am not completely sure if this statement is even true (I started doubting when I read the wikipedia page about positive semi-definite matrices, which states that for real matrices $A^TA$ is positive semi-definite, but there is no such statement on this page regarding complex valued matrices).

$\textbf{Remark: }$ I guess it is unusual to ask a question about an answer I have given, but I am quite worried that I gave a wrong answer, which the OP accepted. Perhaps the most worrying part to me is that he asked this question regarding 'research' he is doing, so I am extremely worried that I might have given a (possible) wrong answer.

• This was the wikipedia page I consulted, but it only states that for real matrices $A$ we have that $A^TA$ is positive definite and since there appears to not be a similar statement for complex matrices, it seemed quite unintuitive to me that we coul make a matrix positive definite by substracting a zero matrix (with one diagonal entry being $1$) to make it positive semi-definite... But so you also think my proof is valid? Mar 16 '17 at 10:59
• In your solution, you haven't used that $A^T A$ is pos. def., so I think your proof is fine. Mar 16 '17 at 11:23
• @Student From that page: "For any matrix A, the matrix $A^*A$ is positive semidefinite" This includes complex matrices. Mar 16 '17 at 11:25
• As for the definition of positive definiteness for complex matrices, indeed there is a stronger and a weaker definition (regarding only the real part of $z^* M z$) - the portion of the Wikipedia page starting at my link elaborates this nicely. But again, this doesn't affect your proof. Mar 16 '17 at 11:26