Relation between trace of matrix $A^*A$ and invertibility of the matrix $A$ Is there any relation between $\mathrm{tr}(A^*A)$ and invertibility of the matrix $A$? What information about the matrix $A$ does $A^*A$ gives ? 
I was confused about this when I came across the following statements:
Is it true that : 


*

*If $A$ is invertible, then $\mathrm{tr}(A^*A)$ is non zero.

*If $|\mathrm{tr}(A^*A)|<n^2$, then $|a_{ij}|<1$ for some $i,j$

*If $|\mathrm{tr}(A^*A)|=0$, then $A$ is a zero matrix.


Can someone help me with the answer? Thank you in advance.
 A: Let $A \in \mathbb{C}^{n \times n}$ be a complex square matrix.
Since for a complex number $z = x + iy$ we have
$$
z^* z = (x - iy)(x + iy) = x^2 + y^2 = |z|^2,
\quad \text{where } |z| := \sqrt{x^2 + y^2}
$$
Now we have
$$
A^* A =
\begin{pmatrix}
a_{1,1}^* & \ldots & a_{n,1}^* \\
\vdots & \ddots & \vdots \\
a_{1,n}^* & \ldots & a_{n,n}^*
\end{pmatrix}
\begin{pmatrix}
a_{1,1} & \ldots & a_{1,n} \\
\vdots & \ddots & \vdots \\
a_{n,1} & \ldots & a_{n,n}
\end{pmatrix}
= \begin{pmatrix}
\sum_{i = 1}^{n} a_{i,1} a_{i,1}^* & & \ast \\
& \ddots & \\
\ast & & \sum_{i = 1}^{n} a_{i,n} a_{i,n}^*
\end{pmatrix}
$$
therefore we have $\operatorname{tr}(A^* A) = \sum_{i,j=1}^{n} |a_{i,j}|^2$


*

*Let let $\operatorname{tr}(A^*A) = 0$ then $|a_{i,j}|=0$ for all so the $A$ is the zero matrix, so $\det(A)=0$, so it's not invertible, proving (1) by contraposition.

*From the above sum formula for the trace, since all summands are positive, when the sum is smaller than $n^2$, which is the number of summands, one of the summands has to be smaller than 1.

*Proven in (1).

