Cfrgtkky

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 :

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


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


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

Can someone help me with the answer? Thank you in advance.

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 $operatorname{tr}(A^* A) = sum_{i,j=1}^{n} |a_{i,j}|^2$

1. 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.


2. 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.


3. Proven in (1).