# 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 :

1. If $$A$$ is invertible, then $$\mathrm{tr}(A^*A)$$ is non zero.
2. If $$|\mathrm{tr}(A^*A)|, then $$|a_{ij}|<1$$ for some $$i,j$$
3. If $$|\mathrm{tr}(A^*A)|=0$$, then $$A$$ is a zero matrix.

• What is $A^*$ in context? Complex conjugate of $A$? Conjugate-transpose of $A$? – Richard Martin Nov 19 '18 at 13:43
• @RichardMartin Conjugate transpose – Jean-Claude Arbaut Nov 19 '18 at 13:44
• Hint: $\mathrm{tr}(A^*A)=\sum_{i,j} |a_{ij}|^2$. @RichardMartin No, it's not the "sum of squares", there is a modulus. – Jean-Claude Arbaut Nov 19 '18 at 13:45

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.