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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.

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