Given a $n \times n$ matrix $A$ with complex entries. And $A^{\ast}$ represents the conjugate transpose of $A$.Then
If $\left | tr{\left ( A^{\ast}A \right )}\right | <n^2$, then $\left |a_{ij} \right| < 1$ for some $i,j$
If $A$ is invertible ,then $ tr{\left ( A^{\ast}A \right )}$ $\neq0$
Solution i tried- The Matrix $A^{\ast}A$ is a Hermitian matrix so all of its eigenvalues will be real.
but further i am not getting what to do ,while thinking about this question some other questions are also coming to my mind, like is $A^{\ast}A$ is a positive definite? how eigenvalues of $A$ is related to eigenvalues of $A^{\ast}A$?.
please help
Thankyou?
If $A=\begin{pmatrix}a_{11} & a_{12}& \cdots&a_{1n}\\a_{21} & a_{22}& \cdots&a_{2n} \\ \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2}& \cdots&a_{nn}\end{pmatrix}$, then $A^*=\begin{pmatrix}\bar a_{11} & \bar a_{21}& \cdots&\bar a_{n1}\\\bar a_{12} & \bar a_{22}& \cdots&\bar a_{n2} \\ \vdots & \vdots & \cdots & \vdots\\ \bar a_{1n} & \bar a_{2n}& \cdots&\bar a_{nn}\end{pmatrix}$
Now $$A^*A=\begin{pmatrix}\bar a_{11} & \bar a_{21}& \cdots&\bar a_{n1}\\\bar a_{12} & \bar a_{22}& \cdots&\bar a_{n2} \\ \vdots & \vdots & \cdots & \vdots\\ \bar a_{1n} & \bar a_{2n}& \cdots&\bar a_{nn}\end{pmatrix} \begin{pmatrix}a_{11} & a_{12}& \cdots&a_{1n}\\a_{21} & a_{22}& \cdots&a_{2n} \\ \vdots & \vdots & \cdots & \vdots\\ a_{n1} & a_{n2}& \cdots&a_{nn}\end{pmatrix}=\begin{pmatrix}\alpha_1 & *& \cdots&*\\* & \alpha_2& \cdots&* \\ \vdots & \vdots & \cdots & \vdots\\ * & *& \cdots&\alpha_n \end{pmatrix}$$ where $$\alpha_1=|a_{11}|^2+|a_{21}|^2+\cdots +|a_{n1}|^2\\ \alpha_2=|a_{12}|^2+|a_{22}|^2+\cdots +|a_{n2}|^2\\ \vdots \\ \alpha_n=|a_{1n}|^2+|a_{2n}|^2+\cdots +|a_{nn}|^2$$
If $|a_{ij}| \geq 1$, then $$\text{trace}(A^*A)=\sum \alpha_i=\sum_{i}\sum_{j} |a_{ij}|^2 \geq n^2$$
Similarly if $\text{trace}(A^*A)=0$ then $|a_{ij}|^2=0$ for each $i$ and $j$ and so $A \equiv 0$