0
$\begingroup$

Let $A$ be an $n\times n$ complex matrix and let $A^*$ denote the conjugate transpose of $A$. Then which of the following are true ?

a) If $A$ is invertible, then $tr(A^*A) \neq0$

b) If $tr(A^*A) \neq0$, then $A$ is invertble

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

d) If $tr(A^*A)=0$, then $A$ is the zero matrix

b) is false, since $A=\begin{bmatrix}i & 0\\0 & 0\end{bmatrix}$ is an example for a $2\times 2$ case.

Another thing I know is $rank(A)=rank(A^*A)$.

So how to prove/disprove others? Any help?

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

For part A take the contrapositive i.e to show$ P \implies Q $ we show ~Q$\implies $~P .Let $Tr(A^*A)=0$ this means that $\sum_{i}^n \sum_{j}^n |a_{ij}|^2=0$ .This means that each $a_{ij}=0\implies A=0$ and hence not invertible.with this you automatically prove (d) as well.hence A and D are true. To prove C again take contapositive Let $|a_{ij}|$ be greater than $1$ for each $i,j $ then $|a_{ij}|^2$ is geater than $1$ for each $i,j $ and since you have $n^2 $ elements hence the sum is greater than $n^2$.Hence c is also true

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.