1
$\begingroup$

Let $A\in\mathbb{C}^{m\times n}$, and $A^{'},A^{*}$ denotes respectively the transpose and conjugate transpose of $A.$ Then,

$(a) $$ rank(AA^{*}A)=rank(A)$

$(b)$$ rank(A)=rank(A^{2})$

$(c)$$ rank(A)=rank(A^{'}A)$

$(d)$$ rank(A^{2})-rank(A)=rank(A^{3})-rank(A^{2}).$

According to me $b,c,d$ are false. For $b,d$ we can take nilpotent matrix and $c$ is not true for complex matrices. But i don't know about $a.$ Please help me to solve this problem. Thanks in advance.

$\endgroup$
  • $\begingroup$ Hint: $(\ker A)^\perp$ is an invariant subspace of $A^\ast A$. $\endgroup$ – user1551 Mar 4 '16 at 22:05
1
$\begingroup$

Noting the following two basic facts : (1) rank$(AB) \le$ rank$(B)$, and (2) rank$(A^*A) =$ rank$(A)$, a simple proof of (a) is as follows.

rank$(A)\overset{(2)}=$ rank$(A^*A) \overset{(2)}=$ rank$(A^*AA^*A) \overset{(1)}\le$ rank$(AA^*A) \overset{(1)}\le$ rank$(A)$.

$\endgroup$
1
$\begingroup$

Let $A=U\Sigma V$ the SVD decomposition of $A$. Then the SVD decomposition of $AA^*A$ is $$AA^*A=U(\Sigma\Sigma^*\Sigma)V$$ Consider now the different cases of $\Sigma$.

Case 1: $\Sigma=[\matrix{\Sigma_1 & 0}]$ with $\Sigma_1$ diagonal. Then $$\Sigma\Sigma^*\Sigma=[\matrix{\Sigma_1\bar{\Sigma}_1\Sigma_1 & 0}]$$ Thus $$rank(A)=rank(\Sigma_1)=rank(\Sigma_1\bar{\Sigma}_1\Sigma_1)=rank(AA^*A)$$

Case 2: $\Sigma=\left[\matrix{\Sigma_1 \\ 0}\right]$ with $\Sigma_1$ diagonal. Then $$\Sigma\Sigma^*\Sigma=\left[\matrix{\Sigma_1\bar{\Sigma}_1\Sigma_1 \\ 0}\right]$$ Thus also in this case $$rank(A)=rank(\Sigma_1)=rank(\Sigma_1\bar{\Sigma}_1\Sigma_1)=rank(AA^*A)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.