# Rank of matrices.

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.

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

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$(A)$.
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 $$rank(A)=rank(\Sigma_1)=rank(\Sigma_1\bar{\Sigma}_1\Sigma_1)=rank(AA^*A)$$