Is $\mathrm {rank}\ (AA^*A)= \mathrm {rank}\ (A)$?

Question

Let $A \in \Bbb C^{m \times n}$ and let $A^*$ denote the conjugate transpose of $A$. Then can we always have $\mathrm {rank}\ (AA^*A)= \mathrm {rank}\ (A)$?

Please give me a counter-example if it doesn't hold at all.

Thank you very much.

Answer 1

Actually its true. I have used the superscript H to denote conjugate transpose.

We will use the result $rank(A^HA)= rank (AA^H)=rank(A)$

To see this, if $A^HAx =O$, then $x^HA^HAx =O \implies |Ax|^2 =O \implies Ax =O$

Thus, $nullity(A^HA) \subset nullity(A)$. The reverse inclusion is trivial.

So, $nullity(A^HA) = nullity(A)\implies n- |nullity(A^HA)| = n- |nullity(A)| \implies rank(A^HA)= rank(A) $

Now, let $AA^HAx =O$

Write $Ax =y$ , then $AA^Hy =O \implies A^Hy =O \implies A^HAx =O $

Thus, $Ax =O \implies nullity (AA^HA) = nullity (A)$

From here you get, $rank(AA^HA) = rank (A)$

Answer 2

If $v\in N(A)$ (the null space of $A$), then $v\in N(AA^*A)$. Suppose $v\in N(AA^*A)$; then also $A^*AA^*Av=0$ and therefore $v^*A^*AA^*Av=0$, which means $$ (A^*Av)^*(A^*Av)=0 $$ so $A^*Av=0$. Then also $v^*A^*Av=0$, which means $$ (Av)^*(Av)=0 $$ so $Av=0$.

Hence $N(A)=N(AA^*A)$. Apply the rank-nullity theorem.