Kernels of Adjoints

Question

Let $A$ be an $m \times n$ matrix. Show that $\mbox{Ker} A = \mbox{Ker} (A^*A)$. To do that you need to prove 2 inclusions, $\mbox{Ker} (A^*A)$ is a subset of $\mbox{Ker} A$ and $\mbox{Ker} A$ is a subset of $\mbox{Ker} (A^*A)$.

Answer 1

It $Ax=0$, I am sure you can prove that $A^*Ax=0$, This yields the latter inclusion.

For the former, all you need is the following observation: $$ \|Ax\|^2=(Ax,Ax)=(x,A^*Ax) $$ where $(x,y)$ denotes the usual inner product on the column vectors defined by $$ (x,y)=x^*y=\sum_{j=1}^n\overline{x_j}y_j $$ and $\|x\|=\sqrt{(x,x)}$ is the resulting Euclidean norm.