How do you get nullspace $N(A)$ to be orthogonal to $C(A^H)$

Question

In the picture below, $C(A)$ is given in number $7$, but I am doing number $8$. I did a gauss jordan whereby I subtracted $R_2-iR_1$ to get $0$ below $1$st pivot and $1$ as the second pivot in column $2$, row $2$. Then I contineued to gauss joradan where I got; rows( $[1,0,-i], [0,1,1]$). So my null space was $N(A) = [i, -1, 1]$ vertically

But then in part be, we have to show that the nullspace is orthogonal to the hermitian; hermitian rows = $( [1, -i,], [-i,0], [0,1])$ But when I do ($(N(A)^T)^* \cdot $(Hermitan) I don't get zero in the 1x2 resultant matrix

Help me out;

webpage for better picture if needed; http://oi57.tinypic.com/ioqg6d.jpg

Answer 1

The notation $A^H$ means the hermitian (or conjugate) transpose of $A$. You want to show that, for any vector $v\in N(A)$ and any vector $w\in C(A^H)$, the (standard) inner product $$ v^Hw=0 $$ The definition of $C(A^H)$ says that $w=A^Hu$ for some $u$; then $$ v^Hw=v^HA^Hu=(Av)^Hu=0 $$ because, by assumption, $v\in N(A)$.