$\ker(A)=\text{Im}(A^*)^\perp$

Question

How do I show that $\ker(A)=\text{Im}(A^*)^\perp$ for any square matrix $A$. I have done this problem before with the linear operator $T$ on a hermitian space but I can't seem to apply what I have done in that problem here.

Answer 1

Let the rows of $A$ be denotes $A_i$. A vector $v$ is in the kernel if $A_i v = 0$ for all $i$, right?

Something is in the image of $A^{*}$ if it's a linear combination of the columns of $A^{*}$, which are the rows of $A$. So the image of $A^{*}$ consists exactly of values of expressions of the form $$ \sum_i c_i A_i. $$

If $w$ is perpendicular to all such expressions, then it's perpendicular to $A_1$ (by choosing $c_1 = 1, c_2 = 0, c_3 = 0, ...$) and similarly for all other $A_i$. In other words, the right hand side is contained in the left.

I think you can probably work out the rest from here; if not, just ask and I'll finish it up.

Answer 2

Let $x\in\ker A$ and $y\in \operatorname{im} A^*$ then there's $z$ such that $y=A^*z$ and then

$$\langle x,y\rangle=\langle x,A^*z\rangle=\langle Ax,z\rangle=\langle 0,z\rangle=0$$ so we proved that $\ker A\subset (\operatorname{im} A^*)^\perp$. Now let $x\in(\operatorname{im} A^*)^\perp$ so for every $y\in \operatorname{im} A^*$ we have $0=\langle x,y\rangle$ so $$\forall z\in V,\quad 0=\langle x,A^*z\rangle=\langle Ax, z\rangle\implies Ax=0\implies x\in\ker A$$ and then we have $(\operatorname{im} A^*)^\perp\subset \ker A$. The equality follows by double inclusion.