Today we were discussing how for an nxn orthogonal projection matrix from $\mathbb{R^{n}}$ onto a subspace W, Ker($A$)=$(Im$A$)^{\perp}$=$W^{\perp}$ and that Ker($A^{T}$) is also $W^{\perp}$. This prompted the question of what conditions are necessary for a matrix so that the kernels of it and its transpose are equal. It looks like it always works when it's an orthogonal projection, but we struggled to find an example of a square matrix for which the aforementioned identities would not hold, and consequently for which Ker($A$)$\neq$Ker($A^{T}$). It looks like we need to find a transformation whose Kernel would not consist only of things perpendicular to its image, but we were wondering if that was possible or if our reasoning was correct at all. Could you please clarify this issue (I know it's convoluted) and try to point out where we were right and wrong, and when do those identities hold?
[Math] A confusion about Ker($A$) and Ker($A^{T}$)
linear algebra
Best Answer
Things always work nicely for projections, because unlike other matrices we have $\text{Range}\oplus\text{kernel}=\text{whole space}$ Let $V=\mathbb{R}^2$ Consider $$A=\left[\begin{array}{cc} 0 & 1\\ 0 & 0\end{array}\right]$$
Then $$\text{ker}(A)=\left\{ \left[\begin{array}{c} a\\ 0\end{array}\right]\ :\ a\in\mathbb{R}\right\},$$ but $$A^{t}=\left[\begin{array}{cc} 0 & 0\\ 1 & 0\end{array}\right]$$ so that $$\text{ker}(A^{t})=\left\{ \left[\begin{array}{c} 0\\ a\end{array}\right]\ :\ a\in\mathbb{R}\right\}.$$ These are definitely different. (In contrast to the previous example, their direct sum is the whole space $V$)