Let $$A=\begin{pmatrix} 1 & 1 & -1 & 2\\ 1 & 1& 0 & 3\\ -1 & 0 & 1 & 0 \end{pmatrix}.$$ I have to calculate $\ker[A]$ and $\ker[A]^{T}$.
I proved that $\dim(\mathrm{Im}[A])=3$ with basis $$\begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix},\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$$ and $\dim(\ker[A])=1$ with basis $$\begin{bmatrix} -1\\ -2\\ -1\\ 1 \end{bmatrix},$$ but i have problems with $\ker[A]^T$. Do i have to calculate $A^T$ and then, with that matrix, finding the $\ker$?
Thanks in advance for any help!
There are a couple of ways to think about this. The most obvious is to take the transpose, then try to find the kernel, but this is the most inefficient way. The comments lay out another path, so let me add one more. If you know about fundamental subspaces, then you might know that $\text{Ker}(A^T)=(\text{Im} (A))^\perp$ (where this orthogonal complement is in $\mathbb{R}^3$). As you stated, the dimension of the image is $3$, which tells you that $\text{Ker}(A^T)=\{0\}.$
As the comments state, we can also think about it in the following way: the transpose is a map $A^T:\mathbb{R}^3\rightarrow\mathbb{R}^4,$ and since $3=\text{rank}(A)=\text{rank}(A^T),$ we know by the rank-nullity theorem that $\text{nullity}(A^T)=\text{dim}(\mathbb{R}^3)-\text{rank}(A^T)=3-3=0.$