Express the kernel of the 1 × 4 matrix A = \begin{bmatrix}1&2&3&4 \end{bmatrix} as the image of a 4 × 3 matrix B.
I understand that the kernel of a matrix is solving the system for A$\vec{x}$ = 0, but I have no idea what this question is asking nor how to do it. Are there any kind souls who can walk me through it?
On
Do you know the Rank-nullity Theorem? It states that, if $L: E \rightarrow F$ is a linear transformation, then $dim(E) = dim(im(L)) + dim(ker(L))$. Let $A:\mathbb{R}^{4} \rightarrow \mathbb{R}$ be the linear transformation such that its matrix in the canonical basis is your matrix $\mathbf{a}$. The image of $A$ is one dimensional, so from the Rank-nullity Theorem we have that $dim(ker(A)) = 3$. So, if you can find three linear independent vectors $\vec{x}_1, \vec{x}_2, \vec{x}_3$ in $\mathbb{R}^{4}$ such that $A\vec{x}_i = 0$, you will have found a basis of $ker(A)$. One such basis can be $\vec{x}_1 = (2, -1, 0, 0)$, $\vec{x}_2 = (3, 0, -1, 0)$, $\vec{x}_3 = (4, 0, 0, -1)$. Then, to construct a matrix $\mathbf{b}$ such that the image of its correspondent linear tranformation (in the canonical basis) $B: \mathbb{R}^{3} \rightarrow \mathbb{R}^{4}$ will be equal to $ker(A)$, it is enough to demand that $\vec{x}_1, \vec{x}_2, \vec{x}_3 \in im(B)$ (can you see why this is true?); we can set, for example, $B\vec{e}_i = \vec{x}_i$, and we will have
$ \mathbf{b} = \begin{bmatrix} 2 & 3 & 4 \\ -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
$Ax=0\Rightarrow [1 \ 2 \ 3 \ 4][x \ y \ z \ v ]^t=0\Rightarrow x+2y+3z+4v=0\Rightarrow x=-2y-3z-4v$
$\Rightarrow \begin{pmatrix} x \\ y \\ z \\ v \end{pmatrix}=\begin{pmatrix} -2y \\ y \\ 0 \\ 0 \end{pmatrix}+\begin{pmatrix} -3z \\ 0 \\ z \\ 0 \end{pmatrix}+\begin{pmatrix} -4v \\ 0 \\ 0 \\ v \end{pmatrix}\Rightarrow ker(A)=<\begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} -3 \\ 0 \\ 1 \\ 0 \end{pmatrix},\begin{pmatrix} -4 \\ 0 \\ 0 \\ 1 \end{pmatrix}> $ Can you continue now?