Basis of null space notation

Question

Suppose $\textsf{W} = \{v\in\Bbb{R}^4 :\, a^tv=0\}$, where $$a=\begin{bmatrix}2\\-1\\3\\-5 \\ \end{bmatrix}$$ I understand that $\textsf{W}$ can be interpreted as the nullspace of a matrix, but I don't understand how to interpret this notation such that I can use it quantitativly.

How would one go about finding the original matrix that $\textsf{W}$ is the nullspace of; additionally, how would one find the basis $S$ of $\textsf{W}$.

I don't understand where to begin when using notation like this. Thanks!

Answer 1

I guess that $a^\top*v$ denotes the usual inner product. If this is the case, then $$W = \{(v_1,\ldots,v_4)\mid 2v_1-v_2+3v_3-5v_4 = 0\}$$

Answer 2

$W$ is $3$-dimensional. One basis is $\{(1,2,0,0),(0,3,1,0),(0,0,5,3)\}$. That's because the three vectors are linearly independent elements of $W$.

There will be infinitely many matrices whose kernel is $W$. That's because you are free to choose what it does (other than zero of course) to a fourth basis vector.

Answer 3

It’s there in the definition of $W$: its defining equation says that it is the null space of $a^t$, which is a $1\times4$ matrix.