Finding basis of the given Linear Transformation?

Question

Find the nullity and a basis of the null space of the linear transformation $A:\mathbb{R}^4\to\mathbb{R}^4$ given by$$ A=\begin{pmatrix} 0&1&-3&-1\\1&0&1&1\\3&1&0&2\\1&1&-2&0\end{pmatrix}. $$

Do I first convert the matrix into a linear transformation (a method that is very time consuming and tedious) and then find the null space?

Or is there a quicker method?

Answer 1

Just solve the linear system $Ax=0$ using elementary row operations.

Find the nullity and a basis of the solution space.

Answer 2

Let $A$ be an $m \times n$ matrix and let $N_{A} = \left \{ \mathbf{x}\in \mathbb{R}^{n} : A\mathbf{x}=\mathbf{0} \right \}$.

The span of the vectors in $N_{A}$ is the kernel of matrix $A$, denoted as $ker(A)$, and $dim(N_{A})$ is the nullity of $A$. If $N_{A}$ contains only the zero vector, then $nullity(A) = 0$.

For the basis, reduce the matrix to row echelon form (not reduced). The non-zero rows form a basis.