Null Space of Matrix (Linear Algebra)

Question

I have the problem in understanding some particular case of Null Space Matrix

1. Given (3x3 Matrix) RREF (A) = $$\pmatrix{0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0}$$ Why the vector of the null space is $$\pmatrix{1 \\ 0 \\ 0}$$

instead of all zero? we know that x2 = 0, x3 = 0

Answer 1

Have a look at the definition of null space (kernel is just a synonym) of a matrix $A$: it's the set of all vectors $x$ such that $Ax=0$ (here, $0$ denotes the all-zero vector). More specifically, and without being too rigorous, the null space $\operatorname{ker}(A)$ is a vector space:

• $0 \in \operatorname{ker}(A)$, since $A\cdot 0 = 0$;
• if $x, y \in \operatorname{ker}(A)$, then $ax + by \in \operatorname{ker}(A)$, for all $a,b \in \mathbb{R}$ (I'm assuming your matrix has real entries, but you can extend the result to other fields): indeed $A(ax+by) = aAx + bAy = a\cdot 0 + b \cdot 0 = 0$.

Once you get more familiar with the theory (the link above is a good starting point), you will see that the null space of your matrix is just $$ \operatorname{ker}(A)=\left\{a\pmatrix{1\\0\\0} : a\in \mathbb{R}\right\} $$ and we say that the null space has dimension 1. And, of course, the null vector is part of it.