Linear Algebra I MT22, Rank-Nullity Theorem
Rank-nullity theorem
When proving the rank-nullity theorem
Let $V, W$ be finite dimensional vector spaces and $T: V \to W$ a linear transformation. Then $\mathrm{rank}(T) + \mathrm{nullity}(T) = \dim V$ .
You start by considering a basis $B _ K = \{v _ 1, \ldots, v _ m\}$ of $\ker T$. What do you extend this basis into?
A basis $B _ V = \{v _ 1, \ldots, v _ m, v _ {m+1}, \ldots v _ n\}$ of $V$.
When proving the rank-nullity theorem
Let $V, W$ be finite dimensional vector spaces and $T: V \to W$ a linear transformation. Then $\mathrm{rank}(T) + \mathrm{nullity}(T) = \dim V$ .
You consider three bases:
- $B _ K = \{v _ 1, \ldots, v _ m\}$ of $\ker T$,
- $B _ V = \{v _ 1, \ldots, v _ m, v _ {m+1}, \ldots v _ n\}$ of $V$, and
- $B _ I = \{ ??? \}$
What do you need to show is a valid basis of $\mathrm{Im}\, T$ for the result to follow?
How is the proof for the dimension formula
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then $\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)$.
and the proof of the rank-nullity theorem
Let $V, W$ be finite dimensional vector spaces and $T: V \to W$ a linear transformation. Then $\mathrm{rank}(T) + \mathrm{nullity}(T) = \dim V$ .
similar?
You do it by showing a set of a known size is a basis.
Suppose that $T : U \to V$ is a linear transformation. What is true about the rank/nullity if and only if $T$ is injective?
Suppose that $T : U \to V$ is a linear transformation. What is true about the rank/nullity if and only if $T$ is surjective?
Bite-sized
Rank-Nullity: for linear $T : V \to W$ with $\dim V < \infty$, $\mathrm{rank}(T) + \mathrm{nullity}(T) = $ $\dim V$. Matrix form: $\mathrm{rank}(A) + \dim(\ker A) = $ $n$ (the number of columns of $A$).
In NLA, what is the rank-nullity theorem used for?
Most commonly, it explains why a full-rank $A \in \mathbb R^{m \times n}$ with $m \ge n$ has trivial nullspace ($\dim \ker A = 0$), so $A^\top A$ is invertible — the cornerstone of normal-equations LS theory. It also underpins the SVD’s “first $r$ columns of $V$ span the row space, last $n - r$ span the nullspace” property.