Linear Algebra I MT22, 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 $\text{rank}(T) + \text{nullity}(T) = \dim V$ .

You start by considering a basis $B _ K = \{v _ 1, \ldots, v _ m\}$ of $\ker T$. What do you extend this basis into?

When proving the rank-nullity theorem

Let $V, W$ be finite dimensional vector spaces and $T: V \to W$ a linear transformation. Then $\text{rank}(T) + \text{nullity}(T) = \dim V$ .

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 $\text{Im } T$ 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 $\text{rank}(T) + \text{nullity}(T) = \dim V$ .

similar?

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?