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The Rank Nullity Theorem

Rank-Nullity theorem

Let \(V,W\) be vector spaces over \(F\) and \(T:V\to W\) be linear.

\[ \dim(V)=\text{rank}(T)+\text{nullity}(T) \]
Theorem

Let \(V,W\) be vector spaces over \(F\) and let \(T:V\to W\) be linear. \(\text{ker}(T)=\{0\}\) if and only if \(T\) is injective.

Theorem

Let \(V,W\) be finite dimensional vector spaces over \(F\) with same dimensions, and let \(T:V\to W\) be linear. TFAE

  1. \(T\) is injective.
  2. \(T\) is surjective.
  3. \(\text{ker}(T)=\{0\}\).
  4. \(\text{range}(T)=W\).
  5. \(T\) is invertible.

Results of the injectivity and surjectivity of linear maps

Theorem

Let \(V\) and \(W\) be vector spaces over \(F\) and suppose \(\dim(V)>\dim(W)\). There is no injective linear map from \(V\) to \(W\).

Why?

If \(T:V\to W\) is linear and injective, then by the rank-nullity theorem

\[ \text{rank}(T)\le\dim(W)<\dim(V)=\text{rank}(T)+\underbrace{\text{nullity}(T)}_{0}=\text{rank}(T), \]

which is a contradiction.

Theorem

Let \(V\) and \(W\) be vector spaces over \(F\) and suppose \(\dim(V)<\dim(W)\). There is no surjective linear map from \(V\) to \(W\).

Why?

If \(T:V\to W\) is linear and surjective, then by the rank-nullity theorem

\[ \dim(W)>\dim(V)=\text{rank}(T)+\text{nullity}(T)=\dim(W)+\text{nullity}(T)\ge\dim(W) \]

which is a contradiction.