TheoremProved

Rank–nullity theorem

Statement

If $T: V \to W$ is a linear map between finite-dimensional vector spaces, then $\dim(\ker T) + \dim(\operatorname{im} T) = \dim V$ .

Why is it true?

Every dimension of the input space $V$ has to go somewhere under $T$ : either it gets crushed to zero (contributing to the kernel) or it survives and shows up as a genuinely new direction in the image. Since those are the only two fates and they never overlap, the dimensions of the two outcomes must add back up to the dimension you started with.

Proof sketch

Pick a basis $v_1,\dots,v_k$ of $\ker T$ and extend it to a basis $v_1,\dots,v_k,v_{k+1},\dots,v_n$ of $V$ . Show that $T(v_{k+1}),\dots,T(v_n)$ form a basis of $\operatorname{im} T$ : they span it because $T(v_1),\dots,T(v_k)$ are zero, and they are linearly independent because any dependence among them would pull back to a dependence relation in $V$ involving a nonzero element of $\ker T$ outside the span of $v_1,\dots,v_k$ , a contradiction. Hence $\dim(\operatorname{im}T)=n-k$ .

Topics that use this theorem

Vector spaces

Related theorems

Cayley–Hamilton theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Sheldon Axler (2015). Linear Algebra Done Right