69. Rank–Nullity Theorem

For a linear transformation 𝑇:𝑉𝑊 between finite-dimensional vector spaces:

dim(𝑉)𝑛=dim(ker(𝑇))nullity+dim(im(𝑇))rank

In matrix form, for an 𝑚×𝑛 matrix 𝐴:

𝑛=dim(null(A))+rank(A)

(See Dimension, Null Space, and Rank.)

The dimension of the input space splits exactly into the part that gets sent to zero (the kernel) and the part that survives as the image.

69.1. Why it’s true (sketch)

Pick a basis {𝑘1,,𝑘𝑝} for ker(𝑇), then extend it to a basis {𝑘1,,𝑘𝑝,𝑣1,,𝑣𝑞} for 𝑉 (so 𝑛=𝑝+𝑞).

Then {𝑇(𝑣1),,𝑇(𝑣𝑞)} is a basis for im(𝑇) — they span (because the 𝑘’s map to 0) and are linearly independent (a non-trivial dependency would land back in ker(𝑇), contradicting linear independence of the 𝑣’s with the 𝑘’s).

So dim(im(𝑇))=𝑞, and dim(ker(𝑇))+dim(im(𝑇))=𝑝+𝑞=𝑛. ∎

69.2. Why it matters

It’s the bookkeeping identity for linear maps. Lots of immediate consequences:

Example

𝐴=[123246]. The two rows are dependent → rank(𝐴)=1.

𝑛=3 → nullity =31=2.

𝐴𝑥=𝟎 has a 2-dimensional solution space (a plane through the origin in 3).

69.3. Equivalent statements

For an 𝑚×𝑛 matrix 𝐴:

69.4. See also