Account for lost and retained dimensions

Try

A matrix with seven columns and three pivots has four free input directions. Rank counts independent output directions; nullity counts independent inputs erased by the map. Their sum is the input dimension, not generally the number of rows.

Predict

Predict the rank bound for an m×n matrix. Its rank cannot exceed either m or n. Rank-nullity says rank A+dim ker A=n. For a map between abstract finite-dimensional spaces, replace n by the dimension of its domain.

Experiment

Row-reduce [[1,2,3],[0,1,1]]. Its two pivot columns show rank 2; one free variable shows nullity 1. Solve for the null vector (−1,−1,1). Select columns 1 and 2 from the original matrix as an image basis; do not substitute the altered columns from echelon form.

Understand

Worked proof of rank-nullity: take a kernel basis k₁,…,kᵣ and extend it to a domain basis by v₁,…,vₛ. Every output is a combination of T(vⱼ), because all T(kᵢ) vanish. If ΣaⱼT(vⱼ)=0, then Σaⱼvⱼ lies in the kernel and is a combination of the kᵢ. Independence of the full basis forces every aⱼ=0. Thus T(vⱼ) form an image basis, of size s, and dim domain=r+s.

Row reduction preserves linear relations among columns because it applies an invertible map to every column. It can change their actual location, so pivot columns used as a basis for the original column space must be taken from the original matrix. Nonzero rows of echelon form can be used for the row space.

Apply

Use the domain and codomain labels before applying a dimension theorem. An onto map R⁵→R³ has rank 3 and nullity 2; it cannot be injective. A map can be neither onto nor injective, so these are not complementary alternatives.

Transfer: how much does composition forget?

Consider linear maps A:V→W and B:W→U on finite-dimensional vector spaces over the same field. A vector may disappear at the first step, or its surviving image may disappear at the second. These are different mechanisms. Simply adding the nullities of A and B counts directions of ker B that A may never reach.

Claim and proof. Restrict A to ker(BA). Its image is exactly im A ∩ ker B: an image Av from this restriction satisfies BAv=0; conversely, if w=Av and Bw=0, then v lies in ker(BA). The kernel of this restricted map is ker A. Rank-nullity for the restriction therefore gives

$$\dim\ker(BA)=\dim\ker A+\dim(\operatorname{im}A\cap\ker B).$$

This is an exact identity, not only a bound. In particular, nullity(BA) is at least nullity(A), but need not equal nullity(A)+nullity(B). If A is onto W, the intersection becomes all of ker B and the simpler addition is valid.

Worked changed case. Let A:R³→R² send (x,y,z) to (x,0), and let B:R²→R send (u,v) to u. Then ker A is the yz-plane, ker B is the vertical axis in R², and im A is the horizontal axis. Their intersection has dimension zero. Thus BA has nullity two, not three. Directly, BA(x,y,z)=x confirms the answer. Notice that the dimensions of kernels refer to different ambient spaces; their vectors cannot be added as if they lived in one space.

Predict, then inspect the certificate

Changed case Image meets the next kernelNext kernel misses the image

A(x,y,z)=(x,y), B(u,v)=u. What is nullity(BA)?

Your prediction Choose a prediction123 Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Write before revealing. Change only B to B(u,v)=v. Compute the intersection and identify the full kernel of BA. Then explain why no basis calculation is needed to prove the general identity.

Worked feedback and a proof checklist

Now im A lies entirely in ker B, so the intersection has dimension one. The formula gives 2+1=3, and BA is the zero map. A complete proof must show both inclusions for the image of the restricted map and identify its kernel; quoting rank-nullity without naming the restriction omits the essential step. The finite-dimensional assumption permits the displayed dimension arithmetic.

Continue

Depth: row rank equals column rank. Elimination exposes the same pivot count in both. In square matrices, full rank ties together invertibility, unique solvability for every right-hand side, injectivity and surjectivity; rectangular maps require more careful statements.