Split a space without losing its kernel

Try

Consider T:R³→R² given by T(x,y,z)=(x+y,z). Its kernel is the line spanned by (1,−1,0). Choose W spanned by (1,0,0) and (0,0,1). Every input has the decomposition (x,y,z)=−y(1,−1,0)+(x+y)(1,0,0)+z(0,0,1). Verify all three coordinates. The first piece is invisible to T; the remaining piece carries exactly the information retained in the output.

Predict

Predict whether a different complement changes the map. Replace W with the plane spanned by (0,1,0) and (0,0,1). This also complements the kernel, but it chooses different representatives for the same output. The map is unchanged. A complement is a useful choice, not generally a distinguished physical component of the data. Orthogonality can later select a canonical complement once an inner product is supplied.

Experiment

For the input (4,1,2), compute its two decompositions using the two complements. With the first complement, the kernel part is (−1,1,0) and the retained part is (5,0,2). With the second, these are (4,−4,0) and (0,5,2). Both retained parts map to (5,2). Make a table listing input, kernel component, complement component, and output; then change the input by adding 3(1,−1,0). Explain which entries change and why the output cannot change.

Understand

An internal direct sum V=U⊕W means V=U+W and U∩W contains only zero. The intersection condition makes the decomposition unique: if u+w=u′+w′, then u−u′=w′−w belongs to both spaces. A finite-dimensional complement exists by extending a basis of U to a basis of V. For rank-nullity, extend a basis of ker T by vectors v₁,…,vₛ. Their images span im T because kernel components vanish. They are independent because a relation among their images would put a combination of the added basis vectors in the kernel, contradicting independence of the full basis. Therefore dim V=dim ker T+dim im T. The restriction of T to any kernel complement is an isomorphism onto im T.

Apply

Apply the same reasoning to two overlapping subspaces. Choose a basis of U∩W and extend it separately to bases of U and W. The shared vectors followed by both sets of added vectors form a basis of U+W: a relation between the two sides lies in the intersection, and independence in each extension eliminates the coefficients. Thus dim(U+W)=dim U+dim W−dim(U∩W). Test it with two distinct planes through zero in R³. Their common line explains exactly why simply adding their dimensions overcounts.

A useful diagnostic is to ask which space each dimension belongs to. In the worked map, the domain dimension is three and image dimension is two; the codomain happens to equal the image. If the codomain were enlarged to R⁴ by appending two zero coordinates, the kernel and rank would stay unchanged. Rank-nullity subtracts from the domain dimension, never automatically from the number of displayed output coordinates.

Continue

A direct sum is about unique representation, not perpendicular angles. The coordinate decompositions above were obtained without an inner product. In data analysis, treating two correlated feature spans as independent contributions can double-count shared directions; the intersection term identifies that algebraic overlap. Next, quotient spaces will retain information without selecting any particular complement. Finite dimension is central to the elementary basis-extension proofs used here; broader versions rely on different existence principles.

Review Linear Algebra I when a prerequisite needs practice. Continue to Make indistinguishable vectors into one vector.