Vector Spaces and Linear Transformations

Vectors are linearly independent if c1v1 + ... + ckvk = 0 implies all ci = 0; otherwise dependent. A basis is a linearly independent set that spans the space; every basis of a space has the same number of vectors = the dimension. In R^n any n linearly independent vectors form a basis, and any set of more than n vectors is dependent. To test independence of n vectors in R^n, form a matrix and check det ≠ 0 (independent) or rank = n. The four fundamental subspaces of an m×n matrix A: column space (dim = rank), row space (dim = rank), null space (dim = n − rank), left null space (dim = m − rank). Memory aid: 'Independent + Spanning = Basis.'

For a linear transformation T: V → W (or matrix A, m×n), the Rank-Nullity Theorem states: rank(T) + nullity(T) = dim(domain) = n. Here rank = dim of image/column space, nullity = dim of kernel/null space. T is injective (one-one) iff nullity = 0; T is surjective (onto) iff rank = dim(W). A linear transformation satisfies T(av + bw) = aT(v) + bT(w). The matrix of T depends on chosen bases. Key shortcut: for a square matrix, injective ⟺ surjective ⟺ bijective ⟺ invertible ⟺ det ≠ 0 (all equivalent in finite dimensions). Remember: nullity counts free variables in Ax = 0.

Are v1 = (1,2,3), v2 = (2,4,6), v3 = (1,0,1) linearly independent in R^3? Form matrix rows and compute det of [[1,2,3],[2,4,6],[1,0,1]]. Notice row 2 = 2 × row 1, so two rows are proportional → det = 0 → the set is linearly DEPENDENT. Rank = 2 (v1 and v3 are independent, v2 is redundant). Thus they span only a 2-dimensional subspace (a plane), not all of R^3. Shortcut: whenever one vector is a scalar multiple of another, the set is immediately dependent without computing the full determinant. Dimension of span = rank = 2.

Cover the answer, recall, then check. 12 cards on the vector-space and linear-map definitions GATE draws from.

Q1. What conditions make a subset W of a vector space a subspace?
A1. W is nonempty, contains 0, and is closed under addition and scalar multiplication. (Closure under both operations is the essence.)

Q2. Define linear independence of vectors v₁,…,vₖ.
A2. The only solution to c₁v₁ + … + cₖvₖ = 0 is c₁ = … = cₖ = 0. Otherwise they are dependent.

Q3. Define a basis and the dimension of a vector space.
A3. A basis is a linearly independent set that spans the space. Dimension = number of vectors in any basis (all bases have the same size).

Q4. What is the span of a set of vectors?
A4. The set of all linear combinations of those vectors — the smallest subspace containing them.

Q5. Define a linear transformation T: V → W.
A5. A map with T(u + v) = T(u) + T(v) and T(cu) = cT(u) — it preserves addition and scalar multiplication. Always T(0) = 0.

Q6. Define the kernel (null space) and range (image) of T.
A6. Kernel = {v : T(v) = 0}; Range = {T(v) : v ∈ V}. Both are subspaces (of V and W respectively).

Q7. State the rank–nullity theorem for T: V → W.
A7. dim(V) = rank(T) + nullity(T), where rank = dim(range), nullity = dim(kernel).

Q8. When is a linear transformation injective (one-to-one)?
A8. ⇔ kernel = {0} (nullity 0). Then T maps independent sets to independent sets.

Q9. How is a linear transformation represented, and how does the matrix change with basis?
A9. By a matrix (once bases are fixed): T(x) = Ax. Changing basis gives a similar matrix P⁻¹AP.

Q10. What is the dimension of the vector space of all m×n real matrices?
A10. m·n (the standard basis is the matrices with a single 1). E.g. 2×2 matrices form a 4-dimensional space.

Q11. Can k > n vectors be linearly independent in an n-dimensional space?
A11. No — any set with more than n vectors in an n-dimensional space is necessarily dependent.

Q12. Relate column space, row rank, and rank of a matrix A.
A12. dim(column space) = dim(row space) = rank(A); row rank always equals column rank.

Vector spaces and linear transformations are the conceptual frame that unifies matrices, systems, and eigenvalues. In GATE CSE the direct questions are usually 1 mark — identify a subspace, compute a dimension, apply rank–nullity, or reason about independence — but the ideas are load-bearing for the rest of Linear Algebra and for topics like error-correcting codes.

Core definitions

A vector space is a set closed under vector addition and scalar multiplication (with the usual axioms). A subspace is a subset that is itself a vector space — check: contains 0, closed under + and scalar ×. Linear independence means no vector is a combination of the others; a basis is an independent spanning set, and its size is the dimension.

A linear transformation T preserves structure: T(u+v)=T(u)+T(v) and T(cu)=cT(u). Fixing bases turns T into a matrix, and the whole theory of rank/kernel applies.

Key results

Concept	Result
Rank–nullity	dim(V) = rank(T) + nullity(T)
Injective T	kernel = {0}
dim of m×n matrices	m·n
Row rank vs column rank	always equal (= rank)
Max independent vectors in dim n	n
Basis of dim-n space	exactly n vectors

Exam Tricks & Tips

• 🎯 Subspace test = 3 checks: contains 0, closed under addition, closed under scalar multiplication. If 0 ∉ W, stop — not a subspace.
• 🎯 Rank–nullity is the workhorse: knowing any two of dim(V), rank, nullity gives the third.
• 🎯 More than n vectors in an n-dim space ⇒ dependent — a one-line elimination in MCQs.
• 🎯 Injective ⇔ nullity 0; surjective ⇔ rank = dim(W). For a square matrix, one implies the other.
• 🎯 Row rank = column rank always — compute whichever is easier.
• ❌ Common mistake: thinking a set that spans must be independent (or vice-versa). Spanning + independent together define a basis; either alone is not enough.

Expected exam pattern

Identify whether a given set is a subspace/basis; compute dimension or rank; apply rank–nullity to find a kernel/image dimension; test independence. Frequently woven into matrix questions via rank.

Quick recap

Subspace ⇔ contains 0 and closed under + and scalar ×. Basis = independent + spanning; its size is the dimension. Linear map: T(u+v)=Tu+Tv, T(cu)=cTu. Rank–nullity: dim(domain)=rank+nullity. Injective ⇔ kernel {0}. Row rank = column rank = rank.