Vector Spaces and Linear Maps

Rank–nullity

For T : V → W with dim V = n: rank T + nullity T = n.

Consequences: for a square matrix, injective ⇔ surjective ⇔ invertible. In infinite dimensions this fails (the shift operator is injective, not surjective).

Rank facts

• rank A = rank Aᵀ = row rank = column rank.
• rank is field-independent: a matrix over $\mathbb{Q}$ has the same rank over $\mathbb{R}$ or $\mathbb{C}$, so nullity computed over either field agrees. (A frequent Part-C item.)
• rank(A + B) ≤ rank A + rank B.
• rank(AB) ≤ min(rank A, rank B).
• Sylvester: rank(AB) ≥ rank A + rank B − n. So two rank-2 matrices in $M_3$ can have product of rank 1, but never 0.
• rank(A) = rank(AᵀA) over $\mathbb{R}$.

Dimension formulas

• dim(U + W) = dim U + dim W − dim(U ∩ W).
• Two subspaces of ${\mathbb{R}}^5$ of dimensions 3 and 3 must intersect in dimension ≥ 1.
• dim of the space of n×n matrices: $n^2$; symmetric n(n+1)/2; skew n(n−1)/2; trace-zero $n^2-1$; upper triangular n(n+1)/2.
• dim P_n (polynomials of degree ≤ n) = n + 1.

Bases

Every vector space has a basis (needs choice for infinite dimensions). Over a finite field counting matters: the number of bases of 𝔽_q^n is $(q^n-1)(q^n-q)\cdots (q^n-$ $q^{n-1})$ = |GL_n(𝔽_q)|.

A vector space over any field is never the union of two proper subspaces; over 𝔽${}_2$ it can be the union of three.

Key takeaways

• Rank–nullity is the first move in almost every question.
• Rank does not change when you extend the field.
• Sylvester's inequality bounds rank(AB) from below — remember both directions.