Bases, dimension, rank–nullity

In one sentence

Dimension is a statement about a set and a field, rank–nullity is the one theorem that constrains everything else, and rank behaves badly under every operation candidates expect it to respect.

Why the exam asks it

Because linear algebra is the part of the syllabus where a candidate is most likely to be confident and wrong. The definitions are familiar from undergraduate work, so the paper does not test them; it tests the base field, the failure of rank to be additive or symmetric, and the places where finite-dimensional intuition silently assumes finite dimension.

The idea

A vector space over a field $F$. A basis is a linearly independent spanning set; every basis of a given space has the same cardinality, and that is the dimension.

Three facts that get used constantly and are worth stating precisely, all in finite dimension:

• Any linearly independent set can be extended to a basis.
• Any spanning set contains a basis.
• In a space of dimension $n$, any $n$ independent vectors form a basis, and so do any $n$ spanning vectors. Getting the count right does half the work.

The base field is part of the question

$\mathbb{C}$ has dimension $1$ over $\mathbb{C}$ and dimension $2$ over $\mathbb{R}$, with basis $\{1, i\}$. $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional. $M_2(\mathbb{C})$ has dimension $4$ over $\mathbb{C}$ and $8$ over $\mathbb{R}$.

This is not a curiosity; it is a standing trap. An option saying "the dimension is $n$" is incomplete until the field is named, and the stem always names it.

Rank–nullity

For $T : V \to W$ linear with $V$ finite-dimensional,

$$\dim V = \operatorname{rank}(T) + \operatorname{nullity}(T).$$

The proof is the picture: choose a basis of $\ker T$, extend it to a basis of $V$, and the added vectors map to a basis of the image. Note the theorem is about $\dim V$, the domain — not the codomain. Options that put $\dim W$ there are a standard misdirection.

Two immediate consequences, both requiring $\dim V = \dim W$ finite:

• $T$ injective $\iff$ $T$ surjective $\iff$ $T$ bijective.

That equivalence is false in infinite dimensions, and the counterexample is worth having. On the space of polynomials, $T(p) = xp(x)$ is injective and not surjective, while differentiation is surjective and not injective. On $\ell^2$ the shift operators do the same job.

Dimensions of sums

$$\dim(U + W) = \dim U + \dim W - \dim(U \cap W).$$

Equality $\dim(U+W) = \dim U + \dim W$ holds exactly when $U \cap W = \{0\}$, that is when the sum is direct. Note also that $U \cup W$ is a subspace only when one contains the other — the union is almost never a subspace, and the sum is what you want.

How rank behaves

• $\operatorname{rank}(A) = \operatorname{rank}(A^T)$: row rank equals column rank.
• $\operatorname{rank}(AB) \le \min(\operatorname{rank}A, \operatorname{rank}B)$.
• Sylvester: $\operatorname{rank}(A) + \operatorname{rank}(B) - n \le \operatorname{rank}(AB)$ for $n \times n$ matrices.
• $\operatorname{rank}(A + B) \le \operatorname{rank}A + \operatorname{rank}B$ — an inequality, never an equality.

And two failures worth committing to memory. With $A = \begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $B = \begin{pmatrix}0&1\\0&0\end{pmatrix}$:

$$AB = \begin{pmatrix}0&1\\0&0\end{pmatrix}, \quad BA = \begin{pmatrix}0&0\\0&0\end{pmatrix},$$

so $\operatorname{rank}(AB) = 1$ while $\operatorname{rank}(BA) = 0$. Rank is not symmetric in the product. And $\begin{pmatrix}1&0\\0&0\end{pmatrix} + \begin{pmatrix}0&0\\0&1\end{pmatrix} = I$ has rank $2$ from two matrices of rank $1$, so rank is not additive either — though here it happens to hit the upper bound.

Where intuition breaks

"The dimension of $\mathbb{C}^n$ is $n$." Over $\mathbb{C}$, yes; over $\mathbb{R}$ it is $2n$. This is the base field dependence trap and it appears in some form most years.

"The union of two subspaces is a subspace." Take the two coordinate axes in $\mathbb{R}^2$: $(1,0)$ and $(0,1)$ are both in the union and their sum is not. It is a subspace exactly when one contains the other.

"$\operatorname{rank}(AB) = \operatorname{rank}(BA)$." The pair above gives $1$ and $0$. Tempting because trace does satisfy $\operatorname{tr}(AB) = \operatorname{tr}(BA)$, and the two facts get merged. Trace is symmetric; rank is not.

"$\operatorname{rank}(A+B) = \operatorname{rank}A + \operatorname{rank}B$." Only an inequality, and it can fail on both sides: $A + (-A) = 0$ has rank $0$.

"$A^2 = 0$, so $A = 0$." $\begin{pmatrix}0&1\\0&0\end{pmatrix}$. There are no zero divisors in a field and plenty among matrices, which is the whole difference.

"$T$ injective, so $T$ is surjective." True when $\dim V = \dim W < \infty$, false otherwise. Multiplication by $x$ on the polynomial space is the standard witness. A hypothesis dropped, and the hypothesis is finite dimensionality — easy to lose because the paper rarely states it in so many words.

"A proper subspace has strictly smaller dimension." Finite-dimensionally, yes. In infinite dimensions a proper subspace can have the same dimension: the even-indexed sequences inside $\ell^2$, or polynomials of even degree inside all polynomials.

"$\dim(U + W) = \dim U + \dim W$." Only when the intersection is trivial. Two distinct planes through the origin in $\mathbb{R}^3$ each have dimension $2$, and their sum is all of $\mathbb{R}^3$, dimension $3$, not $4$.

The exam's angle

1. Read the field first. If the stem says "as a vector space over $\mathbb{R}$" and the space is complex, the dimension doubles, and that is usually the entire question.
2. Count. Most basis questions reduce to comparing a number of vectors against a dimension. $n$ independent vectors in an $n$-dimensional space are a basis; you do not need to check spanning separately.
3. Reach for rank–nullity whenever a kernel or an image appears. It is the only equation available, so it is almost always the intended route.
4. For any claim about rank of a product or sum, test with the two $2\times2$ matrices above. They are small enough to multiply in your head and they break most of what can be claimed.
5. Ask whether finite dimensionality is being used. Injective-implies-surjective, proper subspaces being smaller, and every subspace having a complement of positive codimension all need it.

The night before

• Dimension depends on the field. $\mathbb{C}$ over $\mathbb{R}$ has dimension $2$; $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional.
• Rank–nullity: $\dim V = \operatorname{rank} + \operatorname{nullity}$, with $V$ the domain.
• Finite dimension, equal dimensions: injective $\iff$ surjective $\iff$ bijective. Fails in infinite dimensions — multiplication by $x$ on polynomials.
• $\dim(U+W) = \dim U + \dim W - \dim(U\cap W)$. The union is a subspace only if one contains the other.
• Row rank $=$ column rank. $\operatorname{rank}(AB) \le \min$, and Sylvester bounds it below.
• $\operatorname{rank}(AB) \ne \operatorname{rank}(BA)$ in general — $1$ and $0$ for the standard pair. But $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ always.
• Rank is subadditive, not additive.
• Independent sets extend to bases; spanning sets contain bases.