1. Spaces, maps and dimension

Unless otherwise stated, $F$ is a field and every space in a dimension formula is finite-dimensional.

We take $F$ to be a field and vector spaces to be finite-dimensional unless otherwise stated. The familiar space $F^n$ of column vectors is one model of a vector space, but not the only one: polynomials, matrices and functions can also be vectors. The task is to use their operations, not their appearance.

Definition

Subspaces and direct sums

A nonempty subset $U$ of a vector space $V$ is a subspace if $au+bv\in U$ for all $u,v\in U$ and $a,b\in F$. For subspaces $U,W$, their sum is $U+W=\{u+w:u\in U,w\in W\}$. It is a direct sum, written $U\oplus W$, if every element has a unique such decomposition; equivalently $U\cap W=\{0\}$.

The equivalence follows by subtracting two decompositions; conversely, a nonzero vector in the intersection has two decompositions. If bases of $U$ and $W$ are concatenated, they span the sum. The concatenated list is linearly independent exactly when the sum is direct.

For subspaces $U,W$, their sum is $U+W=\{u+w\}$. It is a direct sum, written $U\oplus W$, precisely when $U\cap W=\{0\}$, equivalently each vector in the sum has a unique decomposition. Extend a basis of $U\cap W$ separately to bases of $U$ and $W$; the combined extended list gives

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

To check independence, move the $U$ and $W$ combinations to opposite sides: the common vector lies in the intersection, so the extra basis coefficients vanish, then the intersection coefficients vanish. Every subspace of a finite-dimensional space has a complement by basis extension, but the complement is usually not unique or canonical.

Definition

Kernel and image

A map $T:V\to W$ is linear if $T(av+bw)=aT(v)+bT(w)$. Its kernel is $\ker T=\{v:T(v)=0\}$ and its image is $\operatorname{im}T=\{T(v):v\in V\}$. Both are subspaces. The map is injective exactly when $\ker T=\{0\}$, and surjective exactly when $\operatorname{im}T=W$.

A map $T:V\to W$ is linear if $T(\alpha v+\beta u)=\alpha Tv+\beta Tu$. Its kernel and image are subspaces. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend to a basis with $v_1,\ldots,v_s$. Then $Tv_1,\ldots,Tv_s$ form a basis of $\operatorname{im}T$: spanning follows by expanding any input; a dependence among these images places the corresponding input combination in the kernel, and independence of the extended basis makes all coefficients zero. Thus

$$\dim V=\dim\ker T+\dim\operatorname{im}T.$$

This is rank–nullity. In equal finite dimensions, injectivity and surjectivity are equivalent, but the equivalence can fail in infinite-dimensional spaces.

Worked example

Differentiation on polynomials

Let $P_2$ be the real polynomials of degree at most two and $D:P_2\to P_2$ be differentiation. For $p=a+bx+cx^2$, $Dp=b+2cx$. Thus $\ker D=\operatorname{span}\{1\}$ and $\operatorname{im}D=\operatorname{span}\{1,x\}$. Their dimensions $1+2=3$ agree with rank–nullity. $D$ is neither injective nor surjective as a map into $P_2$.

Choosing a method and avoiding mistakes

Use the subspace test to prove closure, a basis to prove dimension, and rank–nullity when a kernel or image is easier than the other. For uniqueness of decompositions use the intersection; a sum is not automatically direct. Do not equate $\ker T$ with the zero vector or image with the codomain without proof. Injectivity and surjectivity are equivalent for an endomorphism of a finite-dimensional space, not for arbitrary domain and codomain dimensions.

Exercises

1. In $P_2$, let $U=\operatorname{span}\{1,x\}$ and $W=\operatorname{span}\{x,x^2\}$. Find $U\cap W$ and $\dim(U+W)$.
2. Let $T:P_2\to\mathbb R^2$ send $p$ to $(p(0),p(1))$. Find a basis of its kernel and decide whether it is onto.
3. Prove that if $T:V\to V$ is linear and $V$ is finite-dimensional, injectivity implies surjectivity.

Full solutions

1. A polynomial in both spans has no constant or quadratic term, so $U\cap W=\operatorname{span}\{x\}$. The sum is $P_2$, of dimension $2+2-1=3$.
2. For $p=a+bx+cx^2$, the two conditions are $a=0$ and $b+c=0$. Hence $p=c(x^2-x)$ and the kernel basis is $\{x^2-x\}$. Rank–nullity gives image dimension $3-1=2$, so the image is all of $\mathbb R^2$.
3. Injectivity gives zero kernel. Rank–nullity yields $\dim\operatorname{im}T=\dim V$. A subspace of a finite-dimensional space with the same dimension equals the space, so $T$ is onto.

Connection

Choosing bases turns these structural statements into matrix statements. The kernel and image dimensions do not depend on those choices.