Dimension

Before any formulas, here is the picture to carry in your head: dimension measures how many independent directions a space has, equivalently how many numbers you need to pin down a single point in it.

• A line through the origin has one direction. Once you fix how far along the line you are (one number), the point is determined. It is 1-dimensional.
• A plane through the origin has two independent directions. You need two numbers (how far in each direction) to locate a point. It is 2-dimensional.
• All of $\mathbb{R}^n$ (the set of lists of $n$ real numbers $(x_1,\dots,x_n)$) needs exactly $n$ coordinates, so it is $n$-dimensional.

The subtle word is independent. If you offer three directions in a plane, one of them is redundant — it can be built from the other two — so the plane is still 2-dimensional, not 3. Dimension is the number of genuinely independent directions, with every redundancy removed. A basis is exactly a list of directions that is both non-redundant (linearly independent) and complete (spans the whole space), and the dimension is simply how many vectors are in that list.

Recall that a basis of a vector space $V$ is a set of vectors $B=\{\mathbf{v}_1,\dots,\mathbf{v}_k\}$ that is

1. linearly independent — no vector in $B$ is a linear combination of the others (the only way to write $c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k=\mathbf{0}$ is $c_1=\cdots=c_k=0$), and
2. spanning — every vector in $V$ can be written as some linear combination $c_1\mathbf{v}_1+\cdots+c_k\mathbf{v}_k$.

Definition. The dimension of a (finite-dimensional) vector space $V$, written $\dim(V)$, is the number of vectors in any basis of $V$.

Key facts that follow immediately:

• $\dim(\mathbb{R}^n)=n$. The standard basis $\mathbf{e}_1=(1,0,\dots,0),\ \mathbf{e}_2=(0,1,0,\dots,0),\ \dots,\ \mathbf{e}_n=(0,\dots,0,1)$ has exactly $n$ vectors, is independent, and spans $\mathbb{R}^n$.
• $\dim(\{\mathbf{0}\})=0$. The space containing only the zero vector has the empty set as its basis — zero vectors.
• A space is called finite-dimensional when it has a basis with finitely many vectors; that number is its dimension. (Spaces that need infinitely many basis vectors are infinite-dimensional, but every example in this lesson is finite-dimensional.)

Notice the definition says "any basis." For this to make sense, every basis must have the same number of vectors — otherwise $\dim(V)$ would be ambiguous. That is the content of the next section.

The definition would be useless if one basis of $\mathbb{R}^2$ had 2 vectors and another had 3. Thankfully, this never happens.

Invariance of dimension. In a finite-dimensional vector space, any two bases have exactly the same number of vectors.

The engine behind this is a counting fact you should internalize, because it is useful on its own:

If a space $V$ has a basis of $n$ vectors, then any set of more than $n$ vectors in $V$ is linearly dependent, and any set of fewer than $n$ vectors cannot span $V$.

Intuitively: a basis of size $n$ fixes the number of independent directions at $n$. You can never fit more than $n$ independent vectors (so a bigger set must have a redundancy — it is dependent), and you can never reach every point with fewer than $n$ directions (so a smaller set cannot span). Squeeze these two bounds together and every basis is forced to have exactly $n$ vectors. (The classical argument that makes this precise is the replacement or exchange lemma, which swaps basis vectors for new ones one at a time.)

Why you care in practice: this gives instant answers.

• Five vectors in $\mathbb{R}^4$? Must be dependent — $5>4=\dim(\mathbb{R}^4)$.
• Two vectors spanning $\mathbb{R}^3$? Impossible — $2<3$.
• Exactly $n$ vectors in $\mathbb{R}^n$ that are independent? They automatically span, so they are a basis (and vice versa).

Most problems give you a subspace in one of two forms. Here is how to read off its dimension in each.

Form 1 — a span (a spanning set is given). The subspace is $W=\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_m\}$. Then $\dim(W)=\text{number of linearly independent vectors among the }\mathbf{v}_i=\operatorname{rank}.$ The rank is the maximum number of linearly independent vectors in the set. To find it:

1. Form a matrix using the vectors as its rows.
2. Row reduce to echelon form (Gaussian elimination).
3. Count the nonzero rows (equivalently, the pivots). That count is the rank, and it is $\dim(W)$.

Row operations never change the row space, so the number of nonzero rows is exactly the number of independent directions. Redundant vectors collapse to zero rows and drop out.

Form 2 — a subspace defined by linear equations. If $W\subseteq\mathbb{R}^n$ is the solution set of a system of $k$ independent (non-redundant) homogeneous linear equations, then $\dim(W)=n-k.$ Each independent equation removes one degree of freedom. For example, one nontrivial equation like $x+y+z=0$ in $\mathbb{R}^3$ cuts the dimension from $3$ down to $3-1=2$ (a plane). Two independent equations cut it to $3-2=1$ (a line). Caution: the equations must be independent — two equations that are multiples of each other count as one constraint.

Sanity checks for any answer: a subspace of $\mathbb{R}^n$ always has dimension between $0$ and $n$; dimension $0$ means $W=\{\mathbf{0}\}$, and dimension $n$ means $W=\mathbb{R}^n$.