Linear dependence, independence, and basis

Hand someone three vectors in $\mathbb{R}^3$ and ask: do all three genuinely point in "new" directions, or is one of them secretly built from the other two? That single question — redundant, or independent? — is the seed of some of the most load-bearing machinery in linear algebra. Once you can answer it reliably, you can also answer the next two questions that follow from it: what is the full set of points these vectors can reach (their span), and what is the smallest, most efficient list of vectors — a basis — needed to describe that reach exactly once each? This node makes all three ideas precise and mechanical: a determinant or rank computation for independence, a row-reduction for span and basis, and a single number, the dimension, that every basis of a space agrees on.

The core insight

A set of vectors is linearly independent exactly when the only way to combine them to zero is trivially, which for $n$ vectors in $\mathbb{R}^n$ reduces to checking $\det A \neq 0$ , and a basis is a minimal independent set whose span is the whole space, with every basis of a given space sharing the same size, its dimension.

The core idea

Dependence and independence. A set $\{\mathbf v_1,\dots,\mathbf v_k\}$ in a vector space $V$ is linearly dependent if some non-trivial combination of them vanishes — that is, scalars $c_1,\dots,c_k$ , not all zero, exist with

c_1\mathbf v_1+c_2\mathbf v_2+\cdots+c_k\mathbf v_k=\mathbf 0,

where: $c_i$ are scalars and $\mathbf v_i$ are the given vectors. If the only solution to this equation is $c_1=\cdots=c_k=0$ (the trivial one), the set is linearly independent — no vector in it can be written in terms of the others. This is the entire content of the topic: dependence means "at least one vector is redundant," independence means "none of them is."

Span. The span of a set of vectors is every linear combination you can build from them,

\operatorname{span}\{\mathbf v_1,\dots,\mathbf v_k\}=\{c_1\mathbf v_1+\cdots+c_k\mathbf v_k : c_i\in\mathbb{R}\},

where: the $c_i$ range over all real scalars. The span is always a subspace of $V$ — a line through the origin if there is really only one independent direction among the $\mathbf v_i$ , a plane if there are two, and so on.

Basis and dimension. A basis of $V$ is a set that is simultaneously linearly independent and spans $V$ . Every vector of $V$ then has a unique expression as a combination of basis vectors — uniqueness is exactly what independence buys you, and coverage is what spanning buys you. Every basis of a given finite-dimensional $V$ turns out to have the same number of vectors, called $\dim V$ : for instance $\dim \mathbb{R}^n = n$ (standard basis $\{\mathbf e_1,\dots,\mathbf e_n\}$ ) and $\dim P_n = n+1$ (basis $\{1,x,x^2,\dots,x^n\}$ ). Two facts tie everything together: an independent set can never have more than $\dim V$ vectors, and a spanning set can never have fewer.

Figure

Two vectors from the origin spanning a plane with a third vector lying inside that same plane, shown alongside a genuinely independent triple pointing along three different axes

The determinant/rank test for independence

The goal of this derivation is to justify, with nothing beyond solving a homogeneous linear system, why testing $n$ vectors for independence in $\mathbb{R}^n$ can be reduced entirely to a single number — the determinant of the matrix they form — and why the more general case of $k$ vectors in any dimension reduces to a rank computation instead, since that reduction is exactly the tool used in every worked problem on this topic.

Step 1. Assemble the vectors as the columns of a matrix. If $\mathbf v_1,\dots,\mathbf v_k \in \mathbb{R}^n$ , form $A=[\mathbf v_1\ \mathbf v_2\ \cdots\ \mathbf v_k]$ , an $n\times k$ matrix. The dependence equation $c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf 0$ is then exactly the matrix equation $A\mathbf c=\mathbf 0$ , where $\mathbf c=(c_1,\dots,c_k)^T$ — a linear combination of columns is precisely what matrix-vector multiplication computes.

Step 2. Translate independence into a statement about solutions. The set $\{\mathbf v_1,\dots,\mathbf v_k\}$ is linearly independent exactly when $A\mathbf c=\mathbf 0$ has only the trivial solution $\mathbf c=\mathbf 0$ . From the theory of linear systems, a homogeneous system $A\mathbf c=\mathbf 0$ has only the trivial solution precisely when the columns of $A$ have no free variable after row reduction, which happens exactly when $\operatorname{rank}(A)=k$ , the number of columns.

Step 3. Specialize to the square case. When $k=n$ , $A$ is a square $n\times n$ matrix, and $\operatorname{rank}(A)=n$ (full rank) is equivalent to $A$ being invertible, which is equivalent to $\det A \neq 0$ . So for exactly $n$ vectors in $\mathbb{R}^n$ , independence collapses to the single check $\det A \neq 0$ ; a zero determinant certifies dependence and simultaneously guarantees a non-trivial solution $\mathbf c$ exists, which row reduction then extracts explicitly.

Step 4. Read off the general-shape cases. If $k>n$ (more vectors than dimensions), the columns can never all be independent, since rank is capped at $n<k$ — any $n+1$ or more vectors in $\mathbb{R}^n$ are automatically dependent. If $k<n$ , independence is possible but does not by itself give a basis of $\mathbb{R}^n$ , since the set cannot span the whole space; rank must still equal $k$ for independence to hold.

Worked example

Problem. Determine whether $\mathbf v_1=(1,2,3)$ , $\mathbf v_2=(2,1,3)$ , $\mathbf v_3=(1,-1,0)$ are linearly independent in $\mathbb{R}^3$ , and if not, find the explicit dependency relation.

Step 1. Since there are exactly three vectors in $\mathbb{R}^3$ , form the square matrix of columns and compute its determinant:

A=\begin{pmatrix}1&2&1\\2&1&-1\\3&3&0\end{pmatrix}.

Expanding along the third column, $\det A = 1\cdot\det\begin{pmatrix}2&1\\3&3\end{pmatrix} - (-1)\cdot\det\begin{pmatrix}1&2\\3&3\end{pmatrix} + 0 = 1(6-3)+1(3-6) = 3-3=0$ .

Step 2. Since $\det A=0$ , the set is dependent (by Step 3 of the derivation above). To pin down the actual relation, row-reduce $A$ to solve $A\mathbf c=\mathbf 0$ :

\begin{pmatrix}1&2&1\\2&1&-1\\3&3&0\end{pmatrix}\ \xrightarrow{R_2-2R_1,\ R_3-3R_1}\ \begin{pmatrix}1&2&1\\0&-3&-3\\0&-3&-3\end{pmatrix}\ \xrightarrow{R_3-R_2}\ \begin{pmatrix}1&2&1\\0&-3&-3\\0&0&0\end{pmatrix}.

Step 3. Row 2 gives $-3c_2-3c_3=0 \Rightarrow c_2=-c_3$ . Setting $c_3=1$ gives $c_2=-1$ ; row 1 then gives $c_1+2c_2+c_3=0 \Rightarrow c_1-2+1=0 \Rightarrow c_1=1$ .

Step 4. Check: $1\cdot\mathbf v_1-1\cdot\mathbf v_2+1\cdot\mathbf v_3=(1-2+1,\,2-1-1,\,3-3+0)=(0,0,0)$ . Confirmed: $\mathbf v_1-\mathbf v_2+\mathbf v_3=\mathbf 0$ , so the vectors are dependent and do not form a basis of $\mathbb{R}^3$ .

Competitive stretch

The same row reduction found more than the yes/no answer — it recovered the exact non-trivial relation between the vectors, which is the reduced echelon form's real payoff. If instead you were handed a fourth vector $\mathbf v_4=(0,1,1)\subset\mathbb{R}^3$ and asked for a basis of $\operatorname{span}\{\mathbf v_1,\mathbf v_2,\mathbf v_4\}$ , the same machinery applies: stack them as rows, row-reduce, and discard any row that becomes zero (redundant) — the surviving non-zero rows form a basis, and their count is the dimension.

Common traps

Testing independence "by eye" — checking that no vector is an obvious multiple of another is not enough once there are three or more vectors; example above has no pairwise multiples yet is still dependent.
Confusing "spans $V$ " with "is a basis of $V$ " — a spanning set can have more vectors than $\dim V$ and still be redundant; a basis is the minimal spanning / maximal independent case.
Forgetting the size cap — any set of more than $\dim V$ vectors in $V$ is automatically dependent, no computation needed.
Sign slips while back-substituting the free variable during row reduction — always verify the recovered combination actually sums to $\mathbf 0$ .

Ask the tutor: not sure whether your set of vectors is independent, or want help pulling a clean basis out of a messy spanning set — ask the AI tutor to walk through the row reduction with you.

If det A turns out to be zero, does that tell me which vector is the redundant one?

Not by itself — a zero determinant only certifies that some non-trivial dependency exists among the columns. To find out exactly which combination vanishes, you have to go one step further and row-reduce $A$ to solve $A\mathbf c=\mathbf 0$ , as in the worked example. The resulting free variable hands you the explicit relation, and from it you can see which vector could be dropped while keeping the same span.

Why does rank equal to the number of vectors mean independence, rather than rank equal to n?

Rank counts how many independent directions the columns of $A$ actually produce. If you have $k$ vectors and $\operatorname{rank}(A)=k$ , none of the $k$ columns is expressible via the others — that is independence, regardless of the ambient dimension $n$ . Rank equal to $n$ is a different, stronger statement: it says the columns span all of $\mathbb{R}^n$ . The two coincide only when $k=n$ , which is exactly the square case where independence and spanning become the same yes/no question, testable by a single determinant.

Is there a shortcut for spotting a basis without full row reduction?

For small, simple examples yes: if you already know $\dim V=n$ and you can produce $n$ vectors that you can show are independent (e.g. a square matrix with visibly nonzero determinant, or vectors that are pairwise non-parallel and not coplanar), independence plus the right count automatically gives a basis — you don't need to separately check spanning. This "count trick" is a direct consequence of the fact that an independent set of exactly $\dim V$ vectors must also span $V$ . For anything less clean, row reduction is the reliable method.