Vectors & Spaces · Lesson 3 of 3 · ~6 min

Span and independence

Predict

The instrument can lock $\mathbf{v}$ to the parallel copy $-0.8\,\mathbf{u}$. You still control both scalars $a$ and $b$ freely. Can $a\mathbf{u} + b\mathbf{v}$ then reach a point off the line through $\mathbf{u}$ — say $(1, 2)$?

Hold that thought. Press make v || u in the instrument and try to steer au+bv anywhere off the dotted line.

Try

Try itMove the a · u and b · v sliders and watch the two dashed segments chain tip-to-tail; the heavy arrow au+bv lands where the chain ends. Then press make v || u and try the sliders again, watching the det[u v] readout.

Notice

With the independent pair $\mathbf{u} = (2, 0.5)$ and $\mathbf{v} = (-1, 1.5)$, setting both sliders to $1.00$ lands au+bv on $(1, 2)$, and det[u v] holds at $3.50$. Press make v || u: the faint dot-field collapses to a dotted line, det drops to $0.00$, and no slider setting moves the arrow off that line. The point $(1, 2)$ was reachable a moment ago; now it is not.

Explain

Each slider setting builds a linear combination $a\mathbf{u} + b\mathbf{v}$, and the set of everything you can possibly build is the span:

\operatorname{span}\{\mathbf{u}, \mathbf{v}\} = \{\, a\,\mathbf{u} + b\,\mathbf{v} \;:\; a, b \in \mathbb{R} \,\}

How big the span is depends on one question: is either vector a scalar multiple of the other? If yes, the pair is linearly dependent — $b\mathbf{v}$ only pushes along the line $a\mathbf{u}$ already covers, and the span collapses to that line. If no, the pair is linearly independent, and in $\mathbb{R}^2$ two independent vectors reach everything. The determinant $u_1 v_2 - u_2 v_1$ is the one-number test: up to sign it is the area of the parallelogram the pair frames, and dependence flattens that area to $0$. An independent pair that spans the plane is a basis: every point is then reached by exactly one choice of $(a, b)$ — its coordinates in that basis — and since every basis of $\mathbb{R}^2$ has exactly two vectors, the plane has dimension two.

Apply

Checkpoint — prove it to yourself

For $\mathbf{u} = (2, 0.5)$ and $\mathbf{v} = (-1, 1.5)$ the readout says $\det[u\ v] = 3.50$. What exactly does the nonzero value certify?

With the parallel copy $\mathbf{v} = -0.8\,\mathbf{u}$ locked in, simplify $a\mathbf{u} + b\mathbf{v}$.

Transfer: do $\mathbf{e}_1 = (1, 0)$ and $\mathbf{w} = (3, 0.01)$ form a basis of $\mathbb{R}^2$?

Key idea

Span is everything two moves can build, and independence decides how big it is. An independent pair pins every point of the plane to exactly one $(a, b)$ — a basis turns geometry into bookkeeping — while a dependent pair wastes a direction and collapses the reach to a line.

← Lesson 2