2.3 Subspace or Not

The last two sections kept meeting the same kind of shape. Spans were flat, passed through the origin, and combining two of their points always produced another of their points; the red line of solutions in the previous section was flat but missed the origin, and adding two of its points threw you off it. It is time to name the property. A subspace of $\mathbb{R}^3$ is a set of vectors that contains the zero vector and is closed under the two operations that build spans: adding any two of its vectors lands back in the set, and scaling any of its vectors lands back in the set.

The definition sounds algebraic, but it is a test you can run by eye. The figure below offers three candidates. The first is the plane $z = 0.5x + 0.3y$ , which passes through the marked origin. The second is its translate $z = 0.5x + 0.3y + a$ , floating above or below on the slider $a$ . The third is the curved surface $z = 0.2xy$ — and it is the trap, because it does pass through the origin.

A flat plane passing through a marked origin point, with a second, parallel plane floating above or below it, controlled by a slider named a. A third, saddle-shaped curved surface also passes through the origin, rising in two opposite quadrants and dipping in the other two. Sliding a to 0 drops the floating plane onto the origin plane; the reader compares which of the three surfaces could be a subspace. — The plane $z = 0.5x + 0.3y$ through the marked origin, and its translate $z = 0.5x + 0.3y + a$ on the slider $a$ . The third candidate is the curved surface $z = 0.2xy$ , which also passes through the origin.

Explore the figure

At the default $a = 1$ , orbit the two planes. Only the lower one contains the origin point — the translate already fails the easiest part of the test. Does any point of the translate look "reachable" by scaling another of its points down toward zero?
Drag $a$ to $0$ and watch the translate drop onto the origin plane. Among all values of the slider, exactly one produces a subspace. Which failure returns the instant $a$ leaves $0$ ?
Find the third surface, the curved $z = 0.2xy$ : it rises in two opposite quadrants and dips in the other two. Check by orbiting that it genuinely contains the origin. If containing the origin were the whole test, it would pass.
Now test closure on the curved surface by eye: find the points above $(1, 0)$ and $(0, 1)$ — both sit at height $z = 0$ — and locate their sum $(1, 1, 0)$ . Does the surface pass through it?

Running the test

The origin plane passes in full generality. If $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ both satisfy $z = 0.5x + 0.3y$ , then their sum satisfies $z_1 + z_2 = 0.5(x_1 + x_2) + 0.3(y_1 + y_2)$ — the same equation again, because the equation is linear with zero constant term. Scaling works the same way. In fact this plane is a span: every point of it is $x(1, 0, 0.5) + y(0, 1, 0.3)$ , so it is the span of two vectors, and every span passes the subspace test for exactly the reason above.

The translate fails at $a \ne 0$ , and not only because it misses the origin. Take $a = 1$ : the points $(0, 0, 1)$ and $(2, 0, 2)$ both lie on it, but their sum $(2, 0, 3)$ needs $z = 0.5 \cdot 2 + 1 = 2$ , not $3$ . Closure fails because adding two points doubles the constant $+1$ . A flat sheet in the wrong place is not a subspace — it is a subspace's translate, the shape solution sets of inhomogeneous systems take.

Example 1 The curved surface fails closure twice

The surface $z = 0.2xy$ contains the origin. Show it is still not a subspace, by exhibiting one failure of addition and one failure of scaling.

Show solution

Addition: $(1, 0, 0)$ and $(0, 1, 0)$ are on the surface, since $0.2 \cdot 1 \cdot 0 = 0$ and $0.2 \cdot 0 \cdot 1 = 0$ . Their sum is $(1, 1, 0)$ , but the surface above $(1, 1)$ sits at $z = 0.2 \cdot 1 \cdot 1 = 0.2$ , not $0$ . Scaling: $(1, 1, 0.2)$ is on the surface; doubling gives $(2, 2, 0.4)$ , but the surface requires $z = 0.2 \cdot 2 \cdot 2 = 0.8$ . Doubling the inputs quadrupled the output — the signature of the product $xy$ , which is not linear. Both closure tests fail, so containing the origin was not enough.

The three verdicts assemble into a picture worth keeping: the subspaces of $\mathbb{R}^3$ are exactly the origin alone, the lines through the origin, the planes through the origin, and all of $\mathbb{R}^3$ . Flat, through the origin, closed under the span-building operations — any curvature or any offset breaks the test somewhere.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Chapters 1–6 follow the topic sequence of Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL). Chapter 0's scope (its topics, their order, and the idea each figure shows) follows the Vectors chapter of Ximera's Linear Algebra: An Interactive Introduction. The prose, worked examples and figures were written for this book, which is not affiliated with or endorsed by the authors of either work. License: CC-BY-NC-SA-4.0.