Vector Spaces & Subspaces

You already know how to add arrows and stretch them — that is the whole of working inside a vector space. It is the polite name for any setting where addition and scaling behave the way you expect (the plane R² and 3-D space R³ are the everyday ones).

The interesting question, and the one GATE keeps asking, is which parts of that space are self-contained. Which collections of arrows are “complete”, in the sense that adding or stretching arrows from the collection never throws you outside it? Those complete parts are called subspaces, and they pass or fail on three quick checks.

The subspace test — three conditions

A subset W of a vector space is a subspace exactly when all three hold:

1. It contains the zero vector — the origin 0 is in W.
2. It is closed under addition — if u and v are in W, so is u + v.
3. It is closed under scalar multiplication — if v is in W and c is any scalar, so is cv.

Geometrically in R³, the only things that pass are:

• the origin alone
• every line through the origin
• every plane through the origin
• all of R³

Anything that misses the origin, bends, or is bounded fails.

The fastest screen is condition 1. If the origin is not in the set, stop — it is not a subspace. That one check kills every “offset” set on sight.

Necessary is not sufficient

Do not let that speed run in the other direction, though. This is the classic slip: containing the origin is necessary, never sufficient. Passing check 1 tells you only that you must keep testing.

Take the first quadrant, {(x, y) : x ≥ 0 and y ≥ 0}. It holds the origin ✓ and even survives check 2: adding two points with non-negative coordinates gives another ✓. Then it dies at check 3, because scaling (1, 1) by −1 gives (−1, −1), which is nowhere near the quadrant. All three conditions do independent work, and “every scalar” includes the negative ones.

Drag the two arrows below and watch. As long as both live in your candidate subset, every sum and every scaled copy must too, or closure breaks. Line them up along a line through the origin and notice that addition and scaling never push you off it.

The trap GATE loves

Three flavours of set

The designed difficulty mixes three flavours of set:

• a genuine subspace (a linear equation through the origin)
• an affine set (a linear equation set equal to a non-zero constant, so it misses the origin)
• a set written with squared coordinates — which looks algebraic but breaks closure under scaling

GATE DA 2024 used squared coefficients precisely because they masquerade as linear.

Three sets, three outcomes

Decide which of these subsets of R² are subspaces: (a) {(x, y) : x + y = 0}, (b) {(x, y) : x² + y² ≤ 1}, (c) {(x, y) : x + y = 1}.

Set (a), x + y = 0 — a subspace. The origin gives 0 + 0 = 0 ✓. If two points both satisfy it, so does their sum: (x₁+x₂) + (y₁+y₂) = 0. And scaling by c gives cx + cy = c(x+y) = 0. All three pass — it is the line through the origin of slope −1.

Set (b), the unit disk x² + y² ≤ 1 — NOT a subspace. The origin is inside, so condition 1 passes, but closure fails: (1, 0) is in the disk, yet scaling by 2 gives (2, 0) with 2² + 0² = 4 > 1, outside. The squared terms make the set bounded, and scaling escapes it.

Set (c), x + y = 1 — NOT a subspace. Check the origin first: 0 + 0 = 0 ≠ 1, so it fails immediately. This is the line x + y = 0 shifted off the origin.

Only (a) survives — three sets, three different reasons.

How big is a subspace?

Once a set passes the test, the natural follow-up is how big it is. A subspace’s dimension counts the directions living inside it: 1 for a line, 2 for a plane, n for the whole of Rⁿ.

You can usually read it straight off the defining equation, because each homogeneous linear equation you impose removes exactly one degree of freedom.

Set (a) is the smallest example of that. R² starts with two free coordinates. The equation x + y = 0 forces y = −x, so choosing x decides everything — one free coordinate remains, dimension 1, a line, exactly as we found.

Run the same count in R⁴ and one homogeneous equation leaves 4 − 1 = 3. In general, k independent homogeneous equations in Rⁿ carve out a subspace of dimension n − k.

A question to carry forward

A line through the origin needs just one arrow to generate it; a plane needs two. Here is the thread onward — the word that followed us across from probability: those generating arrows must be independent, each pointing in a genuinely new direction the others cannot reach.

How few independent arrows does it take to build a whole subspace, and how do we tell when one is secretly redundant?

In one breath

• A subspace W passes three tests: (1) contains 0, (2) closed under addition, (3) closed under scalar multiplication (every scalar, including −1).
• Lead with the zero-vector check — if 0 ∉ W, stop (kills every offset/affine set like x+y=1).
• In R³, the only subspaces are:
  • the origin
  • lines through 0
  • planes through 0
  • all of R³
• A subspace equation is linear AND homogeneous (RHS = 0).
• The 2024 trap: squared terms (x²+y² ≤ 1, x²=y) look algebraic but break closure under scaling — scale a member and it leaves the set.
• One homogeneous equation in Rⁿ drops the dimension by 1 (x+y+z=0 in R³ → a 2-D plane).

Practice