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Subspaces &
Subspace Tests

Learn how to identify subspaces of vector spaces using the three-condition test and see why closure under addition and scalar multiplication is the key idea.

Linear Algebra

What Is a Vector Space?

A vector space \(V\) over a field \(\mathbb{F}\) is a set equipped with two operations: vector addition and scalar multiplication, satisfying ten axioms (associativity, commutativity, distributivity, existence of zero, etc.).

\(\mathbb{R}^n\)
The most familiar example: all \(n\)-tuples of real numbers.
\(P_n\)
Polynomials of degree at most \(n\) with real coefficients.
\(M_{m\times n}\)
All \(m \times n\) real matrices under matrix addition and scalar multiplication.

What Is a Subspace?

A subspace \(W\) of a vector space \(V\) is a non-empty subset that is itself a vector space under the same operations inherited from \(V\). In other words, \(W \subseteq V\) and \(W\) "lives inside" \(V\) while obeying all the same rules.

V W 0 u v w

Why Do Subspaces Matter?

Subspaces appear everywhere in linear algebra. Every time you solve a system \(A\mathbf{x} = \mathbf{0}\), the set of solutions forms a subspace. Understanding subspaces helps you analyze null spaces, column spaces, eigenspaces, and more.

Null Space
\(\text{Nul}(A) = \{\mathbf{x} : A\mathbf{x}=\mathbf{0}\}\) is always a subspace of \(\mathbb{R}^n\).
Column Space
\(\text{Col}(A)\) is the span of columns of \(A\): a subspace of \(\mathbb{R}^m\).
Eigenspaces
For each eigenvalue \(\lambda\), the eigenspace \(E_\lambda\) is a subspace of \(\mathbb{R}^n\).

The Subspace Test

To prove that a non-empty subset \(W \subseteq V\) is a subspace, verify three conditions:

1
Contains the zero vector: \(\mathbf{0} \in W\).
2
Closed under addition: If \(\mathbf{u}, \mathbf{v} \in W\), then \(\mathbf{u} + \mathbf{v} \in W\).
3
Closed under scalar multiplication: If \(\mathbf{u} \in W\) and \(c \in \mathbb{F}\), then \(c\mathbf{u} \in W\).

Visualizing the Three Conditions

Contains \(\mathbf{0}\)
\(\mathbf{0}\)
\(\mathbf{u} + \mathbf{v} \in W\)
\(\mathbf{u}\)
\(\mathbf{v}\)
\(\mathbf{u}+\mathbf{v}\)
\(c\mathbf{u} \in W\)
\(\mathbf{u}\)
\(2\mathbf{u}\)

Subspace vs. Subset

Every subspace is a subset, but not every subset is a subspace. The difference is algebraic structure: a subspace must respect the vector space operations.

Subspace Arbitrary Subset
Contains \(\mathbf{0}\) Always Not necessarily
Closed under \(+\) Yes Not guaranteed
Closed under \(c \cdot\) Yes Not guaranteed
Example in \(\mathbb{R}^2\) Line through origin A circle of radius 1

The Two-in-One Shortcut

You can combine conditions 2 and 3 into a single test. If \(W\) is non-empty, then \(W\) is a subspace if and only if for all \(\mathbf{u}, \mathbf{v} \in W\) and all scalars \(c, d \in \mathbb{F}\):

This is called closure under linear combinations. Setting \(c = d = 0\) automatically gives \(\mathbf{0} \in W\), so you don't need to check condition 1 separately.

How to Apply the Test

1
Identify \(V\) and \(W\): What is the ambient vector space? What is the candidate subset?
2
Check \(\mathbf{0} \in W\): Plug in the zero vector and verify it satisfies the defining condition of \(W\).
3
Take two arbitrary elements: Let \(\mathbf{u}, \mathbf{v} \in W\). Write out what this means.
4
Test addition: Compute \(\mathbf{u} + \mathbf{v}\) and show it still satisfies the condition for membership in \(W\).
5
Test scalar multiplication: Compute \(c\mathbf{u}\) for arbitrary scalar \(c\) and show it satisfies the condition.

Example 1: Is This a Subspace?

Let \(W = \{(x, y, z) \in \mathbb{R}^3 : x + 2y - z = 0\}\). Is \(W\) a subspace of \(\mathbb{R}^3\)?

1
Zero vector: \(0 + 2(0) - 0 = 0\). ✓ The zero vector is in \(W\).
2
Addition: Let \(\mathbf{u} = (x_1, y_1, z_1)\) and \(\mathbf{v} = (x_2, y_2, z_2)\) with \(x_i + 2y_i - z_i = 0\). Then \((x_1+x_2) + 2(y_1+y_2) - (z_1+z_2) = 0+0 = 0\). ✓
3
Scalar multiplication: \(c x_1 + 2(c y_1) - c z_1 = c(x_1 + 2y_1 - z_1) = c \cdot 0 = 0\). ✓

Example 2: Not a Subspace

Let \(S = \{(x, y) \in \mathbb{R}^2 : x + y = 1\}\). Is \(S\) a subspace of \(\mathbb{R}^2\)?

x y
\(x+y=1\)
 0 (1,0) (0,1) 0 not on line!

Example 3: A Trickier Case

Let \(W = \{(x, y) \in \mathbb{R}^2 : xy \ge 0\}\). This is the set of all points in the first and third quadrants (including axes). Is \(W\) a subspace?

1
Zero vector: \(0 \cdot 0 = 0 \ge 0\). ✓
2
Scalar multiplication: If \(xy \ge 0\), then \((cx)(cy) = c^2 xy \ge 0\). ✓
3
Addition: Take \(\mathbf{u} = (1,2)\) and \(\mathbf{v} = (-3,-1)\). Both have \(xy \ge 0\). But \(\mathbf{u}+\mathbf{v} = (-2, 1)\) and \((-2)(1) = -2 < 0\). ✕

Classic Subspaces of \(\mathbb{R}^n\)

\(\{\mathbf{0}\}\)
The trivial subspace: just the zero vector. Always a subspace of any \(V\).
\(V\) itself
Every vector space is a subspace of itself. These two are called improper subspaces.
Lines through \(\mathbf{0}\)
Any line through the origin in \(\mathbb{R}^n\) is a 1-dimensional subspace: \(\text{span}(\mathbf{v})\).
Planes through \(\mathbf{0}\)
Any plane through the origin in \(\mathbb{R}^3\) is a 2-dimensional subspace: \(\text{span}(\mathbf{u},\mathbf{v})\).

Span Is Always a Subspace

If \(\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k \in V\), then \(\text{span}(\mathbf{v}_1, \ldots, \mathbf{v}_k)\) is always a subspace of \(V\). Here's why:

1
Zero vector: \(\mathbf{0} = 0\mathbf{v}_1 + \cdots + 0\mathbf{v}_k\) is a linear combination, so \(\mathbf{0} \in \text{span}\). ✓
2
Addition: Sum of two linear combinations is a linear combination. ✓
3
Scalar multiplication: Scalar times a linear combination is a linear combination. ✓

Common Mistakes

✕ Forgetting the zero vector
Checking closure but forgetting to verify \(\mathbf{0} \in W\). Sets like \(\{(x,y):x+y=1\}\) fail here.
✕ Using specific vectors only
Testing with one pair like \((1,0)\) and \((0,1)\) doesn't prove closure for all vectors in \(W\).
✕ Ignoring negative scalars
Checking only \(c > 0\) for scalar closure. You must also check \(c < 0\) (and \(c=0\)).

Summary Table

Candidate Set \(\mathbf{0} \in W\)? Closed \(+\)? Closed \(c \cdot\)? Subspace?
\(\{(x,y,z):x+2y-z=0\}\) ✓ ✓ ✓ Yes
\(\{(x,y):x+y=1\}\) ✕ -- -- No
\(\{(x,y):xy \ge 0\}\) ✓ ✕ ✓ No
\(\text{span}(\mathbf{v}_1,\ldots,\mathbf{v}_k)\) ✓ ✓ ✓ Always

Question 1 of 5

Which of the following is always a subspace of \(\mathbb{R}^n\)?

Question 2 of 5

Let \(W = \{(x,y,z) \in \mathbb{R}^3 : 3x - y + 4z = 0\}\). Which condition does not need a separate check if you use the "closure under linear combinations" shortcut?

Question 3 of 5

The set \(S = \{(x,y) \in \mathbb{R}^2 : y = x^2\}\) is not a subspace. Which condition fails first?

Question 4 of 5

Which of these is a subspace of \(\mathbb{R}^3\)?

Question 5 of 5

To disprove that a set \(W\) is a subspace, what is the most efficient strategy?