📂Linear Algebra

Subspace of Vector Space

Definition¹

Let $W$ be a non-empty subset of the vector space $V$. If $W$ satisfies the definition of a vector space with respect to the addition and scalar multiplication defined in $V$, then $W$ is called a subspace of the vector space $V$, and is denoted as follows:

$$ W \le V $$

Explanation

To determine whether a subset $W$ of a vector space $V$ is a subspace of $V$, it must satisfy all 10 rules required for being a vector space. It would be quite cumbersome and difficult to check all 10 rules every time we consider a subset of a vector space. Fortunately, due to being a subset of some vector space, some rules are trivially satisfied.

For instance, if $\mathbf{u},\mathbf{v}$ is an element of $W$, it is also an element of $V$, hence (A2), (A3), (M2)-(M5) are naturally satisfied. Therefore, it is enough to verify the closure under addition (A1), the existence of the zero vector (A4), the existence of additive inverses (A5), and the closure under scalar multiplication (M1) to conclude that $W$ is a subspace. However, in practice, it’s even simpler. Satisfying conditions (A1) and (M1) is a necessary and sufficient condition for being a subspace.

Examples

Examples of subspaces of the vector space $V$ include:

• Itself $V$
• Cosets $v + W$

For a linear transformation $T : V \to W$,

• The null space of $T$ $N(T) \le V$
• The range of $T$ $R(T) \le W$

For a linear transformation $T : V \to V$,

• Eigenspace $E_{\lambda}$
• $T$-Invariant space
• $T$-Cyclic space

Theorem: Subspace Test

Let $W$ be a non-empty subset of the vector space $V$. It is a necessary and sufficient condition for $W$ to be a subspace of $V$ if $W$ satisfies the following two conditions:

(A1) The subset $W$ is closed under addition as defined in $V$.

(M1) The subset $W$ is closed under scalar multiplication as defined in $V$.

Proof

• $(\implies)$

  Assume that $W$ is a subspace of $V$. If $W$ is a subspace, it is trivial that $W$ satisfies (A1) and (M1) by the definition of a vector space.

• $(\impliedby)$

  Assume that $W$ satisfies $(A1)$ and $(M1)$. Let $\mathbf{u} \in W$. Then, $W$ is closed under scalar multiplication, and since $0\mathbf{u}=\mathbf{0}$, the following is true:

  $$ 0 \mathbf{u} = \mathbf{0} \in W $$

  For the same reason, the following is true by $(-1)\mathbf{u}=-\mathbf{u}$:

  $$ (-1)\mathbf{u} = -\mathbf{u} \in W $$

  Therefore, $W$ satisfies (A1)-(M5), thus it is a subspace of $V$.

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Theorem: The Intersection of Subspaces is a Subspace²

Let $W_{1}, W_2$ be subspaces of the vector space $V$. Then, $W_{1} \cap W_2$ is also a subspace of $V$.

Proof

By the Subspace Test, we need to check if $W_{1} \cap W_2$ satisfies (A1) and (M1). Let $W= W_{1} \cap W_2$.

• (A1)

  Since $W = W_{1} \cap W_2$, any two vectors $\mathbf u,\mathbf v$ inside $W$ are also contained in $W_{1}$ and $W_2$. Since $W_{1}, W_2$ is a subspace, it is closed under addition. Therefore, the following is true:

  $$ \mathbf u + \mathbf v \in W_{1}, \quad \mathbf u + \mathbf v \in W_2 $$

  Hence, by the definition of intersection, the following is true:

  $$ \mathbf u + \mathbf v \in W $$

  Any two vectors $\mathbf u,\ \mathbf v$ inside $W$ implies that $\mathbf u + \mathbf v$ is also an element of $W$, so $W$ is closed under addition and satisfies (A1).

• (M1)

  The proof follows similarly to the above case.

  Since $W = W_{1} \cap W_2$, any vector $\mathbf u$ inside $W$ is also contained in $W_{1}$ and $W_2$. Since $W_{1},\ W_2$ is a subspace, it is closed under scalar multiplication. Hence, for any scalar $k$, the following is true:

  $$ k\mathbf{u} \in W_{1} \quad k \mathbf{u} \in W_2 $$

  Hence, by the definition of intersection, the following is true:

  $$ k\mathbf u \in W $$

  Any vector $\mathbf u$ inside $W$ implies that $k\mathbf u$ is also an element of $W$, so $W$ is closed under scalar multiplication and satisfies (M1).

• Conclusion

  If $W_{1}, W_{2}$ is a subspace, then $W = W_{1} \cap W_2$ satisfies (A1) and (M1), thus $W$ is also a subspace.

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Corollary

If $W_{1}, W_{2}, \dots W_{n}$ are subspaces of the vector space $V$, then $W = W_{1} \cap \cdots \cap \dots W_{n}$ is also a subspace of $V$.