A linear subspace or vector subspace is a vector space that is a subset of some larger vector space.

To be considered a linear subspace, a vector set needs to meet the following 3 requirements.

- (1) The set includes the zero vector.
- (2) The set is closed (or, closure) under scalar multiplication.
- (3) The set is closed (or, closure) under addition.

For instance, if we say that \( V \) is a subset of \( \mathbb{R}^n \) and in order to make sure \( V \) is a subspace, it needs to meet the 3 requirements mentioned above.

First, it needs to include the n-dimension zero vector. For instance, if \( \mathbb{R}^n \) is \( \mathbb{R}^2 \), then \( V \) should include the 2 dimension 0 vector \( \vec{v} = \left[\begin{array}{ccc} 0 \\ 0 \end{array} \right]\).

Second, the set \( V \) is closed under scalar multiplication. This means that, for any in the vector set \( \vec{v} \), \( c\vec{v} \) must also be in \( V \), in which \( c\) is any real-number scalar.

Third, the set has to be closed under addition. This means that, if \( c_1 \vec{v_1} \) and \( c_2 \vec{v_2} \) are both vectors in the set \( V \), the vector \( c_1 \vec{v_1} + c_2 \vec{v_2} \) must also be in \( V \) as well. If you combine with the 2nd requirement, here, we are saying that any linear combination of the vectors in *\( V \)* will also be in \( V \).

**Example 1**

We can test if \( V_1 \) is a linear subspace.

\( V_1 = \{ \vec{v_1}=\left[\begin{array}{ccc} 0 \\ 0 \end{array} \right] \} \)

First, since it includes 0 vector \( \vec{v_1} \), it meets the first requirement. Second, any scalar \( c \), it will lead to a 0 vector. Third, since it is only one vector, it goes back to the second requirement.

**Example 2**

We can test if \( V_2 \) is a subspace.

\( V_2 = \{ \vec{v_1}=\left[\begin{array}{ccc} 0 \\ 0 \end{array} \right], \vec{v_2}=\left[\begin{array}{ccc} 2 \\ 3 \end{array} \right] \} \)

First, since it includes 0 vector \( \vec{v_1} \), it meets the first requirement. Second, if using the scalar 2 to time a vector \( 2\vec{v_2} = 2 \left[\begin{array}{ccc} 2 \\ 3 \end{array} \right] \), \( 2 \left[\begin{array}{ccc} 2 \\ 3 \end{array} \right] \) is not the set of \( V_2 \).

**Example 3**

Note that, a span is all of the linear combinations of the vector. We can also test if \( V_3 \), the span of a vector, is a linear subspace.

\( V_3 = span( \left[\begin{array}{ccc} 2 \\ 3 \end{array} \right]) \)

We can plot the line as follows, which goes through point (0, 0) and extends both sides.

First, since it goes through \( \left[\begin{array}{ccc} 0 \\ 0 \end{array} \right] \), it meets the first requirement. Second, we time the vector using the scalar 2, namely \( 2\vec{v_2} = 2 \left[\begin{array}{ccc} 2 \\ 3 \end{array} \right] \), which is still on the span of the vector. Third, go back to requirement 2.

**Reference**

You can refer to the following two articles on linear subspaces.

Linear subspaces (Khan Academy)

Definition of A Linear Subspace

**Further Reading**

You can also read my other tutorials on the difference between Space and Subspace. Importantly, I also have tutorials on how to view mean as a projection and how to view linear regression as a projection as well.