# Linear Subspace

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.