# Subspaces - Examples with Solutions

## Definiiton of Subspaces

If W is a subset of a vector space V and if W is itself a vector space under the inherited operations of addition and scalar multiplication from V, then W is called a subspace.1 , 2

To show that the W is a subspace of V, it is enough to show that

- W is a subset of V
- The zero vector of V is in W
- For any vectors \( \textbf{u} \) and \( \textbf{v} \) in W, \( \textbf{u} + \textbf{v} \) is in W. (closure under additon)
- For any vector \( \textbf{u} \) and scalar \( r \), \( r \cdot \textbf{u} \) is in W. (closure under scalar multiplication).

## Examples of Subspaces

Example 1

The set W of vectors of the form \( (x,0) \) where \( x \in \mathbb{R} \) is a subspace of \( \mathbb{R}^2 \) because:

W is a subset of \( \mathbb{R}^2 \) whose vectors are of the form \( (x,y) \) where \( x \in \mathbb{R} \) and \( y \in \mathbb{R} \)

The zero vector \( (0,0)\) is in W

\( (x_1,0) + (x_2,0) = (x_1 + x_2 , 0) \) , closure under addition

\( r \cdot (x,0) = (r x , 0) \) , closure under scalar multiplication

Example 2

The set W of vectors of the form \( (x,y) \) such that \( x \ge 0 \) and \( y \ge 0 \) is not a subspace of \( \mathbb{R}^2 \) because it is not closed under scalar multiplication.

Vector \( \textbf{u} = (2,2) \) is in W but its negative \( -1(2,2) = (-2,-2) \) is not in W.

Example 3

The set W of vectors of the form \( W = \{ (x,y,z) | x + y + z = 0 \} \) is a subspace of \( \mathbb{R}^3 \) because

1) It is a subset of \( \mathbb{R}^3 = \{ (x,y,z) \} \)

2) The vector \( (0,0,0) \) is in W since \( 0 + 0 + 0 = 0 \)

3) Let \( \textbf{u} = (x_1 , y_1 , z_1) \) and \( \textbf{v} = (x_2 , y_2 , z_2) \) be vectors in W. Hence

\( x_1 + y_1 + z_1 = 0 \) and \( x_2 + y_2 + z_2 = 0 \)

\( (x_1 , y_1 , z_1) + (x_2 , y_2 , z_2) \\\\ \quad = (x_1+x_2 , y_1+y_2 , z_1+z_2) \\\\ \quad = (x_1+x_2) + (y_1+y_2) + (z_1+z_2) \\\\ \quad = (x_1+y_1+z_1) + (x_2+y_2+z_2) = 0 + 0 = 0 \)

hence closure under addition.

4) Let \( r \) be a real number

\( r (x_1 , y_1 , z_1) = (r x_1 , r y_1 , r z_1) \)

\( r x_1 + r y_1 + r z_1 \\\\ \quad = r( x_1 + y_1 + z_1 ) \\\\ \quad = r \cdot 0 = 0 \)

hence closure under scalar multiplication

### More References and links

- Vector Spaces - Questions with Solutions
- Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
- Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres