Subspaces

Subspaces of \(\mathbb{R}^n\) include lines, planes and hyperplanes through the origin. A basis of a subspace is a linearly independent set of spanning vectors. The Rank-Nullity Theorem describes the dimensions of the nullspace and range of a matrix.

Subspaces

A subset \(U \subseteq \mathbb{R}^n\) is a subspace if:

1. \(U\) contains the zero vector \(\boldsymbol{0}\)

2. \(\boldsymbol{u}_1 + \boldsymbol{u}_2 \in U\) for all \(\boldsymbol{u}_1,\boldsymbol{u}_2 \in U\)

3. \(c \boldsymbol{u} \in U\) for all \(c \in \mathbb{R},\boldsymbol{u} \in U\)

Condition 2 is called closed under addition. Condition 3 is called closed under scalar multiplication. Condition 3 with \(c=0\) implies Condition 1.

• The zero subspace \(\{ \boldsymbol{0} \}\) and the entire space \(\mathbb{R}^n\) are both subspaces of \(\mathbb{R}^n\).

• Subspaces of \(\mathbb{R}^2\) include any line through the origin.

• Subspaces of \(\mathbb{R}^3\) include any line or plane through the origin.

• In general, subspaces of \(\mathbb{R}^n\) are hyperplanes of any dimension through the origin.

Consider the set

\[\begin{split} U = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} : y \geq 0 \right\} \end{split}\]

Is \(U \subset \mathbb{R}^2\) a subspace?

Solution

Then \(U\) contains the zero vector

\[\begin{split} \boldsymbol{0} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \in U \end{split}\]

and \(U\) is closed under addition since

\[\begin{split} \boldsymbol{u}_1 + \boldsymbol{u}_2 = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} + \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \\ y_1 + y_2 \end{bmatrix} \in U \end{split}\]

because \(y_1 + y_2 \geq 0\) since \(y_1 \geq 0\) and \(y_2 \geq 0\). However \(U\) is not closed under scalar multiplication because

\[\begin{split} (-1) \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \end{bmatrix} \not\in U \end{split}\]

Therefore \(U\) is not a subspace of \(\mathbb{R}^2\).

Linear Independence and Span

A linear combination of vectors \(\boldsymbol{u}_1,\dots,\boldsymbol{u}_m \in \mathbb{R}^n\) is a vector

\[ c_1 \boldsymbol{u}_1 + \cdots + c_m \boldsymbol{u}_m \]

where \(c_1,\dots,c_m \in \mathbb{R}\). The span of vectors \(\boldsymbol{u}_1,\dots,\boldsymbol{u}_m \in \mathbb{R}^n\) is the set of all linear combinations

\[ \mathrm{span} \{ \boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \} = \{ c_1 \boldsymbol{u}_1 + \cdots + c_m \boldsymbol{u}_m \in \mathbb{R}^n : c_1,\dots,c_m \in \mathbb{R} \} \]

Let \(\boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \in \mathbb{R}^n\). Then \(\mathrm{span} \{ \boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \}\) is a subspace of \(\mathbb{R}^n\).

The span of a single nonzero vector \(\boldsymbol{u}\) is a line with direction \(\boldsymbol{u}\). The span of two nonzero vectors \(\boldsymbol{u}\) and \(\boldsymbol{v}\) is a plane as long as \(\boldsymbol{u}\) and \(\boldsymbol{v}\) are not colinear.

A set of vectors \(\{ \boldsymbol{u}_1,\dots,\boldsymbol{u}_m \} \subset \mathbb{R}^n\) forms a linearly independent set if the vectors satisfy the property:

\[ c_1 \boldsymbol{u}_1 + \cdots + c_m \boldsymbol{u}_m = \boldsymbol{0} \hspace{5mm} \text{if and only if} \hspace{5mm} c_1 = \cdots = c_m = 0 \]

In other words, \(\{ \boldsymbol{u}_1,\dots,\boldsymbol{u}_m \}\) is a linearly independent set if no vector in the set can be expressed as a linear combination of the others.

How do we know if a set of vectors \(\{ \boldsymbol{u}_1,\dots,\boldsymbol{u}_m \}\) is linearly independent? Create a matrix where the columns are the given vectors

\[\begin{split} A = \begin{bmatrix} & & \\ \boldsymbol{u_1} & \cdots & \boldsymbol{u_m} \\ & & \end{bmatrix} \end{split}\]

Then \(\{ \boldsymbol{u}_1,\dots,\boldsymbol{u}_m \}\) is a linearly independent set if and only if the linear system \(A \boldsymbol{x} = \boldsymbol{0}\) has only the trivial solution \(\boldsymbol{x} = \boldsymbol{0}\).

Basis and Dimension

Let \(U \subseteq \mathbb{R}^n\) be a subspace. A set of vectors \(\{ \boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \}\) forms a basis of \(U\) if:

1. \(\{ \boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \}\) is a linearly independent set

2. \(\mathrm{span} \{ \boldsymbol{u}_1 , \dots , \boldsymbol{u}_m \} = U\)

The dimension of \(U\) is the number \(m\) of vectors in a basis.

Note that the first condition in the preceding definition guarantees that basis are minimal i.e. contain the smallest number of vectors which span the vectorspace \(U\).

Exercises

Determine whether or not the set

\[\begin{split} U = \left\{ \begin{bmatrix} a \\ b \\ c \end{bmatrix} : abc = 0 \right\} \end{split}\]

is a subspace of \(\mathbb{R}^3\).

Solution

\(U\) is not a subspace because it is not closed under vector addition.

Determine whether \(\mathrm{span} \{ \boldsymbol{u}_1 , \boldsymbol{u}_2 \} = \mathrm{span} \{ \boldsymbol{u}_3 , \boldsymbol{u}_4 \}\) where

\[\begin{split} \boldsymbol{u}_1 = \left[ \begin{array}{r} 2 \\ -3 \\ 1 \\ -1 \end{array} \right] \hspace{10mm} \boldsymbol{u}_2 = \left[ \begin{array}{r} -5 \\ 1 \\ 2 \\ -2 \end{array} \right] \hspace{10mm} \boldsymbol{u}_3 = \left[ \begin{array}{r} -1 \\ -5 \\ 4 \\ -4 \end{array} \right] \hspace{10mm} \boldsymbol{u}_4 = \left[ \begin{array}{r} 3 \\ -11 \\ 6 \\ -10 \end{array} \right] \end{split}\]

Solution

\(\mathrm{span} \{ \boldsymbol{u}_1 , \boldsymbol{u}_2 \} \ne \mathrm{span} \{ \boldsymbol{u}_3 , \boldsymbol{u}_4 \}\) since \(\boldsymbol{u}_1,\boldsymbol{u}_2,\boldsymbol{u}_4\) are linearly independent.

Let \(U = \mathrm{span} \{ \boldsymbol{u}_1 , \boldsymbol{u}_2 , \boldsymbol{u}_3 , \boldsymbol{u}_4 \} \subseteq \mathbb{R}^4\) where

\[\begin{split} \boldsymbol{u}_1 = \left[ \begin{array}{r} 2 \\ 4 \\ 4 \\ 2 \end{array} \right] \hspace{10mm} \boldsymbol{u}_2 = \left[ \begin{array}{r} 3 \\ 5 \\ 3 \\ 1 \end{array} \right] \hspace{10mm} \boldsymbol{u}_3 = \left[ \begin{array}{r} 3 \\ 3 \\ -1 \\ -11 \end{array} \right] \hspace{10mm} \boldsymbol{u}_4 = \left[ \begin{array}{r} 0 \\ 3 \\ 11 \\ -2 \end{array} \right] \end{split}\]

• Find a basis and the dimension of \(U\).

• Is \(\{ \boldsymbol{u}_1 , \boldsymbol{u}_3 , \boldsymbol{u}_4 \}\) a basis of \(U\)? Explain.

Solution

\(\dim(U) = 3\) and \(\{ \boldsymbol{u}_1 , \boldsymbol{u}_3 , \boldsymbol{u}_4 \}\) also forms a basis of \(U\).