15  Vector Spaces and Subspaces

Recall the definition of a subspace:

A subspace \(\mathcal{H}\) of a vector space \(\mathcal{V}\) is a subset of \(\mathcal{V}\) such that

  • \(\mathcal{H}\) contains the zero vector – \(\mathbf{0} \in \mathcal{H}\)

  • \(\mathcal{H}\) is closed under vector addition. Therefore, for \(\mathbf{u}\) and \(\mathbf{v}\) in \(\mathcal{H}\), the sum \(\mathbf{u} + \mathbf{v}\) is in \(\mathcal{H}\)

  • \(\mathcal{H}\) is closed under scalar multiplication. Therefore, for \(\mathbf{u}\) in \(\mathcal{H}\) and a scalar \(a\), the product \(a \mathbf{u}\) is in \(\mathcal{H}\)

A consequence of this definition is that a subspace \(\mathcal{H}\) is closed under linear combinations.

15.1 Null space and column space

Also, recall the special subspaces of the column space and the null space.

15.1.1 Null space

The null space null(\(\mathbf{A}\)) of an \(m \times n\) \(\mathbf{A}\) is the set of all solutions of the homogeneous equation \(\mathbf{A} \mathbf{x} = \mathbf{0}\)

Another way to write null(\(\mathbf{A}\)) is

\[ \begin{aligned} \mbox{null}(\mathbf{A}) = \{\mathbf{x} : \mathbf{x} \in \mathcal{R}^n \mbox{ and } \mathbf{A} \mathbf{x} = \mathbf{0} \} \end{aligned} \]

Theorem 15.1 The null space of an \(m \times n\) matrix \(\mathbf{A}\) is a subspace of \(\mathcal{R}^n\).

To show that the null space of \(\mathbf{A}\), denoted null(\(\mathbf{A})\), is a subspace we need to show the following

  1. The zero vector \(\mathbf{0}\) is in the null space of \(\mathbf{A}\)

  2. The null space of \(\mathbf{A}\) is closed under addition

  3. The null space of \(\mathbf{A}\) is closed under scalar multiplication

The null space of \(\mathbf{A}\) is defined as the set of vectors \(\mathbf{x}\) such that \(\mathbf{A} \mathbf{x} = \mathbf{0}\).

  1. First, we show that the zero vector is in the subspace by setting \(\mathbf{x} = \mathbf{0}\). Thus, because \(\mathbf{A} \mathbf{0} = \mathbf{0}\), the zero vector \(\mathbf{0}\) is in the null space of \(\mathbf{A}\) .

  2. Next, let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in the null space of \(\mathbf{A}\). Thus, by the definition of the null space we have \(\mathbf{A} \mathbf{u} = \mathbf{0}\) and \(\mathbf{A} \mathbf{v} = \mathbf{0}\). Consider the vector \(\mathbf{u} + \mathbf{v}\) and consider \(\mathbf{A} (\mathbf{u} + \mathbf{v}) = \mathbf{A} \mathbf{u} + \mathbf{A} \mathbf{v} = \mathbf{0} + \mathbf{0} = \mathbf{0}\). Thus \(\mathbf{u} + \mathbf{v}\) is in the null space of \(\mathbf{A}\).

  3. Finally, let \(\mathbf{u}\) be a vector in the null space of \(\mathbf{A}\) and let \(c\) be a scalar. Thus, by the definition of the null space we have \(\mathbf{A} \mathbf{u} = \mathbf{0}\). Then, consider \(\mathbf{A} (c \mathbf{u}) = c \mathbf{A} \mathbf{u} = c \mathbf{0} = \mathbf{0}\)

Because the three requirements for a subspace are met, this gives us that the null space of \(\mathbf{A}\) is a subspace.

As a consequence, there will exist a set of vectors that span the null space null(\(\mathbf{A}\)). However, the null space of \(\mathbf{A}\) is defined implicitly. This means that the null space of \(\mathbf{A}\) is not obvious given the vectors of \(\mathbf{A}\) and must be checked/calculated.

Example 15.1  

  • In class

Find a spanning set for null(\(\mathbf{A}\)) where

\[ \begin{aligned} \mathbf{A} = \begin{pmatrix} 7 & -2 & 7 & -4 & 5 \\ 2 & 0 & 3 & 3 & 9 \\ -5 & 2 & -5 & 7 & -2 \end{pmatrix} \end{aligned} \]

  • Find solution to system of homogeneous system of equations \(\mathbf{A} \mathbf{x} = \mathbf{0}\)
A <- matrix(c(7, 2, -5, -2, 0, 2, 7, 3, -5, -4, 3, 7, 5, 9, -2), 3, 5)
rref(cbind(A, 0))
     [,1] [,2] [,3] [,4]  [,5] [,6]
[1,]    1    0    0 1.50 -4.50    0
[2,]    0    1    0 7.25  2.75    0
[3,]    0    0    1 0.00  6.00    0

  • Take the general solution and write as a linear combination of vectors where the coefficients are the free variables.

  • general solution \(x_1 = -1.5 x_4 + 4.5 x_5\), \(x_2 = -7.25 x_4 - 2.75 x_5\), \(x_3 = -6 x_5\) and both \(x_4\) and \(x_5\) are free. Write out the general solution in vector form.

\[ \begin{aligned} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} = \begin{pmatrix} -1.5 x_4 + 4.5 x_5\\ -7.25 x_4 - 2.75 x_5 \\ -6 x_5 \\ x_4 \\ x_5 \end{pmatrix} = x_4 \begin{pmatrix} -1.5 \\ -7.25 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_5 \begin{pmatrix} 4.5 \\ 2.75 \\ -6 \\ 0 \\ 1 \end{pmatrix} \end{aligned} \]

  • From above, the free variables \(x_4\) and \(x_5\) are multiplied by the vectors \(\mathbf{u} = \begin{pmatrix} -1.5 \\ -7.25 \\ 0 \\ 1 \\ 0 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} 4.5 \\ 2.75 \\ -6 \\ 0 \\ 1 \end{pmatrix}\) where \(\{ \mathbf{u}, \mathbf{v} \}\) are a spanning set for the null(\(\mathbf{A}\))

Example 15.2  

  • In class – do another Find a spanning set for null(\(\mathbf{A}\)) where

15.1.2 Column space

The columns space col(\(\mathbf{A}\)) of an \(m \times n\) \(\mathbf{A}\) is the set of all linear combinations of the columns of \(\mathbf{A}\).

If \(\{ \mathbf{a}_1, \ldots, \mathbf{a}_n\}\) are the columns of \(\mathbf{A}\), then

\[ \begin{aligned} \mbox{col}(\mathbf{A}) = \mbox{span}(\mathbf{A}) \end{aligned} \]

this can be written in set notation as

\[ \begin{aligned} \mbox{col}(\mathbf{A}) = \{ \mathbf{b} : \mathbf{A} \mathbf{x} = \mathbf{b} \mbox{ for some } \mathbf{x} \in \mathcal{R}^n \} \end{aligned} \]

Theorem 15.2 The column space of an \(m \times n\) matrix \(\mathbf{A}\) is a subspace of \(\mathcal{R}^n\).

Do in class

  • \(\mathbf{0}\) vector

  • sum of vectors

  • scalar multiplication

Compared to the null space, the column space is defined explicitly–it is the span of the columns of \(\mathbf{A}\). The definition of the column space results in the fact that col(\(\mathbf{A}\)) is the range of the linear transformation \(\mathbf{x} \rightarrow \mathbf{A} \mathbf{x}\).

Example 15.3  

  • In class

Find a spanning set for col(\(\mathbf{A}\)) where

\[ \begin{aligned} \mathbf{A} = \begin{pmatrix} 6 & 0 & 4 \\ 5 & -1 & -9 \\ -4 & 7 & 4 \\ 6 & 2 & 9 \end{pmatrix} \end{aligned} \]

15.1.3 Understanding the differerneces between the column space and the null space