Vector Spaces

The Essentials

A vector space is made up of component vectors. It is defined by every vector that can be made by a linear combination of its component vectors. Therefore, this vector, x, would be a part of the space, S:

\[ x= \begin{bmatrix} 2\\ 4\\ 6\\ 8 \end{bmatrix}, S= \begin{Bmatrix} \begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 2\\ 3 \end{bmatrix} \end{Bmatrix} \]

This is because x is a linear combination of the component vectors of S ( \( 2S_1+2S_2=x \)).

A group of vectors is called a span of the space if every vector in the space can be made by a linear combination of the vectors in the span. Note that the span would not be unique to the space. A basis is a linearly independent span. That is, a basis is a span that only has just enough information to make up the space. Bases are not unique either. The dimension of a space is the number of vectors in a basis of the space.

Four frequently talked about vector spaces are the row space, column space, null space, and the left null space of a matrix. The row space is a basis of the space that can be made by a linear combination of the rows of the matrix. Likewise, the column space is a basis of the space made by the column vectors of the matrix. These two spaces are just the row or column vectors of the matrix if the matrix is linearly independent. In any case, the components of these spaces are the vectors (row or column, respectively) that correspond to the vectors that have pivots when the matrix is reduced to row echelon form. Therefore, the dimension of both the row and column space is equal to the rank of the matrix (the number of pivots):

\[ \dim\mathcal{R}(A)=r \]
\[ \dim\mathcal{C}(A)=r \]

The null space of a matrix is a basis of all vectors that can't be made by a linear combination of the rows of the matrix. In other words, it is a basis of every vector that solves this equation: \( A\vec{x}=\vec{0} \). The left null space of a matrix is the null space of the transpose of that matrix. It represents all vectors that can't be made by a linear combination of the columns of the matrix. The dimension of the null space of a matrix is equal to the number of columns of the matrix minus the rank of the matrix; Likewise, the dimension of the left null space of a matrix is equal to the number of rows minus the rank of the matrix:

\[ \dim\mathcal{N}(A)=m-r \]
\[ \dim\mathcal{N}(A^T)=n-r \]

The null space of a matrix can be found reducing the matrix to reduced row echelon form. The pivot variables are written in terms of the free variables and then the solution vector is written with just free variables, which can be split into one vector for each free variable. Those vectors form the null space.

Example

Find bases for the row space, the column space, and the null space of this matrix:

\[ A= \begin{bmatrix} 1 & 0 & 1 & 2 & -1\\ -1 & 2 & 3 & -2 & 1\\ 1 & 1 & 1 & 1 & 1 \end{bmatrix} \]

Finding the row and column spaces begins with finding where the pivots will be. This means reducing the matrix to row echelon form:

\[ E= \begin{bmatrix} 1 & 0 & 1 & 2 & -1\\ 0 & 2 & 4 & 0 & 0\\ 0 & 0 & -2 & -1 & 2 \end{bmatrix} \]

All three rows have a pivot. That means that a basis of the row space can be made by the three vectors in the matrix. They can be the three row vectors from either the original matrix A or the reduced matrix E since the row operations that were performed don't change the row space of the matrix. Usually, the row vectors are taken from the reduced matrix:

\[ \mathcal{R}(A)= \begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 1\\ 2\\ -1 \end{bmatrix}, \begin{bmatrix} 0\\ 2\\ 4\\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ -2\\ -1\\ 2 \end{bmatrix} \end{Bmatrix} \]

A basis for the column space can be made from the columns of the original matrix corresponding with the columns with pivots in the reduced matrix. The first three columns of the reduced matrix have pivots, therefore the first three columns of A form a basis of the column space:

\[ \mathcal{C}(A)= \begin{Bmatrix} \begin{bmatrix} 1\\ -1\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 2\\ 1 \end{bmatrix}, \begin{bmatrix} 1\\ 3\\ 1 \end{bmatrix} \end{Bmatrix} \]

To find the null space, the matrix will be reduced to reduced row echelon form:

\[ R= \begin{bmatrix} 1 & 0 & 0 & 3/2 & 0\\ 0 & 1 & 0 & -1 & 2\\ 0 & 0 & 1 & 1/2 & -1 \end{bmatrix} \]

Recall that the vectors in the null space solve this equation: \( A\vec{x}=\vec{0} \). Therefore, we can write this:

\[ \begin{bmatrix} 1 & 0 & 0 & 3/2 & 0\\ 0 & 1 & 0 & -1 & 2\\ 0 & 0 & 1 & 1/2 & -1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5 \end{bmatrix}= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} \]

If we look at this matrix multiplication, the pivots line up the first three variable, \( x_1,x_2,x_3 \), which are called pivot variables. The variables \( x_4,x_5 \) are called free variables, because their columns don't have pivots. If we multiply the matrix by the vector we get a system of three equations:

\[ \begin{cases} x_1+3/2x_4=0\\ x_2-x_4+x_5=0\\ x_3+1/2x_4-x_5=0 \end{cases} \]

Each of which has exactly one pivot variable. Now we will solve each equation for the pivot variable in terms of the free variables:

\[ \begin{cases} x_1=-3/2x_4\\ x_2=x_4-x_5\\ x_3=-1/2x_4+x_5 \end{cases} \]

Now we substitute these expressions into the solution vector \( \vec{x} \):

\[ \begin{bmatrix} -3/4x_4\\ x_4-x_5\\ -1/2x_4+x_5\\ x_4\\ x_5 \end{bmatrix} \]

This vector can be split into a number of vectors, one for each free variable:

\[ x_4 \begin{bmatrix} -3/4\\ 1\\ -1/2\\ 1\\ 0 \end{bmatrix}+x_5 \begin{bmatrix} 0\\ -1\\ 1\\ 0\\ 1 \end{bmatrix} \]

A basis of the null space is these two vectors:

\[ \mathcal{N}(A)= \begin{Bmatrix} \begin{bmatrix} -3/4\\ 1\\ -1/2\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} 0\\ -1\\ 1\\ 0\\ 1 \end{bmatrix} \end{Bmatrix} \]

Practice

Find a basis for the row space, the column space, the null space, and the left null space of this matrix:

\[ A= \begin{bmatrix} 1 & 0 & 1 & 1\\ 1 & 1 & 1 & 1\\ 3 & 2 & 3 & 3 \end{bmatrix} \]

Solution:

\[ \mathcal{R}(A)= \begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 1\\ 1 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix} \end{Bmatrix} \]
\[ \mathcal{C}(A)= \begin{Bmatrix} \begin{bmatrix} 1\\ 1\\ 3 \end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 2 \end{bmatrix} \end{Bmatrix} \]
\[ \mathcal{N}(A)= \begin{Bmatrix} \begin{bmatrix}-1\\ 0\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix}-1\\ 0\\ 0\\ 1 \end{bmatrix} \end{Bmatrix} \]
\[ \mathcal{N}(A^T)= \begin{Bmatrix} \begin{bmatrix} -1\\ -2\\ 0 \end{bmatrix} \end{Bmatrix} \]