Column and Row Spaces

and are linear systems of equations, with an identical $3\times 4$ coefficient matrix, which we call $A$ here. However, is consistent, while is not. We can explain this difference by employing the column space of the matrix $A$.

The column vector of constants, $\vect{b}$, in is given below, and one solution listed for $\linearsystem{A}{\vect{b}}$ is $\vect{x}$, \vect{b}&=\colvector{8\\-12\\4} & \vect{x}&=\colvector{7\\8\\1\\3}

By , we can summarize this solution as a linear combination of the columns of $A$ that equals $\vect{b}$, 7\colvector{2\\-3\\1}+ 8\colvector{1\\4\\1}+ 1\colvector{7\\-5\\4}+ 3\colvector{-7\\-6\\-5}= \colvector{8\\-12\\4}=\vect{b}.

This equation says that $\vect{b}$ is a linear combination of the columns of $A$, and then by , we can say that $\vect{b}\in\csp{A}$.

On the other hand, is the linear system $\linearsystem{A}{\vect{c}}$, where the vector of constants is \vect{c}=\colvector{2\\3\\2} and this system of equations is inconsistent. This means $\vect{c}\not\in\csp{A}$, for if it were, then it would equal a linear combination of the columns of $A$ and would lead us to a solution of the system $\linearsystem{A}{\vect{c}}$.

Consider the $5\times 7$ matrix $A$, \begin{bmatrix} 2 & 4 & 1 & -1 & 1 & 4 & 4 \\ 1 & 2 & 1 & 0 & 2 & 4 & 7 \\ 0 & 0 & 1 & 4 & 1 & 8 & 7 \\ 1 & 2 & -1 & 2 & 1 & 9 & 6 \\ -2 & -4 & 1 & 3 & -1 & -2 & -2 \end{bmatrix}

According to the definition (), the column space of $A$ is \csp{A}= \spn{\set{ \colvector{ 2 \\ 1 \\ 0 \\ 1 \\ -2 },\, \colvector{ 4 \\ 2 \\ 0 \\ 2 \\ -4 },\, \colvector{ 1 \\ 1 \\ 1 \\ -1 \\ 1 },\, \colvector{ -1 \\ 0 \\ 4 \\ 2 \\ 3 },\, \colvector{ 1 \\ 2 \\ 1 \\ 1 \\ -1 },\, \colvector{ 4 \\ 4 \\ 8 \\ 9 \\ -2 },\, \colvector{ 4 \\ 7 \\ 7 \\ 6 \\ -2 } }}

While this is a concise description of an infinite set, we might be able to describe the span with fewer than seven vectors. This is the substance of . So we take these seven vectors and make them the columns of a matrix, which is simply the original matrix $A$ again. Now we row-reduce, \begin{bmatrix} 2 & 4 & 1 & -1 & 1 & 4 & 4 \\ 1 & 2 & 1 & 0 & 2 & 4 & 7 \\ 0 & 0 & 1 & 4 & 1 & 8 & 7 \\ 1 & 2 & -1 & 2 & 1 & 9 & 6 \\ -2 & -4 & 1 & 3 & -1 & -2 & -2 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 2 & 0 & 0 & 0 & 3 & 1 \\ 0 & 0 & \leading{1} & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & \leading{1} & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & \leading{1} & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}

The pivot columns are $D=\set{1,\,3,\,4,\,5}$, so we can create the set T=\set{ \colvector{ 2 \\ 1 \\ 0 \\ 1 \\ -2 },\, \colvector{ 1 \\ 1 \\ 1 \\ -1 \\ 1 },\, \colvector{ -1 \\ 0 \\ 4 \\ 2 \\ 3 },\, \colvector{ 1 \\ 2 \\ 1 \\ 1 \\ -1 } } and know that $\csp{A}=\spn{T}$ and $T$ is a linearly independent set of columns from the set of columns of $A$.

The coefficient matrix in is $A$, which row-reduces to $B$, A&=&B&=

Columns 1 and 2 are pivot columns, so by we can write \csp{A}=\spn{\set{\vect{A}_1,\,\vect{A}_2}}=\spn{\set{\colvector{1\\2\\1},\,\colvector{-1\\1\\1}}}.

We want to show in this example that $\csp{A}\neq\complex{3}$. So take, for example, the vector $\vect{b}=\colvector{1\\3\\2}$. Then there is no solution to the system $\linearsystem{A}{\vect{b}}$, or equivalently, it is not possible to write $\vect{b}$ as a linear combination of $\vect{A}_1$ and $\vect{A}_2$. Try one of these two computations yourself. (Or try both!). Since $\vect{b}\not\in\csp{A}$, the column space of $A$ cannot be all of $\complex{3}$. So by varying the vector of constants, it is possible to create inconsistent systems of equations with this coefficient matrix (the vector $\vect{b}$ being one such example).

In we wished to show that the coefficient matrix from was not invertible as a first example of a matrix without an inverse. Our device there was to find an inconsistent linear system with $A$ as the coefficient matrix. The vector of constants in that example was $\vect{b}$, deliberately chosen outside the column space of $A$.

The coefficient matrix in , call it $B$ here, is known to be nonsingular (see ). By , the linear system $\linearsystem{B}{\vect{b}}$ has a (unique) solution for every choice of $\vect{b}$. then says that $\vect{b}\in\csp{B}$ for all $\vect{b}\in\complex{3}$. Stated differently, there is no way to build an inconsistent system with the coefficient matrix $B$, but then we knew that already from .

The coefficient matrix in is I=

To build the row space, we transpose the matrix, \transpose{I}= \begin{bmatrix} 1 & 2 & 0 & -1\\ 4 & 8 & 0 & -4\\ 0 & -1 & 2 & 2\\ -1 & 3 & -3 & 4\\ 0 & 9 & -4 & 8\\ 7 & -13 & 12 & -31\\ -9 & 7 & -8 & 37 \end{bmatrix}

Then the columns of this matrix are used in a span to build the row space, \rsp{I}=\csp{\transpose{I}}= \spn{\set{ \colvector{1\\4\\0\\-1\\0\\7\\-9},\, \colvector{2\\8\\-1\\3\\9\\-13\\7},\, \colvector{0\\0\\2\\-3\\-4\\12\\-8},\, \colvector{-1\\-4\\2\\4\\8\\-31\\37} }}.

However, we can use to get a slightly better description. First, row-reduce $\transpose{I}$, \begin{bmatrix} \leading{1} & 0 & 0 & -\frac{31}{7}\\ 0 & \leading{1} & 0 & \frac{12}{7}\\ 0 & 0 & \leading{1} & \frac{13}{7}\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}.

Since the pivot columns have indices $D=\set{1,\,2,\,3}$, the column space of $\transpose{I}$ can be spanned by just the first three columns of $\transpose{I}$, \rsp{I}=\csp{\transpose{I}}= \spn{\set{ \colvector{1\\4\\0\\-1\\0\\7\\-9},\, \colvector{2\\8\\-1\\3\\9\\-13\\7},\, \colvector{0\\0\\2\\-3\\-4\\12\\-8} }}.