From Columns to Every Target Vector
A matrix with \(m\) rows produces vectors in \(\mathbb{R}^m\). Its columns span \(\mathbb{R}^m\) when every vector in that target space can be formed as a linear combination of those columns. This is a stronger requirement than being able to reach one particular target: the same columns must be able to produce every possible right-hand side.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\). For a coefficient vector \(\mathbf{x}=(x_1,\dots,x_n)^\top\), matrix multiplication gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). Thus, asking whether a particular \(\mathbf{b}\in\mathbb{R}^m\) is a combination of the columns is the same as asking whether \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution.
The phrase “for every \(\mathbf{b}\)” is essential. A solution for one right-hand side only shows that one vector belongs to the column space. In contrast, consistency for every right-hand side says that the column space contains every vector in the target space. The criterion does not require that the solution be unique: a system may have several solutions for a particular \(\mathbf{b}\) and still be consistent.
A Useful Coordinate Test
The standard basis of \(\mathbb{R}^m\) is \(\mathbf{e}_1,\dots,\mathbf{e}_m\), where \(\mathbf{e}_i\) has a \(1\) in position \(i\) and \(0\) in every other position. These vectors give a convenient way to think about spanning the whole target space.
Proof. Let the list be \(\mathbf{v}_1,\dots,\mathbf{v}_p\). If it spans \(\mathbb{R}^m\), then each \(\mathbf{e}_i\), being a vector in \(\mathbb{R}^m\), is a linear combination of the list. Conversely, suppose that each \(\mathbf{e}_i\) is a linear combination of the list. For any \(\mathbf{b}=(b_1,\dots,b_m)^\top\in\mathbb{R}^m\), the standard-coordinate expression is \(\mathbf{b}=b_1\mathbf{e}_1+\cdots+b_m\mathbf{e}_m\). Substituting a linear combination of the list for each \(\mathbf{e}_i\) expresses \(\mathbf{b}\) as a linear combination of \(\mathbf{v}_1,\dots,\mathbf{v}_p\). Since \(\mathbf{b}\) was arbitrary, the list spans \(\mathbb{R}^m\). \(\square\)
This gives a finite collection of target vectors to check conceptually, though directly testing arbitrary right-hand sides or using row reduction can be more efficient in a particular problem. The theorem concerns the original columns: row reduction is useful for studying the equations, but the rows of a row-reduced matrix are not a replacement list of the original columns.
Worked Examples
Worked Example: A Square Matrix That Reaches Every Target
Show that the columns of \(\mathbf{A}=\begin{bmatrix}1&2\\3&5\end{bmatrix}\) span \(\mathbb{R}^2\) by solving \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for an arbitrary right-hand side \(\mathbf{b}=(r,s)^\top\).
The equations are \(x_1+2x_2=r\) and \(3x_1+5x_2=s\). Multiply the first equation by \(3\), giving \(3x_1+6x_2=3r\). Subtract the second equation from this result: \(x_2=3r-s\). Substituting into the first equation gives \(x_1=r-2(3r-s)=2s-5r\). Thus, for every real \(r,s\), a solution is
Check: Multiplication gives \(\mathbf{A}\mathbf{x}=\begin{bmatrix}(2s-5r)+2(3r-s)\\3(2s-5r)+5(3r-s)\end{bmatrix}=\begin{bmatrix}r\\s\end{bmatrix}=\mathbf{b}\). Since \(r\) and \(s\) were arbitrary, every target is reached, so the columns span \(\mathbb{R}^2\).
Worked Example: Three Columns Spanning \(\mathbb{R}^2\)
Consider \(\mathbf{A}=\begin{bmatrix}1&0&2\\0&1&-1\end{bmatrix}\). Its columns are \((1,0)^\top\), \((0,1)^\top\), and \((2,-1)^\top\). Solve \(\mathbf{A}\mathbf{x}=(r,s)^\top\) in a way that displays the available choices.
The equations are \(x_1+2x_3=r\) and \(x_2-x_3=s\). Choose \(x_3=t\), where \(t\) is any real number. Then \(x_1=r-2t\) and \(x_2=s+t\), so a solution for every target is
Check: \(\mathbf{A}\mathbf{x}=\begin{bmatrix}(r-2t)+2t\\(s+t)-t\end{bmatrix}=\begin{bmatrix}r\\s\end{bmatrix}\). For example, take \(\mathbf{b}=(3,-2)^\top\) and \(t=1\). Then \(\mathbf{x}=(1,-1,1)^\top\), and \(\mathbf{A}\mathbf{x}=(1+2,-1-1)^\top=(3,-2)^\top\). The arbitrary-target calculation, not just this one check, establishes that the columns span \(\mathbb{R}^2\).
Worked Example: A Matrix That Misses a Target
Determine whether the columns of \(\mathbf{A}=\begin{bmatrix}1&2\\2&4\end{bmatrix}\) span \(\mathbb{R}^2\). Write the augmented matrix for the arbitrary right-hand side \((r,s)^\top\):
The second row requires \(s-2r=0\) for consistency. Therefore, this matrix does not give a solution for every right-hand side. For a specific failure, choose \(\mathbf{b}=(0,1)^\top\), so \(r=0\) and \(s=1\). The reduced equation becomes \(0=1\), which is impossible.
Check: For any coefficient vector \((x_1,x_2)^\top\), \(\mathbf{A}\mathbf{x}=(x_1+2x_2,\,2x_1+4x_2)^\top\). Its second coordinate is always twice its first, whereas the coordinates of \((0,1)^\top\) do not satisfy that condition. Thus the chosen target is genuinely outside the column space, and the columns do not span \(\mathbb{R}^2\).
Two Consequences of the Definition
Some quick conclusions can be proved directly from what it means for columns to span. They are useful for recognizing matrices that cannot reach every target, or for building a spanning list from one that is already known to work.
Proof. Suppose row \(k\) of \(\mathbf{A}\) is zero. Every linear combination of the columns then has coordinate \(k\) equal to \(0\), because that coordinate is a linear combination of zero entries. But \(\mathbf{e}_k\in\mathbb{R}^m\) has coordinate \(k\) equal to \(1\), so \(\mathbf{e}_k\) is not a linear combination of the columns. Hence the columns do not span \(\mathbb{R}^m\). \(\square\)
There is also a simple way to preserve spanning when adding more columns.
Proof. Let the original list span \(\mathbb{R}^m\), and take any \(\mathbf{b}\in\mathbb{R}^m\). By the spanning assumption, \(\mathbf{b}\) is a linear combination of the original vectors. The appended list contains every original vector, so the same combination, with coefficient \(0\) on each appended vector, expresses \(\mathbf{b}\) using the longer list. This holds for every \(\mathbf{b}\), so the longer list spans \(\mathbb{R}^m\). \(\square\)
Common Pitfall: One Solvable System Is Not Enough
A common error is to solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for one chosen \(\mathbf{b}\), find a solution, and conclude that the columns span the whole target space. For the matrix in the previous example, \(\mathbf{b}=(1,2)^\top\) is reachable: choose \(\mathbf{x}=(1,0)^\top\), and \(\mathbf{A}\mathbf{x}=(1,2)^\top\). The incorrect conclusion would be “the columns span \(\mathbb{R}^2\) because this system is consistent.”
The correction is to check the requirement in the criterion: every right-hand side must be reachable. For the same matrix, the target \((0,1)^\top\) leads to \(0=1\), so the columns fail to span \(\mathbb{R}^2\). A single successful target answers a membership question about that vector; it does not settle whether the entire space is spanned.
In practice, keep track of the distinction between the number of columns and the dimension of the target. A matrix with more columns than rows can still span its target, as the \(2\times3\) example showed. The decisive question is not whether the matrix is square, but whether every target in \(\mathbb{R}^m\) can be expressed using its columns.
Check Your Understanding
For each question, distinguish a statement about one target from a statement about every target.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), give a solution to \(\mathbf{A}\mathbf{x}=(r,s)^\top\) for arbitrary real \(r,s\), and state what this implies about the columns.
- For \(\mathbf{A}=\begin{bmatrix}1&1\\1&1\end{bmatrix}\), find a right-hand side for which \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is inconsistent.
- Explain why finding one solution to \(\mathbf{A}\mathbf{x}=\mathbf{b}\) does not by itself prove that the columns of \(\mathbf{A}\) span \(\mathbb{R}^m\).
- Can a matrix with three columns and two rows have columns that span \(\mathbb{R}^2\)? Justify your answer using an example or the definition.
- Prove that if row \(k\) of a matrix is zero, then its columns cannot produce \(\mathbf{e}_k\).
