From Column Combinations to Systems
When we ask whether the columns of a matrix span \(\mathbb{R}^m\), we ask whether every vector in \(\mathbb{R}^m\) can be made by combining those columns. A system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) asks whether a particular vector \(\mathbf{b}\) can be made in exactly that way. The connection between these questions is not a coincidence: the entries of \(\mathbf{x}\) are the coefficients in the column combination.
Let \(\mathbf{A}\in\mathbb{R}^{m\times n}\), with columns \(\mathbf{a}_1,\dots,\mathbf{a}_n\), and let \(\mathbf{x}=(x_1,\dots,x_n)^\top\in\mathbb{R}^n\). By the definition of matrix-vector multiplication, each entry of \(\mathbf{A}\mathbf{x}\) is the corresponding row’s sum of products. Equivalently, collecting the columns gives
This identity is the bridge for the proof. A solution \(\mathbf{x}\) to \(\mathbf{A}\mathbf{x}=\mathbf{b}\) supplies coefficients expressing \(\mathbf{b}\) as a combination of the columns. Conversely, such coefficients, when put into \(\mathbf{x}\), give a solution of the system.
“At least one” matters: consistency does not assert that a solution is unique. The criterion below concerns the existence of a solution for each right-hand side. It does not require that the same \(\mathbf{x}\) work for different right-hand sides; the coefficients may depend on \(\mathbf{b}\).
The Every-Right-Hand-Side Criterion
Proof. First suppose the columns of \(\mathbf{A}\) span \(\mathbb{R}^m\). Take any \(\mathbf{b}\in\mathbb{R}^m\). By the definition of span, there are real numbers \(x_1,\dots,x_n\) such that \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\). Set \(\mathbf{x}=(x_1,\dots,x_n)^\top\). The column formula for matrix-vector multiplication gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n=\mathbf{b}\), so the system is consistent. Because \(\mathbf{b}\) was arbitrary, the system is consistent for every \(\mathbf{b}\in\mathbb{R}^m\).
Now suppose that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b}\in\mathbb{R}^m\). Take any \(\mathbf{b}\in\mathbb{R}^m\). By the assumption, there is an \(\mathbf{x}=(x_1,\dots,x_n)^\top\) with \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Applying the same column formula gives \(\mathbf{b}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), so \(\mathbf{b}\) belongs to the span of the columns. This holds for every \(\mathbf{b}\in\mathbb{R}^m\); therefore every vector in \(\mathbb{R}^m\) belongs to that span, and the columns span \(\mathbb{R}^m\). \(\square\)
The proof uses only the definition of span and the column formula for \(\mathbf{A}\mathbf{x}\). In the forward direction, spanning provides coefficients for an arbitrary target, and those coefficients form a solution. In the reverse direction, consistency provides a solution for an arbitrary target, and its entries are coefficients showing that the target lies in the span. The two directions are separate because one starts with a representation and produces a solution, while the other starts with a solution and produces a representation.
Using the Criterion in Both Directions
Worked Example: Showing That Two Columns Span \(\mathbb{R}^2\)
Let \(\mathbf{A}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), with columns \(\mathbf{a}_1=(1,1)^\top\) and \(\mathbf{a}_2=(1,-1)^\top\). We will show directly that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b}\in\mathbb{R}^2\), and then apply the theorem.
Write an arbitrary right-hand side as \(\mathbf{b}=(b_1,b_2)^\top\). The system is
Adding the equations gives \(2x_1=b_1+b_2\), so take \(x_1=(b_1+b_2)/2\). Subtracting the second equation from the first gives \(2x_2=b_1-b_2\), so take \(x_2=(b_1-b_2)/2\). Thus one candidate solution for every \(\mathbf{b}\) is \(\mathbf{x}=((b_1+b_2)/2,(b_1-b_2)/2)^\top\).
Check: multiplying back gives
Both coordinates match the arbitrary target. Therefore every right-hand side gives a consistent system, and the theorem says that the columns span \(\mathbb{R}^2\).
Worked Example: One Inconsistent Target Rules Out Spanning
Consider \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\), whose columns are \((1,0,0)^\top\) and \((0,1,0)^\top\) in \(\mathbb{R}^3\). To show that these columns do not span \(\mathbb{R}^3\), it is enough to find one right-hand side for which the system is inconsistent. Choose \(\mathbf{b}=(0,0,1)^\top\).
For \(\mathbf{x}=(x_1,x_2)^\top\), the equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) becomes
The first two coordinates require \(x_1=0\) and \(x_2=0\), while the third requires \(0=1\). No \(\mathbf{x}\) can satisfy all three equations, so this system is inconsistent. Equivalently, every column combination has the form \(x_1(1,0,0)^\top+x_2(0,1,0)^\top=(x_1,x_2,0)^\top\), which cannot equal \((0,0,1)^\top\). Since at least one right-hand side fails, the theorem implies that the columns do not span \(\mathbb{R}^3\).
Check: the proposed target really is excluded: the third entry of every column combination is \(x_1\cdot0+x_2\cdot0=0\), whereas the third entry of \(\mathbf{b}\) is \(1\).
Proof. If the columns fail to span, the theorem’s “if and only if” statement rules out consistency for every right-hand side. Therefore at least one right-hand side has an inconsistent system. Conversely, if there is an inconsistent right-hand side, the system is not consistent for every right-hand side; by the theorem, the columns do not span \(\mathbb{R}^m\). \(\square\)
An Explicit Solution for an Arbitrary Target
Worked Example: Verifying a Three-Column Spanning Claim
Let \(\mathbf{A}=\begin{bmatrix}1&0&1\\0&1&1\\1&1&0\end{bmatrix}\), with columns \(\mathbf{a}_1=(1,0,1)^\top\), \(\mathbf{a}_2=(0,1,1)^\top\), and \(\mathbf{a}_3=(1,1,0)^\top\). Rather than testing selected right-hand sides, find coefficients for an arbitrary \(\mathbf{b}=(b_1,b_2,b_3)^\top\).
The equations of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) are \(x_1+x_3=b_1\), \(x_2+x_3=b_2\), and \(x_1+x_2=b_3\). Adding the first and third equations and subtracting the second gives \(2x_1=b_1+b_3-b_2\). Adding the second and third and subtracting the first gives \(2x_2=-b_1+b_2+b_3\). Adding the first and second and subtracting the third gives \(2x_3=b_1+b_2-b_3\). Consequently, choose
Check: the first coordinate of \(\mathbf{A}\mathbf{x}\) is \(x_1+x_3=((b_1+b_3-b_2)+(b_1+b_2-b_3))/2=b_1\). The second is \(x_2+x_3=((-b_1+b_2+b_3)+(b_1+b_2-b_3))/2=b_2\). The third is \(x_1+x_2=((b_1+b_3-b_2)+(-b_1+b_2+b_3))/2=b_3\). Thus \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for every choice of \(b_1,b_2,b_3\), proving consistency for every right-hand side. The columns therefore span \(\mathbb{R}^3\).
This example illustrates a useful proof strategy: take a general target, solve in terms of its entries, and substitute the proposed solution back into the matrix equation. A solution computed for one particular target proves consistency only for that target. A formula that works for an arbitrary target proves the universal statement required by the theorem.
The Quantifier That Commonly Goes Wrong
The phrase “for every \(\mathbf{b}\)” cannot be replaced by “for some \(\mathbf{b}\)” or “for several tested right-hand sides.” For example, consider \(\mathbf{A}=\begin{bmatrix}1\\0\end{bmatrix}\), a matrix with one column in \(\mathbb{R}^2\). The systems with right-hand sides \((1,0)^\top\) and \((2,0)^\top\) are both consistent: choosing \(x=1\) and \(x=2\), respectively, gives the requested products. A mistaken conclusion would be: “Several right-hand sides work, so the column spans \(\mathbb{R}^2\).”
The correction is to check the quantifier. For \(\mathbf{b}=(0,1)^\top\), the equation is \((x,0)^\top=(0,1)^\top\). Its second coordinate requires \(0=1\), so it is inconsistent. The column therefore does not span \(\mathbb{R}^2\). Success for a few targets cannot establish that every target works; one failure, however, is enough to disprove the every-target claim.
There is also a common wording trap in the reverse direction. If every right-hand side has a solution, the solution may depend on the right-hand side. The theorem does not claim that one fixed coefficient vector \(\mathbf{x}\) produces all vectors \(\mathbf{b}\). Rather, for each chosen \(\mathbf{b}\), consistency supplies a suitable \(\mathbf{x}\), and that \(\mathbf{x}\) gives a column combination equal to that particular \(\mathbf{b}\).
| Statement about \(\mathbf{A}\) | What it means | Conclusion |
|---|---|---|
| The columns span \(\mathbb{R}^m\) | Every target is a column combination | \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is consistent for every \(\mathbf{b}\in\mathbb{R}^m\) |
| One target has no solution | That target is not a column combination | The columns do not span \(\mathbb{R}^m\) |
| Some tested targets have solutions | Those targets are column combinations | This alone does not establish spanning |
Why the Criterion Is Useful
The criterion lets us translate between two ways of asking the same question. “Do the columns span the target space?” is a statement about all vectors in \(\mathbb{R}^m\). “Is \(\mathbf{A}\mathbf{x}=\mathbf{b}\) consistent for every \(\mathbf{b}\)?” is a statement about all right-hand sides of a family of systems. The proof identifies the precise translation: the entries of a solution are the coefficients of the corresponding column combination.
When proving spanning, begin with an arbitrary target and show it can be expressed as a combination of the columns. When disproving spanning, find a target that cannot be so expressed, or equivalently a right-hand side that makes the system inconsistent. These approaches follow directly from the two directions of the theorem; they do not require a unique solution, and they do not rely on checking only selected targets.
Check Your Understanding
For each question, pay close attention to whether the claim concerns one target or every target.
- In the theorem, where does the coefficient vector \(\mathbf{x}\) come from when the columns are assumed to span?
- Why does a solution to \(\mathbf{A}\mathbf{x}=\mathbf{b}\) show that \(\mathbf{b}\) belongs to the span of the columns?
- If one right-hand side makes \(\mathbf{A}\mathbf{x}=\mathbf{b}\) inconsistent, what can you conclude about the columns?
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), find coefficients that solve \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for arbitrary \(\mathbf{b}=(b_1,b_2,b_3)^\top\), or identify a condition on \(\mathbf{b}\) required for consistency.
- Explain why checking consistency for several chosen right-hand sides does not prove that the columns span the whole target space.
