Which Vectors Can a Matrix Produce?
For a fixed matrix \(\mathbf{A}\), the equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) asks whether some input vector \(\mathbf{x}\) produces the specified output \(\mathbf{b}\). The matrix determines which outputs are possible. For example, if \(\mathbf{A}\) has three rows, every output \(\mathbf{A}\mathbf{x}\) has three entries, but it need not be true that every vector in \(\mathbb{R}^3\) can occur as an output.
Write an \(m\times n\) matrix as a list of its columns, \(\mathbf{A}=\begin{bmatrix}\mathbf{a}_1&\cdots&\mathbf{a}_n\end{bmatrix}\), where each column belongs to \(\mathbb{R}^m\). The input \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\) has one coordinate for each column. By the definition of matrix multiplication, the output is the weighted sum of those columns:
Thus the set of possible outputs is determined by the linear combinations of the columns.
The dimensions matter: the column space is a subset of \(\mathbb{R}^m\), because the columns have \(m\) entries and every product \(\mathbf{A}\mathbf{x}\) has \(m\) entries. The input \(\mathbf{x}\), in contrast, belongs to \(\mathbb{R}^n\). In general, the number of columns and the number of entries in a right-hand side are different.
Why the Column Space Is the Set of Outputs
Proof. First, take any output \(\mathbf{A}\mathbf{x}\), where \(\mathbf{x}=(x_1,\dots,x_n)^{\top}\). Matrix multiplication gives \(\mathbf{A}\mathbf{x}=x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n\), which is a linear combination of the columns. Therefore every output belongs to \(\operatorname{Col}\mathbf{A}\). Conversely, take any \(\mathbf{b}\in\operatorname{Col}\mathbf{A}\). By the definition of span, there are scalars \(x_1,\dots,x_n\) 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 matrix product formula then gives \(\mathbf{A}\mathbf{x}=\mathbf{b}\), so \(\mathbf{b}\) is an output. This proves both containments and the stated equivalence. \(\square\)
This result turns a question about possible outputs into a question about a span. It also says exactly what a solution of \(\mathbf{A}\mathbf{x}=\mathbf{b}\) represents: its entries are coefficients expressing \(\mathbf{b}\) as a linear combination of the columns. Different coefficient vectors can sometimes produce the same output; the output set remains the column space either way.
Proof. Write the columns of \(\mathbf{A}\) as \(\mathbf{a}_1,\dots,\mathbf{a}_n\). The zero vector belongs to the column space because \(\mathbf{0}=0\mathbf{a}_1+\cdots+0\mathbf{a}_n\). Now let \(\mathbf{u}\) and \(\mathbf{v}\) belong to the column space. There are coefficient vectors \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\) with \(\mathbf{u}=\mathbf{A}\mathbf{x}\) and \(\mathbf{v}=\mathbf{A}\mathbf{y}\). For any scalars \(c,d\in\mathbb{R}\), matrix multiplication gives \(c\mathbf{u}+d\mathbf{v}=c\mathbf{A}\mathbf{x}+d\mathbf{A}\mathbf{y}=\mathbf{A}(c\mathbf{x}+d\mathbf{y})\). This is an output of \(\mathbf{A}\), so it belongs to the column space by the first theorem. The column space contains zero and is closed under linear combinations, which proves it is a subspace of \(\mathbb{R}^m\). \(\square\)
When the Outputs Fill a Plane or a Line
Worked Example: A Plane of Reachable Vectors in \(\mathbb{R}^3\)
Let \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\). Describe all reachable right-hand sides and decide whether \(\mathbf{b}=(2,-1,1)^{\top}\) and \(\mathbf{c}=(1,0,0)^{\top}\) are reachable.
The columns are \(\mathbf{a}_1=(1,0,1)^{\top}\) and \(\mathbf{a}_2=(0,1,1)^{\top}\). For an arbitrary input \(\mathbf{x}=(x_1,x_2)^{\top}\), multiply row by row:
Consequently, every reachable vector has the form \((x_1,x_2,x_1+x_2)^{\top}\). Equivalently, a vector \((r,s,t)^{\top}\) is reachable exactly when \(t=r+s\): its first two entries force \(x_1=r\) and \(x_2=s\), and then its third entry must be \(r+s\).
For \(\mathbf{b}=(2,-1,1)^{\top}\), the required relation holds because \(1=2+(-1)\). Choose \(\mathbf{x}=(2,-1)^{\top}\). Checking the product gives
For \(\mathbf{c}=(1,0,0)^{\top}\), the condition fails: its third entry is \(0\), whereas the sum of the first two entries is \(1+0=1\). More directly, the first two coordinates would require \(x_1=1\) and \(x_2=0\), which produce third coordinate \(x_1+x_2=1\), not \(0\). Thus \(\mathbf{c}\) is not reachable. The column space is a plane through the origin, described by \(t=r+s\).
This example shows why listing the columns is not enough: the right-hand side must be a combination of them. In this matrix the two input coordinates set the first two output entries, while the third output entry is constrained by those choices. A matrix with \(m\) rows can therefore produce only a proper subspace of \(\mathbb{R}^m\).
Worked Example: Dependent Columns and a Line of Outputs
Let \(\mathbf{B}=\begin{bmatrix}1&2&-1\\2&4&-2\end{bmatrix}\). Describe its column space, then determine whether \(\mathbf{d}=(3,6)^{\top}\) and \(\mathbf{e}=(1,0)^{\top}\) are reachable.
The columns are \((1,2)^{\top}\), \((2,4)^{\top}\), and \((-1,-2)^{\top}\). The second is twice the first, and the third is negative one times the first. Thus every linear combination of the columns is a multiple of \((1,2)^{\top}\). Conversely, every multiple \(s(1,2)^{\top}\) is a column combination, since it equals \(s\) times the first column. Therefore
For an input \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\), the full product is
The vector \(\mathbf{d}\) is reachable because it has the form \((s,2s)^{\top}\) with \(s=3\). Taking \(\mathbf{x}=(3,0,0)^{\top}\) gives \(\mathbf{B}\mathbf{x}=(3,6)^{\top}=\mathbf{d}\). The vector \(\mathbf{e}\) is not reachable: if \((1,0)^{\top}=(s,2s)^{\top}\), its first coordinate requires \(s=1\), while its second requires \(2s=0\). These requirements contradict each other.
There are three input coordinates but only one direction of possible output. Extra columns do not necessarily create new reachable directions: here each column is a multiple of the first.
When Every Right-Hand Side Is Reachable
Worked Example: Solving for an Arbitrary Right-Hand Side
Let \(\mathbf{C}=\begin{bmatrix}1&2\\3&1\end{bmatrix}\). Show directly that every vector \((r,s)^{\top}\in\mathbb{R}^2\) is reachable, and find an input that produces it.
Write \(\mathbf{x}=(x_1,x_2)^{\top}\). The equation \(\mathbf{C}\mathbf{x}=(r,s)^{\top}\) means
Multiply the first equation by \(3\) to get \(3x_1+6x_2=3r\). Subtract the second equation: \(5x_2=3r-s\), so \(x_2=(3r-s)/5\). Substituting into the first equation gives \(x_1=r-2(3r-s)/5=(2s-r)/5\). Thus for every real \(r,s\), the input
produces \((r,s)^{\top}\). Check both entries by multiplication:
Since \(r\) and \(s\) were arbitrary, every vector in \(\mathbb{R}^2\) is reachable, and \(\operatorname{Col}\mathbf{C}=\mathbb{R}^2\).
The calculation establishes reachability without needing to test particular right-hand sides one at a time. In contrast, the first two examples describe restricted output sets: a plane in \(\mathbb{R}^3\) and a line in \(\mathbb{R}^2\). The relevant question is always whether the proposed right-hand side belongs to the span of the columns.
Common Pitfall: Checking Coordinates Separately
A coordinate appearing somewhere among the columns does not guarantee that a right-hand side with that coordinate can be produced. The coefficients used for one output coordinate also affect the other coordinates. For the matrix in the first worked example, the first column contributes \(1\) to both the first and third entries. It is therefore wrong to argue that \((1,0,0)^{\top}\) is reachable just because the first column has a \(1\) in its first entry.
The attempted choice \(x_1=1,\ x_2=0\) gives
not \((1,0,0)^{\top}\). In fact, the first two desired entries force \(x_1=1\) and \(x_2=0\), leaving no choice that can change the third entry to \(0\). The correction is to test the whole vector as a linear combination of columns, not to match its coordinates independently.
Check Your Understanding
For each question, use the columns and the definition of the column space to justify your answer.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\1&1\\0&1\end{bmatrix}\), describe the form of every vector in \(\operatorname{Col}\mathbf{A}\).
- Is \((2,3,1)^{\top}\) in the column space of the matrix in question 1? If so, give an input vector and check the product.
- Describe \(\operatorname{Col}\begin{bmatrix}2&-4\\1&-2\end{bmatrix}\) as a set of vectors.
- For the matrix in question 3, is \((2,1)^{\top}\) reachable? Explain using the relation between the coordinates of vectors in its column space.
- Explain why the column space of an \(m\times n\) matrix is a subset of \(\mathbb{R}^m\), not \(\mathbb{R}^n\).
