Four Different Questions Require Four Careful Checks
Arguments about linear maps often go wrong not because the calculations are difficult, but because a correct calculation is used to answer the wrong question. Basis images belong in columns, composition order follows the order in which maps act, and pivot columns and pivot rows answer different questions for a rectangular matrix. Likewise, finding an input for one chosen output does not prove that every output can be reached.
Throughout, maps are linear maps between real coordinate spaces, and vectors are columns. For a linear map \(T:\mathbb{R}^n\to\mathbb{R}^m\), its standard matrix is the matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\) satisfying \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) for every \(\mathbf{x}\in\mathbb{R}^n\). A map is one-to-one if distinct inputs have distinct outputs; it is onto if every vector in its codomain is an output.
Basis Images Go in Columns
Proof. Write any \(\mathbf{x}\in\mathbb{R}^n\) as \(\mathbf{x}=x_1\mathbf{e}_1+\cdots+x_n\mathbf{e}_n\). Linearity gives \(T(\mathbf{x})=x_1T(\mathbf{e}_1)+\cdots+x_nT(\mathbf{e}_n)\). By the definition of matrix-vector multiplication, this is exactly the product of the matrix whose columns are \(T(\mathbf{e}_1),\dots,T(\mathbf{e}_n)\) with \(\mathbf{x}\). Thus that matrix represents \(T\). If another matrix represented \(T\), its \(j\)th column would be its product with \(\mathbf{e}_j\), which must equal \(T(\mathbf{e}_j)\); hence the matrix is unique. \(\square\)
Worked Example: Correcting a Rows-for-Columns Error
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be given by \(T(x,y)=(x+2y,\;3x-y)\). Find its standard matrix and check the result on a nonstandard input.
The standard basis vectors are \(\mathbf{e}_1=(1,0)\) and \(\mathbf{e}_2=(0,1)\). Their images are
These image vectors are columns, so the standard matrix is
For \(\mathbf{x}=(2,-1)\), direct evaluation gives \(T(2,-1)=(2-2,\;6+1)=(0,7)\). Matrix multiplication agrees:
A common wrong calculation places the same image coordinates in rows, producing \(\begin{bmatrix}1&3\\2&-1\end{bmatrix}\). That matrix sends \((2,-1)\) to \((-1,5)\), not to \((0,7)\). The correction is to put each output \(T(\mathbf{e}_j)\) in column \(j\), then check by applying the matrix to a basis vector or another input.
Composition Order Follows the Actions
If \(T:\mathbb{R}^n\to\mathbb{R}^m\) has standard matrix \(\mathbf{A}\), and \(S:\mathbb{R}^m\to\mathbb{R}^k\) has standard matrix \(\mathbf{B}\), then \(S\circ T\) means “apply \(T\), then apply \(S\).” The input is acted on first by \(\mathbf{A}\), then by \(\mathbf{B}\), so the product is \(\mathbf{B}\mathbf{A}\), not \(\mathbf{A}\mathbf{B}\).
Proof. For every \(\mathbf{x}\in\mathbb{R}^n\), the definition of standard matrix gives \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) and \(S(\mathbf{y})=\mathbf{B}\mathbf{y}\) for every \(\mathbf{y}\in\mathbb{R}^m\). Substituting \(\mathbf{y}=T(\mathbf{x})\) yields \((S\circ T)(\mathbf{x})=S(T(\mathbf{x}))=\mathbf{B}(\mathbf{A}\mathbf{x})=(\mathbf{B}\mathbf{A})\mathbf{x}\). Thus \(\mathbf{B}\mathbf{A}\) represents the composite map. \(\square\)
Worked Example: Catching a Reversed Composition
Let \(T\) and \(S\) have standard matrices \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\), respectively. Find the matrix for \(S\circ T\), and compare it with the reversed order.
For \(S\circ T\), apply \(T\) first, so multiply \(\mathbf{B}\mathbf{A}\):
For the reversed composite \(T\circ S\), the product is \(\mathbf{A}\mathbf{B}\):
The two products differ. As a check, start with \(\mathbf{e}_1=(1,0)\). First \(T(\mathbf{e}_1)=(1,0)\), then \(S(1,0)=(1,1)\), which is the first column of \(\mathbf{B}\mathbf{A}\). In the reversed order, \(S(\mathbf{e}_1)=(1,1)\), then \(T(1,1)=(2,1)\), the first column of \(\mathbf{A}\mathbf{B}\). Reversing the matrix order changes the transformation in this example.
Pivot Columns and Pivot Rows Answer Different Questions
For a matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), the system \(\mathbf{A}\mathbf{x}=\mathbf{b}\) describes inputs mapped to output \(\mathbf{b}\). A pivot in a column concerns whether a corresponding input variable is free in the homogeneous system. A pivot in a row concerns whether an equation can impose a restriction on the possible right-hand sides. These are not interchangeable tests.
Proof. Row operations on a homogeneous system preserve its solutions. Therefore \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\operatorname{rref}(\mathbf{A})\mathbf{x}=\mathbf{0}\) have the same solutions. The map is one-to-one exactly when the only input mapped to \(\mathbf{0}\) is \(\mathbf{0}\): indeed, if two inputs have the same output, their difference maps to \(\mathbf{0}\), and conversely a nonzero input mapped to \(\mathbf{0}\) gives two distinct inputs with the same output. In RREF, if every column has a pivot, there are no free variables and the homogeneous system has only the zero solution. If some column has no pivot, set one free variable to \(1\), the other free variables to \(0\), and solve for the pivot variables; this gives a nonzero solution. This proves the one-to-one criterion.
For the onto criterion, row operations can be represented by multiplication on the left by an invertible matrix \(\mathbf{P}\), so \(\operatorname{rref}(\mathbf{A})=\mathbf{P}\mathbf{A}\). Multiplication by \(\mathbf{P}\) maps \(\mathbb{R}^m\) bijectively to itself, so \(\mathbf{A}\) is onto exactly when \(\mathbf{P}\mathbf{A}\) is onto. If RREF has a pivot in every row, then each system \(\operatorname{rref}(\mathbf{A})\mathbf{x}=\mathbf{c}\) has a solution for every \(\mathbf{c}\in\mathbb{R}^m\): set free variables to zero and solve for the pivot variables, starting from the last pivot row. If a row has no pivot, that row in RREF is zero. Choosing \(\mathbf{c}\) with a nonzero entry in that row makes the transformed system inconsistent. Thus the map is onto exactly when every row has a pivot. \(\square\)
Worked Example: Two Pivot Columns Do Not Mean Onto
Classify the map \(T:\mathbb{R}^2\to\mathbb{R}^3\) with matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\).
This matrix is already in RREF. Both columns contain pivots, so the pivot test says \(T\) is one-to-one. Indeed, \(\mathbf{A}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\) gives \(x=0\) and \(y=0\). But the third row has no pivot, so the map is not onto \(\mathbb{R}^3\). Its outputs have the form \((x,y,0)\); for example, \((0,0,1)\) is not an output.
Check this last claim by solving for that target:
The contradiction confirms that this particular target has no preimage. Here the pivot columns establish one-to-one behavior, while the missing pivot row rules out being onto.
One Consistent Right-Hand Side Is Not Surjectivity
To show that \(T\) is onto, one must show that every \(\mathbf{b}\) in the codomain has at least one solution to \(\mathbf{A}\mathbf{x}=\mathbf{b}\). Solving the system for one selected \(\mathbf{b}\) establishes only that this one output is reached. It cannot establish a statement about all outputs.
Worked Example: A Solvable Target and a Missing Target
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\). Check whether the target \(\mathbf{b}_1=(2,0)\) has a preimage, and decide whether this proves that \(T\) is onto.
For \(\mathbf{b}_1\), the system is
Taking \(x=2\) and \(y=0\) gives a solution. Substitution checks it: \(\mathbf{A}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}2\\0\end{bmatrix}\). But for \(\mathbf{b}_2=(0,1)\), the system includes the equation \(0=1\), so it has no solution. Thus \(T\) reaches \(\mathbf{b}_1\) but is not onto. In RREF, the second row has no pivot, agreeing with the pivot-row criterion.
A Reliable Diagnostic Routine
When checking a standard-matrix or pivot argument, first identify the map’s domain and codomain. This fixes the matrix size: a map from \(\mathbb{R}^n\) to \(\mathbb{R}^m\) has \(n\) columns and \(m\) rows. Then match the claim to the correct evidence. The basis-image formula checks matrix construction; the order of actions checks composition; pivot columns test one-to-one behavior; pivot rows test being onto. To establish onto directly, the claim must cover every target, not just a convenient example.
For a standard matrix, column \(j\) must be \(T(\mathbf{e}_j)\).
In \(S\circ T\), the matrix of \(T\) is the right-hand factor.
Every column pivot tests one-to-one behavior; every row pivot tests onto behavior.
A solution for one right-hand side is not a solution for every right-hand side.
A useful safeguard is to verify a matrix on the standard basis vectors: the products with \(\mathbf{e}_j\) are exactly its columns. For composition, test the order on a basis vector if the products seem easy to reverse. For pivot claims, write down the matrix dimensions before counting pivots. These checks expose mismatches between the statement being made and the calculation offered to support it.
Check Your Understanding
For each question, identify which calculation or criterion addresses the claim.
- If \(T(\mathbf{e}_1)=(2,1)\) and \(T(\mathbf{e}_2)=(0,3)\), write the standard matrix of \(T:\mathbb{R}^2\to\mathbb{R}^2\).
- If \(T\) has matrix \(\mathbf{A}\) and \(S\) has matrix \(\mathbf{B}\), which product represents \(S\circ T\)? Explain the order.
- For a \(3\times2\) matrix in RREF with a pivot in each column but not each row, what can you conclude about one-to-one behavior and being onto?
- Why does finding a solution to \(\mathbf{A}\mathbf{x}=\mathbf{b}\) for one chosen \(\mathbf{b}\) not prove that the map is onto?
- Give one direct check that can reveal whether proposed columns really are the standard matrix of a map.
