Two Different Questions About a Transformation
A linear transformation can preserve enough information to distinguish every input without reaching every possible output. It can also reach every output while sending different inputs to the same place. These are separate properties: one-to-one concerns whether outputs identify their inputs, while onto concerns whether every vector in the stated codomain is reached. Rectangular matrices make the distinction especially visible because the input and output spaces may have different dimensions.
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be a linear transformation with standard matrix \(\mathbf{A}\in\mathbb{R}^{m\times n}\), so \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\). The dimensions matter: the codomain is \(\mathbb{R}^m\), whether or not every vector in it is an output. The set of outputs is the column space \(\operatorname{Col}\mathbf{A}\), the span of the columns of \(\mathbf{A}\).
For a matrix transformation, being one-to-one means no two different inputs produce the same output. Subtracting the inputs and using linearity shows that this is equivalent to \(\mathbf{A}\mathbf{x}=\mathbf{0}\) having only the solution \(\mathbf{x}=\mathbf{0}\). Being onto means that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) has a solution for every \(\mathbf{b}\) in the codomain. In other words, the columns of \(\mathbf{A}\) must span \(\mathbb{R}^m\).
A Rectangular Map That Is One-to-One but Not Onto
Worked Example: An Injective Map from R² to R³
Consider \(T:\mathbb{R}^2\to\mathbb{R}^3\) with standard matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}\). Determine which property holds.
For an input \(\mathbf{x}=\begin{bmatrix}s\\t\end{bmatrix}\), matrix multiplication gives
To test whether \(T\) is one-to-one, solve the homogeneous equation:
The only input mapped to the zero vector is \(\begin{bmatrix}0\\0\end{bmatrix}\), so \(T\) is one-to-one. To test whether it is onto, take a target \(\begin{bmatrix}a\\b\\c\end{bmatrix}\). Every output has third coordinate \(0\), so the target \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) has no preimage. Therefore \(T\) is not onto. Indeed, its image is the plane consisting of vectors whose third coordinate is zero.
Check: For every \(\begin{bmatrix}s\\t\end{bmatrix}\), multiplying gives \(\begin{bmatrix}s\\t\\0\end{bmatrix}\), as claimed. The output \(\begin{bmatrix}0\\0\\1\end{bmatrix}\) cannot equal this because its third coordinate is \(1\), not \(0\).
A Rectangular Map That Is Onto but Not One-to-One
Worked Example: A Surjective Map from R³ to R²
Let \(S:\mathbb{R}^3\to\mathbb{R}^2\) have standard matrix \(\mathbf{B}=\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}\). Determine which property holds.
For a target \(\begin{bmatrix}a\\b\end{bmatrix}\in\mathbb{R}^2\), choose the input \(\begin{bmatrix}a\\b\\0\end{bmatrix}\). Then
This construction supplies a preimage for every target, so \(S\) is onto. To test whether it is one-to-one, solve \(\mathbf{B}\mathbf{x}=\mathbf{0}\): the equations are \(x+z=0\) and \(y+z=0\). Setting \(z=1\) gives \(x=-1\), \(y=-1\), so \(\begin{bmatrix}-1\\-1\\1\end{bmatrix}\) is a nonzero input mapped to zero. Thus \(S\) is not one-to-one.
Check: Multiplying that nonzero input back gives
Since \(S(\mathbf{0})=\mathbf{0}\) as well, these are two distinct inputs with the same output.
These examples show why it is important to name the codomain when asking whether a map is onto. The first transformation reaches every point in a plane inside \(\mathbb{R}^3\), but it is not onto \(\mathbb{R}^3\). The second reaches every point in \(\mathbb{R}^2\), though some of those outputs have more than one preimage.
Why the Properties Coincide in the Square Case
For a general \(m\times n\) matrix, the two tests ask for pivots in different directions. Injectivity requires a pivot in every column; surjectivity requires a pivot in every row. When the matrix is square, it has the same number of rows and columns, so these conditions become equivalent. The row-reduction criteria below also explain why the distinction in the rectangular examples is possible.
Proof. Let \(\mathbf{R}\) be the RREF of \(\mathbf{A}\). Row operations preserve the solutions of a homogeneous system, so \(\mathbf{A}\mathbf{x}=\mathbf{0}\) and \(\mathbf{R}\mathbf{x}=\mathbf{0}\) have the same solutions. In the reduced system, a column without a pivot corresponds to a free variable. If every column has a pivot, there are no free variables, and the homogeneous system has only its zero solution. If a column lacks a pivot, set one free variable to \(1\), set the other free variables to \(0\), and solve for the pivot variables. This gives a nonzero solution. Thus the kernel contains only zero exactly when every column has a pivot, proving the one-to-one criterion.
For the onto criterion, row operations amount to multiplying \(\mathbf{A}\) on the left by an invertible matrix \(\mathbf{E}\), so the equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) is equivalent to \(\mathbf{R}\mathbf{x}=\mathbf{E}\mathbf{b}\). If every row of \(\mathbf{R}\) has a pivot, each system \(\mathbf{R}\mathbf{x}=\mathbf{E}\mathbf{b}\) is consistent: assign the free variables zero and solve for the pivot variables. Thus every \(\mathbf{b}\) has a preimage. If some row has no pivot, that row of the RREF is zero. Choose \(\mathbf{b}=\mathbf{E}^{-1}\mathbf{e}_i\), where \(i\) is the index of a zero row and \(\mathbf{e}_i\) is the corresponding standard basis vector in \(\mathbb{R}^m\). Then \(\mathbf{E}\mathbf{b}=\mathbf{e}_i\), making the zero row of \(\mathbf{R}\mathbf{x}=\mathbf{E}\mathbf{b}\) the inconsistent equation \(0=1\). So \(T\) is not onto. This proves both directions of the onto criterion. \(\square\)
Proof. In the RREF of an \(n\times n\) matrix, having a pivot in every column means there are \(n\) pivots. Each pivot is in a different row, and there are \(n\) rows, so there is a pivot in every row as well. Conversely, \(n\) pivots in every row must occupy \(n\) different columns, so there is a pivot in every column. By the theorem, having pivots in every column is equivalent to being one-to-one, and having pivots in every row is equivalent to being onto. Hence the two properties are equivalent. \(\square\)
Bijective Square Matrices and Inverses
A square matrix \(\mathbf{A}\) is called invertible if there is a square matrix \(\mathbf{B}\) such that \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}=\mathbf{I}_n\). For a square matrix, this is exactly the situation in which the associated transformation is both one-to-one and onto. The inverse matrix represents the rule that recovers the unique input from each output.
Worked Example: A Square Invertible Transformation
Consider \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). Verify that \(\mathbf{B}\) is an inverse and interpret the transformation.
Multiplying in both orders gives
The transformation sends \(\begin{bmatrix}x\\y\end{bmatrix}\) to \(\begin{bmatrix}x+y\\y\end{bmatrix}\). Given any target \(\begin{bmatrix}a\\b\end{bmatrix}\), the input \(\mathbf{B}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}a-b\\b\end{bmatrix}\) maps back to that target. This input is unique because multiplying an equation \(\mathbf{A}\mathbf{x}=\mathbf{b}\) by \(\mathbf{B}\) forces \(\mathbf{x}=\mathbf{B}\mathbf{b}\). Thus the map is onto and one-to-one.
Check: Substituting the proposed input gives \(\mathbf{A}\begin{bmatrix}a-b\\b\end{bmatrix}=\begin{bmatrix}a-b+b\\b\end{bmatrix}=\begin{bmatrix}a\\b\end{bmatrix}\), and the two matrix products above verify the inverse in both orders.
The link between invertibility and the two properties can be established directly, without using determinants.
Proof. Suppose first that \(\mathbf{A}\) is invertible with inverse \(\mathbf{B}\). If \(\mathbf{A}\mathbf{x}_1=\mathbf{A}\mathbf{x}_2\), multiply both sides by \(\mathbf{B}\) to obtain \(\mathbf{x}_1=\mathbf{x}_2\), so \(T\) is one-to-one. For any \(\mathbf{b}\in\mathbb{R}^n\), choose \(\mathbf{x}=\mathbf{B}\mathbf{b}\). Then \(\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{B}\mathbf{b}=\mathbf{b}\), so \(T\) is onto.
Conversely, suppose \(T\) is bijective. For each standard basis vector \(\mathbf{e}_j\), surjectivity gives an input \(\mathbf{x}_j\) with \(\mathbf{A}\mathbf{x}_j=\mathbf{e}_j\). Form the matrix \(\mathbf{B}\) whose columns are \(\mathbf{x}_1,\dots,\mathbf{x}_n\). The \(j\)th column of \(\mathbf{A}\mathbf{B}\) is \(\mathbf{A}\mathbf{x}_j=\mathbf{e}_j\), so \(\mathbf{A}\mathbf{B}=\mathbf{I}_n\). For any \(\mathbf{x}\in\mathbb{R}^n\), this implies \(\mathbf{A}(\mathbf{B}\mathbf{A}\mathbf{x})=(\mathbf{A}\mathbf{B})\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{x}\). Since \(T\) is one-to-one, the inputs \(\mathbf{B}\mathbf{A}\mathbf{x}\) and \(\mathbf{x}\) must be equal. This holds for every \(\mathbf{x}\), so \(\mathbf{B}\mathbf{A}=\mathbf{I}_n\). Therefore \(\mathbf{B}\) is an inverse of \(\mathbf{A}\). \(\square\)
Common Pitfall: Treating One Property as the Other
A frequent mistake is to assume that a transformation is onto just because it is one-to-one, or vice versa. For example, the square transformation \(\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}x\\0\end{bmatrix}\) is not one-to-one: both \(\begin{bmatrix}0\\0\end{bmatrix}\) and \(\begin{bmatrix}0\\1\end{bmatrix}\) map to \(\begin{bmatrix}0\\0\end{bmatrix}\). It is also not onto \(\mathbb{R}^2\), because \(\begin{bmatrix}0\\1\end{bmatrix}\) is not an output. The false calculation “the matrix has two columns, so it must be onto \(\mathbb{R}^2\)” confuses the number of columns with whether the columns span the codomain. In fact, its matrix \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\) has no pivot in its second row or second column.
Use the domain and codomain, then ask the appropriate question: injectivity asks whether two inputs can share an output; surjectivity asks whether a specified target has a preimage. For a square matrix the answers must agree, but for a rectangular matrix they need not.
| Property | Matrix test | Question to ask |
|---|---|---|
| One-to-one | Pivot in every column | Does \(\mathbf{A}\mathbf{x}=\mathbf{0}\) have only the zero solution? |
| Onto | Pivot in every row | Can every vector in the codomain be reached? |
| Invertible, for a square matrix | Pivots in every row and column | Can each output be reached from exactly one input? |
Check Your Understanding
For each matrix transformation, identify whether it is one-to-one, onto, both, or neither. Support your answer with the relevant system or a preimage argument.
- For \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), decide whether \(T:\mathbb{R}^2\to\mathbb{R}^3\) is one-to-one and whether it is onto.
- For \(\mathbf{B}=\begin{bmatrix}1&2&0\\0&1&1\end{bmatrix}\), find a nonzero vector in \(\operatorname{Nul}\mathbf{B}\), or show that none exists, and test whether \(\mathbf{B}\) is onto \(\mathbb{R}^2\).
- Explain why a pivot in every column is the test for one-to-one, while a pivot in every row is the test for onto.
- Prove that if a square matrix transformation has an inverse matrix, then it is both one-to-one and onto.
- Why does the statement “one-to-one if and only if onto” require a square matrix in this discussion? Give one of the rectangular examples as evidence.
