Dimensions Set Different Pivot Requirements
An \(m\times n\) matrix \(\mathbf{A}\) defines a transformation \(T_{\mathbf{A}}:\mathbb{R}^n\to\mathbb{R}^m\) by \(T_{\mathbf{A}}(\mathbf{x})=\mathbf{A}\mathbf{x}\). The number of columns, \(n\), is the number of input coordinates; the number of rows, \(m\), is the number of output coordinates. For a rectangular matrix these numbers differ, so the pivot requirements for being injective and surjective may not both be possible.
A transformation is injective if different inputs always have different outputs: whenever \(T_{\mathbf{A}}(\mathbf{x})=T_{\mathbf{A}}(\mathbf{z})\), we have \(\mathbf{x}=\mathbf{z}\). It is surjective if every vector in the codomain is an output: for every \(\mathbf{y}\in\mathbb{R}^m\), there is some \(\mathbf{x}\in\mathbb{R}^n\) such that \(\mathbf{A}\mathbf{x}=\mathbf{y}\).
The criteria ask for different things. Injectivity requires a pivot for each input coordinate, so it counts columns. Surjectivity requires a pivot for each output coordinate, so it counts rows. In either case, a pivot is a leading entry in the reduced row echelon form, and no two pivots occupy the same row or the same column.
Why Rectangular Dimensions Rule Out One Property
Proof. Write \(r\) for the number of pivots in \(\operatorname{rref}(\mathbf{A})\). Each pivot uses a distinct row and a distinct column, so \(r\le m\) and \(r\le n\). If \(m<n\), injectivity would require a pivot in every one of the \(n\) columns, which would require \(r=n>m\); this contradicts \(r\le m\). Surjectivity in this case requires a pivot in each of the \(m\) rows, exactly as the pivot-row criterion states, and the number of columns does not rule out that requirement. If \(m>n\), surjectivity would require \(r=m>n\), contradicting \(r\le n\). Injectivity requires pivots in all \(n\) columns, which is possible and is exactly the pivot-column criterion. \(\square\)
This proposition gives a limitation, not a guarantee. For example, when \(m<n\), the dimensions do not prevent surjectivity, but a particular matrix may still fail to have a pivot in every row. Likewise, when \(m>n\), the dimensions permit injectivity, but the matrix must actually have a pivot in every column.
Proof. If both properties hold, the pivot-column criterion gives \(n\) pivots, while the pivot-row criterion gives \(m\) pivots. When \(m=n\), these conditions are satisfied exactly when every row and every column has a pivot. Conversely, if \(m=n\) and every row and column has a pivot, the two pivot criteria give surjectivity and injectivity, respectively. \(\square\)
A Wide Matrix Can Be Onto
Worked Example: A Surjective Transformation from R³ to R²
Consider \(\mathbf{A}\in\mathbb{R}^{2\times3}\) and \(T_{\mathbf{A}}:\mathbb{R}^3\to\mathbb{R}^2\), where
The matrix is already in reduced row echelon form. There is a pivot in each of its two rows, so the pivot-row criterion says that \(T_{\mathbf{A}}\) is surjective. There is no pivot in column 3, so it is not injective.
We can check surjectivity directly. For any output \(\mathbf{y}=(a,b)^{\top}\in\mathbb{R}^2\), choose \(\mathbf{x}=(a,b,0)^{\top}\). Then
To check the failure of injectivity, use the nonzero input \(\mathbf{x}=(1,1,-1)^{\top}\):
Thus two distinct inputs, \(\mathbf{0}\) and \((1,1,-1)^{\top}\), produce the same output. This agrees with the dimension restriction: since \(3>2\), this transformation cannot be injective.
A Tall Matrix Can Be Injective
Worked Example: An Injective Transformation from R² to R³
Consider \(T_{\mathbf{B}}:\mathbb{R}^2\to\mathbb{R}^3\), with
Subtract \(R_1\) from \(R_3\), then subtract \(R_2\) from the new \(R_3\):
The reduced matrix has a pivot in each of its two columns, so \(T_{\mathbf{B}}\) is injective. It has no pivot in row 3, so it is not surjective. Directly, if \(\mathbf{x}=(s,t)^{\top}\), then
If \(\mathbf{B}\mathbf{x}=\mathbf{0}\), the first two coordinates give \(s=0\) and \(t=0\); hence the only input that maps to zero is \(\mathbf{0}\), confirming injectivity. But \((0,0,1)^{\top}\in\mathbb{R}^3\) is not an output: matching the first two coordinates would force \(s=t=0\), and then the third coordinate \(s+t\) would be \(0\), not \(1\). So the transformation is not surjective.
When the Matrix Does Not Meet Either Requirement
Worked Example: Neither Injective Nor Surjective
Consider \(T_{\mathbf{C}}:\mathbb{R}^3\to\mathbb{R}^2\), with
Subtract \(2R_1\) from \(R_2\), then divide the first row by \(1\) (leaving it unchanged):
There is only one pivot. Column 3 has no pivot, so the map is not injective; row 2 has no pivot, so it is not surjective. Both failures can also be verified directly. For \(\mathbf{x}=(x_1,x_2,x_3)^{\top}\),
The nonzero input \((-2,1,0)^{\top}\) maps to \((0,0)^{\top}\), because \(-2+2=0\) and \(2(-2)+4(1)=0\). This confirms the failure of injectivity. Every output has the form \((q,2q)^{\top}\), since the second coordinate is twice the first. In particular, \((0,1)^{\top}\) is not an output, confirming the failure of surjectivity.
A Reliable Way to Classify the Transformation
An \(m\times n\) matrix maps inputs from \(\mathbb{R}^n\) to outputs in \(\mathbb{R}^m\). Compare the number of rows and columns before doing any calculations.
Record the pivot rows and pivot columns in its RREF. Row reduction identifies the pivot requirements; the entries of the RREF should not be mistaken for a new matrix defining the same transformation.
The map is injective exactly when every column has a pivot. If \(m<n\), this requirement is impossible, because there are more columns than rows.
The map is surjective exactly when every row has a pivot. If \(m>n\), this requirement is impossible, because there are more rows than columns.
These tests keep the roles of input and output dimensions visible. A missing pivot column signals that some nonzero input can be lost, so injectivity fails. A missing pivot row signals that some output coordinate cannot be reached in the required way, so surjectivity fails. When dimensions are square, both tests ask for a pivot in every row and every column; with a rectangular matrix, only one of the two requirements can even be possible.
Common Pitfall: Counting the Wrong Pivots
A common mistake is to say, “There are three columns and only two rows, so this map cannot be onto.” That conclusion applies the injectivity requirement—pivots in all columns—to a question about surjectivity, which requires pivots in all rows. The \(2\times3\) matrix from the first example is a counterexample to the claim: for every \((a,b)^{\top}\in\mathbb{R}^2\), the input \((a,b,0)^{\top}\) satisfies \(\mathbf{A}(a,b,0)^{\top}=(a,b)^{\top}\). Thus that map is onto despite having more columns than rows.
The converse counting error is also worth avoiding. When \(m>n\), having fewer columns than rows rules out a pivot in every row, and therefore rules out surjectivity. It does not, by itself, rule out injectivity. In the tall-matrix example, both columns have pivots, so injectivity holds even though one row has none.
Check Your Understanding
For each question, distinguish the pivot-column condition from the pivot-row condition and verify any claimed input or output by multiplication.
- For a \(3\times5\) matrix, which of injectivity and surjectivity is ruled out by the dimensions? State the pivot condition for the other property.
- For a \(4\times2\) matrix, which property is impossible? What must its RREF show for the other property to hold?
- Can a \(2\times4\) matrix be injective? Explain using the number of possible pivots.
- Give a nonzero input that maps to zero under \(\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}\), and verify the multiplication.
- For \(\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\), identify the missing pivot row and give an output in \(\mathbb{R}^3\) that is not attained.
