Two fundamental properties of transformations are whether they are one-to-one (injective) and onto (surjective). For linear transformations, these properties can be determined from the pivot structure of the standard matrix.

One-to-One Transformations

Definition (One-to-One):
A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is one-to-one (or injective) if: $T(\vec{x}_1) = T(\vec{x}_2) \implies \vec{x}_1 = \vec{x}_2$

Equivalently: Different inputs produce different outputs.

Geometric intuition: No two distinct vectors map to the same output—the transformation doesn't "collapse" any information.

Example 1: One-to-One Transformation

Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be defined by $T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \\ x + y \end{bmatrix}$ .

Standard matrix: $A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{bmatrix}$

Is this one-to-one? Suppose $T(\vec{x}_1) = T(\vec{x}_2)$ : $\begin{bmatrix} x_1 \\ y_1 \\ x_1 + y_1 \end{bmatrix} = \begin{bmatrix} x_2 \\ y_2 \\ x_2 + y_2 \end{bmatrix}$

From first two components: $x_1 = x_2$ and $y_1 = y_2$ . Thus $\vec{x}_1 = \vec{x}_2$ .

Answer: Yes, $T$ is one-to-one.

Example 2: Not One-to-One

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be projection onto the $x$ -axis: $T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix}$ .

Standard matrix: $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

Is this one-to-one? Consider: $T\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = T\begin{bmatrix} 1 \\ 5 \end{bmatrix}$

Different inputs give the same output!

Answer: No, $T$ is not one-to-one.

One-to-One Test via Homogeneous Equations

Theorem (One-to-One Test):
A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ with standard matrix $A$ is one-to-one if and only if: $A\vec{x} = \vec{0} \text{ has only the trivial solution}$

Proof:

( $\Rightarrow$ ) Suppose $T$ is one-to-one. Since $T(\vec{0}) = \vec{0}$ , and $T$ is one-to-one, any $\vec{x}$ with $T(\vec{x}) = \vec{0}$ must equal $\vec{0}$ .
( $\Leftarrow$ ) Suppose $T(\vec{x}_1) = T(\vec{x}_2)$ . Then $T(\vec{x}_1 - \vec{x}_2) = T(\vec{x}_1) - T(\vec{x}_2) = \vec{0}$ . By assumption, $\vec{x}_1 - \vec{x}_2 = \vec{0}$ , so $\vec{x}_1 = \vec{x}_2$ .

The columns of $A$ are linearly independent if and only if $A\vec{x} = \vec{0}$ has only the trivial solution.

Therefore: $T$ is one-to-one $\Leftrightarrow$ columns of $A$ are linearly independent.

This makes sense geometrically: if the columns are dependent, multiple input combinations produce the same output.

One-to-One and Pivot Columns

Theorem (Pivot Column Criterion for One-to-One):
Let $T: \mathbb{R}^n \to \mathbb{R}^m$ have standard matrix $A$ . Then $T$ is one-to-one if and only if: Every column of $A$ is a pivot column.

Why?

Every column is pivot $\Leftrightarrow$ No free variables in $A\vec{x} = \vec{0}$
No free variables $\Leftrightarrow$ Only trivial solution
Only trivial solution $\Leftrightarrow$ $T$ is one-to-one

Example 3: Testing One-to-One

Determine if $T: \mathbb{R}^3 \to \mathbb{R}^3$ with standard matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix}$ is one-to-one.

Solution: Row reduce to find pivot columns (already in echelon form):

Pivot columns: 1 and 2
Column 3 is not a pivot column (free variable)

Since not every column is a pivot column, $T$ is not one-to-one.

Verification: The homogeneous system $A\vec{x} = \vec{0}$ has nontrivial solutions (column 3 is free).

Example 4: One-to-One Transformation

Is $T: \mathbb{R}^2 \to \mathbb{R}^3$ with $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$ one-to-one?

Testing One-to-One

Step 0 of 3

Both columns are pivot columns!

Answer: Yes, $T$ is one-to-one.

Onto Transformations

Definition (Onto):
A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is onto (or surjective) if: For every $\vec{b}$ in $\mathbb{R}^m$ , there exists $\vec{x}$ in $\mathbb{R}^n$ such that $T(\vec{x}) = \vec{b}$ .

Equivalently: The range (image) of $T$ is all of $\mathbb{R}^m$ .

Geometric intuition: Every possible output is achieved—the transformation "reaches" all of the target space.

Example 5: Onto Transformation

Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be defined by $T\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + y \\ y + z \end{bmatrix}$ .

Is this onto? For any $\vec{b} = \begin{bmatrix} a \\ b \end{bmatrix}$ , can we find $\vec{x}$ ?

Choose $x = a, y = 0, z = b$ : $T\begin{bmatrix} a \\ 0 \\ b \end{bmatrix} = \begin{bmatrix} a + 0 \\ 0 + b \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}$

For any target $\begin{bmatrix} a \\ b \end{bmatrix}$ , we found an input!

Answer: Yes, $T$ is onto.

Example 6: Not Onto

Let $T: \mathbb{R}^2 \to \mathbb{R}^3$ be $T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ y \\ 0 \end{bmatrix}$ .

Is this onto? Can we reach $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ ?

We'd need: $\begin{bmatrix} x \\ y \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

But the third component is always 0, never 1!

Answer: No, $T$ is not onto. The range is only the $xy$ -plane in $\mathbb{R}^3$ .

Onto Test via Span

Theorem (Onto Test):
A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ with standard matrix $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \cdots & \vec{a}_n \end{bmatrix}$ is onto if and only if: $\text{Span}\{\vec{a}_1, \vec{a}_2, \ldots, \vec{a}_n\} = \mathbb{R}^m$

Equivalently: The columns of $A$ span $\mathbb{R}^m$ .

Why? For any $\vec{b} \in \mathbb{R}^m$ , we need $\vec{b} = A\vec{x}$ for some $\vec{x}$ . But $A\vec{x}$ is a linear combination of the columns, so this is possible if and only if $\vec{b}$ is in the span of the columns.

Onto and Pivot Rows

Theorem (Pivot Row Criterion for Onto):
Let $T: \mathbb{R}^n \to \mathbb{R}^m$ have standard matrix $A$ . Then $T$ is onto if and only if: Every row of $A$ has a pivot (equivalently: $A$ has $m$ pivot rows).

Why?

$T$ onto $\Leftrightarrow$ $A\vec{x} = \vec{b}$ is consistent for all $\vec{b}$
Consistent for all $\vec{b}$ $\Leftrightarrow$ No pivot in augmented column for any $\vec{b}$
This happens $\Leftrightarrow$ Every row has a pivot in the coefficient matrix

Example 7: Testing Onto

Determine if $T: \mathbb{R}^3 \to \mathbb{R}^2$ with $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \end{bmatrix}$ is onto.

Solution: Count pivot rows (already in echelon form):

Row 1 has pivot
Row 2 has pivot
Total: 2 pivot rows = $m$ (number of rows)

Answer: Yes, $T$ is onto.

Example 8: Not Onto

Is $T: \mathbb{R}^2 \to \mathbb{R}^3$ with $A = \begin{bmatrix} 1 & 2 \\ 3 & 6 \\ 1 & 2 \end{bmatrix}$ onto?

Testing Onto

Step 0 of 2

Only 1 pivot row, but $m = 3$ rows.

Answer: No, $T$ is not onto. The range is only 1-dimensional (a line in $\mathbb{R}^3$ ).

Summary Table

Practice Problems

Determine if $T: \mathbb{R}^3 \to \mathbb{R}^2$ with standard matrix $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ is one-to-one.

Check if every column is a pivot column.

The matrix is already in RREF:

Pivot columns: 1 and 2
Column 3 is not a pivot column (no pivot in that column)

Not every column is a pivot column.

Answer: No, $T$ is not one-to-one.

Verification: $A\vec{x} = \vec{0}$ has nontrivial solutions (with $x_3$ free).

Is $T: \mathbb{R}^3 \to \mathbb{R}^2$ with $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix}$ onto?

Check if every row has a pivot.

Row 1 has pivot (in column 1)
Row 2 has pivot (in column 2)
Total pivot rows: 2 = $m$ (number of rows)

Every row has a pivot!

Answer: Yes, $T$ is onto.

Can a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ be both one-to-one and onto?

Think about the number of pivot columns and pivot rows needed.

For $T: \mathbb{R}^2 \to \mathbb{R}^3$ :

One-to-one requires: 2 pivot columns (every column a pivot)
Onto requires: 3 pivot rows (every row has a pivot)

But a $3 \times 2$ matrix can have at most 2 pivot columns, which means at most 2 pivot rows (not 3).

Answer: No, it's impossible for $T: \mathbb{R}^2 \to \mathbb{R}^3$ to be both one-to-one and onto.

General principle: For $T: \mathbb{R}^n \to \mathbb{R}^m$ to be both one-to-one and onto, we need $n = m$ .

Give an example of a linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ that is:

One-to-one but not onto
Onto but not one-to-one
Neither one-to-one nor onto

Use the pivot criteria to construct appropriate matrices.

1. One-to-one but not onto: Impossible for $T: \mathbb{R}^3 \to \mathbb{R}^2$ !

One-to-one needs 3 pivot columns
A $2 \times 3$ matrix can have at most 2 pivot rows
If there are 3 pivot columns, there must be at least 3 pivots, but we only have 2 rows
So we can't have 3 pivot columns in a $2 \times 3$ matrix

2. Onto but not one-to-one: $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix}$

2 pivot rows (onto ✓)
Only 2 pivot columns out of 3 (not one-to-one ✓)

3. Neither: $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{bmatrix}$ Row reduces to: $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{bmatrix}$

Only 1 pivot row (not onto ✗)
Only 1 pivot column (not one-to-one ✗)

Summary

One-to-one (injective): Different inputs → different outputs
- Test: $A\vec{x} = \vec{0}$ has only trivial solution
- Criterion: Every column is a pivot column
- Equivalent: Columns are linearly independent
Onto (surjective): Every output is achieved
- Test: Columns span $\mathbb{R}^m$
- Criterion: Every row has a pivot
- Equivalent: $A\vec{x} = \vec{b}$ consistent for all $\vec{b}$
For $T: \mathbb{R}^n \to \mathbb{R}^m$ :
- One-to-one $\Rightarrow$ $n \leq m$ (at least as many outputs as inputs)
- Onto $\Rightarrow$ $n \geq m$ (at least as many inputs as outputs)
- Both $\Rightarrow$ $n = m$ (must be square matrix, though not all square matrices are both)

One-to-One and Onto Transformations