Two natural questions about a linear map $T: \mathbb{R}^n \to \mathbb{R}^m$: does it ever send two different inputs to the same output, and does it reach every vector in $\mathbb{R}^m$? Both answers come from the pivots, just like existence and uniqueness for systems.
$T$ is one-to-one if each $\mathbf{b}$ in $\mathbb{R}^m$ is the image of at most one $\mathbf{x}$. $T$ is onto $\mathbb{R}^m$ if each $\mathbf{b}$ is the image of at least one $\mathbf{x}$.
Key idea
Why it works
Worked example: A map into a bigger space
- Row reduce: $R_3 \to R_3 - 2R_1$ gives $(0, -1)$; $R_3 \to R_3 + R_2$ gives $(0, 0)$. Echelon form: $\begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}$.
- Pivots in both columns: one-to-one.
- Row 3 has no pivot: not onto. The range is a plane in $\mathbb{R}^3$ (spanned by the two columns).
Answer: One-to-one, not onto.
Worked example: A map into a smaller space
- Standard matrix: $\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{bmatrix}$, already in echelon form.
- Pivots in both rows: onto.
- Column 3 has no pivot: not one-to-one. For example, $T(1, -2, 1)$ $= (0, 0)$ $= T(0, 0, 0)$.
Answer: Onto, not one-to-one.
| Size of A | One-to-one possible? | Onto possible? |
|---|---|---|
| $m > n$ (tall) | Yes | No: at most $n < m$ pivots |
| $m < n$ (wide) | No: at most $m < n$ pivots | Yes |
| $m = n$ (square) | Yes | Yes, and then both hold together |
A practical way to think about it: a one-to-one map loses no information, since you can always tell which input produced an output. An onto map leaves nothing out, since every possible output really occurs. A map from a smaller space into a bigger one can be one-to-one but never onto; a map from a bigger space onto a smaller one can be onto but never one-to-one, because it must merge some inputs.