Surjective, Injective, Bijective

61. Surjective, Injective, Bijective

Three properties a function $𝑇 : 𝑉 \to 𝑊$ might (or might not) have, with clean characterizations for linear transformations.

Different inputs produce different outputs — no two distinct inputs collide:

{\vec{𝑥}}_{1} \neq {\vec{𝑥}}_{2} \Rightarrow 𝑇 ({\vec{𝑥}}_{1}) \neq 𝑇 ({\vec{𝑥}}_{2})

For a linear $𝑇$ , equivalently:

ker (𝑇) = {𝟎}

(See Kernel — only the zero vector maps to zero.)

Every element of the codomain is hit by some input:

\forall \vec{𝑦} \in 𝑊, \exists \vec{𝑥} \in 𝑉 with 𝑇 (\vec{𝑥}) = \vec{𝑦}

For a linear $𝑇 : ℝ^{𝑛} \to ℝ^{𝑚}$ , equivalently:

im (𝑇) = ℝ^{𝑚}, i.e. rank (𝐴) = 𝑚

(See Image / Rank.)

Both injective and surjective. Each output has exactly one preimage. Bijective linear transformations are invertible.

For a linear $𝑇 : ℝ^{𝑛} \to ℝ^{𝑚}$ , bijective requires $𝑚 = 𝑛$ (square matrix) and $𝐴$ is invertible.

Property	Equivalent (rank)	Equivalent (kernel/image)	Possible only if
Injective	$rank (𝐴) = 𝑛$	$ker (𝑇) = {𝟎}$	$𝑚 \geq 𝑛$ (full column rank)
Surjective	$rank (𝐴) = 𝑚$	$im (𝑇) = ℝ^{𝑚}$	$𝑛 \geq 𝑚$ (full row rank)
Bijective	$rank (𝐴) = 𝑚 = 𝑛$	both above	$𝑚 = 𝑛$ and $𝐴$ invertible

Example

$𝐴 = [\begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix}]$ — $𝑇 : ℝ^{2} \to ℝ^{3}$ .

A linear $𝑇$ has an inverse iff it’s bijective iff $𝐴$ is square and invertible.

When invertible, $𝑇^{- 1}$ corresponds to $𝐴^{- 1}$ (see Matrix Inverse).