Image, kernel, and linear isomorphism

Let $T:X\to Y$ be a linear operator between vector spaces.

The image of $T$ is $\operatorname{im}T:=T(X)=\{T(x)\mid x\in X\}\subset Y.$

The kernel of $T$ is $\ker T:=T^{-1}(\{0\})=\{x\in X\mid T(x)=0\}\subset X.$

If $T$ is bijective (one-to-one and onto), then $T$ is a linear isomorphism, and we say that $X$ and $Y$ are isomorphic, written $X\cong Y$ .

Remarks

Both $\ker T$ and $\operatorname{im}T$ are linear subspaces (see images and preimages of subspaces). The quotient $X/\ker T$ is canonically isomorphic to $\operatorname{im}T$ (see the isomorphism theorem).

Examples

If $T:\mathbb{R}^2\to\mathbb{R}$ is $T(x,y)=x+y$ , then $\ker T=\{(t,-t):t\in\mathbb{R}\}$ and $\operatorname{im}T=\mathbb{R}$ .
If $T=\mathrm{id}_X$ , then $\ker T=\{0\}$ and $\operatorname{im}T=X$ ; $T$ is an isomorphism.