TheoremBase

Kernel and Image of a Linear Map Are Subspaces

The kernel and the image of a linear map are subspaces; the map is injective exactly when its kernel is zero, and surjective exactly when its image is the whole codomain.

Statement

In the setting of Linear Algebra: Standing Notation, let KK be a field, let VV and WW be vector spaces over KK, and let T∈L(V,W)T\in\mathcal{L}(V,W).

Linear combinations. T0=0T0=0 and T(−v)=−TvT(-v)=-Tv for every v∈Vv\in V; and for n∈N0n\in\mathbb{N}_{0}, λ1,…,λn∈K\lambda_{1},\dots,\lambda_{n}\in K and v1,…,vn∈Vv_{1},\dots,v_{n}\in V,

T(∑i=1nλivi)=∑i=1nλi Tvi.T\Bigl(\sum_{i=1}^{n}\lambda_{i}v_{i}\Bigr)=\sum_{i=1}^{n}\lambda_{i}\,Tv_{i}.

Subspaces. ker⁡T\ker T is a linear subspace of VV, and im⁡T\operatorname{im}T is a linear subspace of WW.

Injectivity. TT is injective if and only if ker⁡T={0}\ker T=\{0\}.

Surjectivity. TT is surjective if and only if im⁡T=W\operatorname{im}T=W.

Proofs

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