Kernel and image

Both sets are subspaces: the kernel lives in the domain and the image in the codomain. If \(T(u)=0\) and \(T(v)=0\), then \(T(au+bv)=0\); and a combination of values \(T(u)\) and \(T(v)\) is the value \(T(au+bv)\). Hence both always contain the zero vector, since \(T(0)=0\).

When \(T(x)=Ax\), the kernel is the set of solutions of \(Ax=0\) and the image is the set of linear combinations of the columns of \(A\). Example: for \(T(x,y,z)=(x+y,\,y+z)\), solving \(x+y=0\), \(y+z=0\) gives \(x=z=-y\), so the kernel is the line spanned by \((1,-1,1)\); the columns \((1,0)\), \((1,1)\) and \((0,1)\) span all of \(\mathbb{R}^2\), so the image is \(\mathbb{R}^2\).

The kernel detects injectivity: \(T\) is injective exactly when its kernel is \(\{0\}\), because \(T(u)=T(v)\) is equivalent to \(u-v\in\ker T\). The image detects surjectivity: \(T\) is surjective when the image is the whole codomain.

Typical mistake: believing the kernel is empty when \(T\) is injective. It is never empty, since it always contains zero; injective means it contains only zero. The dimensions of the kernel and the image are tied together precisely, and that is exactly the rank–nullity theorem.