Exercise 1207Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Basis of a kernel

#kernel#basis#linear map

Statement

Let $f:\mathbb{R}^3\to\mathbb{R}$ be the linear form $f(x,y,z)=x+2y+3z$ . Find a basis of $\ker f$ .

Hint

The kernel is a plane. Solve $x+2y+3z=0$ and express the solutions in terms of two free parameters.

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Equation. $x=-2y-3z$ . The solutions are
$(x,y,z)=(-2y-3z,y,z)=y(-2,1,0)+z(-3,0,1).$

Basis. $((-2,1,0),(-3,0,1))$ .

Dimension. $\dim\ker f=2$ .

Verification. $f(-2,1,0)=-2+2+0=0$ and $f(-3,0,1)=-3+0+3=0$ , and the two vectors are independent.