Exercise 1194Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

A basis of a hyperplane in $\mathbb{R}^4$

#basis#hyperplane#dimension

Statement

Find a basis of the hyperplane $H=\{(x,y,z,t)\in\mathbb{R}^4 : x+2y+3z+4t=0\}$ .

Hint

The hyperplane is the kernel of a nonzero linear form. Solve the equation with three free parameters.

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Solve. $x=-2y-3z-4t$ . The solutions are
$(x,y,z,t)=(-2y-3z-4t, y, z, t) = y(-2,1,0,0)+z(-3,0,1,0)+t(-4,0,0,1).$

Basis. $((-2,1,0,0),(-3,0,1,0),(-4,0,0,1))$ .

Independence. The three vectors are independent (their second, third, fourth coordinates form the identity).

Conclusion. $\dim H=3$ and the family above is a basis.