Exercise 1916Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Intersection and sum of subspaces

#intersection#sum#dimension formula

Statement

In $\mathbb{R}^3$ , let $F=\{(x,y,z): x+y+z=0\}$ and $G=\{(x,y,z): x-y=0\}$ . Compute $\dim F$ , $\dim G$ , $\dim(F\cap G)$ , and $\dim(F+G)$ .

Hint

Use the formula $\dim(F+G)=\dim F+\dim G-\dim(F\cap G)$ , and find a basis of each subspace.

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Solution

Dimensions. $F$ is a plane through the origin, so $\dim F=2$ . $G$ is a plane through the origin, so $\dim G=2$ .

Intersection. $F\cap G$ is given by the system $x+y+z=0$ , $x=y$ . Substituting $y=x$ : $2x+z=0$ , so $z=-2x$ . A basis is $(1,1,-2)$ . So $\dim(F\cap G)=1$ .

Sum. $\dim(F+G)=2+2-1=3$ , so $F+G=\mathbb{R}^3$ .

Conclusion. $\dim F=2$ , $\dim G=2$ , $\dim(F\cap G)=1$ , $\dim(F+G)=3$ .

Intersection and sum of subspaces

Statement

Solution

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