Exercise 1901Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Direct sum of subspaces

#direct sum#supplementary subspaces#dimension

Statement

In $\mathbb{R}^3$ , let $F=\{(x,y,z): x+y+z=0\}$ and $G=\operatorname{span}((1,1,1))$ . Show that $F$ and $G$ are supplementary: $\mathbb{R}^3=F\oplus G$ .

Hint

Show that $F\cap G=\{0\}$ and $\dim F+\dim G=3$ .

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Solution

Dimension. $\dim F=2$ (a plane through the origin) and $\dim G=1$ (a line). Their sum is $3=\dim\mathbb{R}^3$ .

Intersection. A vector in $G$ has the form $\lambda(1,1,1)$ . If it is also in $F$ , then $\lambda+\lambda+\lambda=3\lambda=0$ , so $\lambda=0$ . Thus $F\cap G=\{0\}$ .

Conclusion. By the direct sum criterion, $\mathbb{R}^3=F\oplus G$ .

Interpretation. Every vector $v\in\mathbb{R}^3$ decomposes uniquely as $v=v_F+v_G$ with $v_F\in F$ and $v_G\in G$ .

Direct sum of subspaces

Statement

Solution

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