Exercise 1902Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Decomposing a vector in a direct sum

#direct sum#decomposition#projection

Statement

With $F=\{(x,y,z): x+y+z=0\}$ and $G=\operatorname{span}((1,1,1))$ , decompose $v=(3,1,2)$ as $v=v_F+v_G$ with $v_F\in F$ and $v_G\in G$ .

Hint

Write $v_G=\lambda(1,1,1)$ and impose $v-v_G\in F$ .

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Solution

Setup. Let $v_G=\lambda(1,1,1)$ . Then $v_F=v-v_G=(3-\lambda,1-\lambda,2-\lambda)$ . We need $v_F\in F$ :
$(3-\lambda)+(1-\lambda)+(2-\lambda)=0 \implies 6-3\lambda=0 \implies \lambda=2.$

Decomposition. $v_G=2(1,1,1)=(2,2,2)$ and $v_F=(3,1,2)-(2,2,2)=(1,-1,0)$ .

Verification. $(1,-1,0)\in F$ since $1-1+0=0$ . And $v=v_F+v_G$ .

Conclusion. $v=(1,-1,0)+(2,2,2)$ .

Decomposing a vector in a direct sum

Statement

Solution

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