Exercise 1922Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Orthogonal complement

#orthogonal complement#dimension#inner product

Statement

In $\mathbb{R}^3$ with the standard inner product, let $F=\operatorname{span}((1,1,0),(0,1,1))$ . Compute $F^\perp$ and verify $\dim F+\dim F^\perp=3$ .

Hint

$F^\perp$ is the set of vectors orthogonal to a generating set of $F$ .

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Solution

Equations. $v=(x,y,z)\in F^\perp$ iff $v\cdot(1,1,0)=0$ and $v\cdot(0,1,1)=0$ :
$x+y=0, \qquad y+z=0.$
So $y=-x$ and $z=-y=x$ . Thus $v=x(1,-1,1)$ .

Conclusion. $F^\perp=\operatorname{span}((1,-1,1))$ , dimension $1$ .

Check. $\dim F=2$ (the two generators are independent) and $\dim F^\perp=1$ , so $2+1=3=\dim\mathbb{R}^3$ . This confirms $\mathbb{R}^3=F\oplus F^\perp$ .

Orthogonal complement

Statement

Solution

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