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Let $W$ be a given subspace of $V$. Find the basis of orthogonal complement $W^{\perp}$.

If $V=R^5$ and $W = lin\lbrace(2,2,-1,0,1),(-1,-1,2,-3,1),(1,1,-2,0,1),(0,0,1,1,1)\rbrace$.

I assume that $W^{\perp}$ will 1 dimensional and I have to find vector that will be orthogonal to all of these four vectors from basis W, but how to find it? Is there a way to do it with Gram-Schmidt method or maybe there is smarter way?

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1 Answer 1

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The orthogonal complement $W^{\perp}$ consists of all vectors perpendicular to the vectors of $W$. Expressing that a vector $(a,b,c,d,e)$ is perpendicular to each of the vectors that span $W$, is the same as finding the null space of the following matrix: $$A=\begin{pmatrix} 2 & 2 & -1 & 0 & 1 \\ -1 & -1 & 2 & -3 & 1 \\ 1 & 1 & -2 & 0 & 1 \\ 0 & 0 & 1 & 1 & 1 \end{pmatrix}$$ So you need to solve $A \vec x= \vec 0$ and find a basis for its solution set; this will be a basis for $W^{\perp}$. Does that help?

Spoiler:

You should find the space spanned by $(1,-1,0,0,0)$.

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