- #1
Sudharaka
Gold Member
MHB
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Hi everyone, :)
I just want to confirm my answer to this question.
Question:
Find the distance between a vector \(v\) and a subspace \(U\) in a Euclidean space \(V\).
Answer:
Here what we have to find essentially, is the length of the projection of \(v\) to the orthogonal compliment, \(U^\perp\) of \(U\). Hence if we can find a orthogonal basis of \(U^\perp\); say \(\{e_1,\,e_2,\,\cdots,\,e_n\}\) then the distance is given by,
\[d=\left|\frac{v.e_1}{e_1. e_1}e_1+\cdots+\frac{v.e_n}{e_n. e_n}e_n\right|\]
To find the orthogonal basis we might want to use the Gram-Schimdt orthogonalization procedure.
Let us take an example. Let \(v=(1,\,2,\,3,\,4,\,5)\) and the subspace \(U\subset \mathbb{R}^5\) given by,
\[x_1+2x_2+3x_3+4x_4=0\]
\[5x_1+6x_2+7x_3+8x_4=0\]
Therefore we have that \(v_1=(1,\,2,\,3,\,4,\,0)\mbox{ and }v_2=(5,\,6,\,7,\,8,\,0)\) as linearly independent vectors of \(U^\perp\). Now using the Gram-Schimdt orthogonalization procedure we get,
\[e_1=v_1=(1,\,2,\,3,\,4,\,0)\]
\[e_2=v_2-\frac{v_2 . e_1}{e_1. e_1}e_1=\left(\frac{8}{3},\,\frac{4}{3},\,0,\,-\frac{4}{3},\,0\right)\]
Now that we have found a orthogonal basis for \(U^\perp\) we can find the distance as,
\[d=\left|\frac{v.e_1}{e_1. e_1}e_1+\frac{v.e_2}{e_2. e_2}e_2\right|=|e_1|=\sqrt{30}\]
Am I correct? :)
I just want to confirm my answer to this question.
Question:
Find the distance between a vector \(v\) and a subspace \(U\) in a Euclidean space \(V\).
Answer:
Here what we have to find essentially, is the length of the projection of \(v\) to the orthogonal compliment, \(U^\perp\) of \(U\). Hence if we can find a orthogonal basis of \(U^\perp\); say \(\{e_1,\,e_2,\,\cdots,\,e_n\}\) then the distance is given by,
\[d=\left|\frac{v.e_1}{e_1. e_1}e_1+\cdots+\frac{v.e_n}{e_n. e_n}e_n\right|\]
To find the orthogonal basis we might want to use the Gram-Schimdt orthogonalization procedure.
Let us take an example. Let \(v=(1,\,2,\,3,\,4,\,5)\) and the subspace \(U\subset \mathbb{R}^5\) given by,
\[x_1+2x_2+3x_3+4x_4=0\]
\[5x_1+6x_2+7x_3+8x_4=0\]
Therefore we have that \(v_1=(1,\,2,\,3,\,4,\,0)\mbox{ and }v_2=(5,\,6,\,7,\,8,\,0)\) as linearly independent vectors of \(U^\perp\). Now using the Gram-Schimdt orthogonalization procedure we get,
\[e_1=v_1=(1,\,2,\,3,\,4,\,0)\]
\[e_2=v_2-\frac{v_2 . e_1}{e_1. e_1}e_1=\left(\frac{8}{3},\,\frac{4}{3},\,0,\,-\frac{4}{3},\,0\right)\]
Now that we have found a orthogonal basis for \(U^\perp\) we can find the distance as,
\[d=\left|\frac{v.e_1}{e_1. e_1}e_1+\frac{v.e_2}{e_2. e_2}e_2\right|=|e_1|=\sqrt{30}\]
Am I correct? :)