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[SOLVED] Distance Between a Subspace and a Vector

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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? :)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,869
Yup. All correct.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621