Find the closest point to y in the subspace W spanned by v1v and v2 y=[3151],v1=[31−11],v2=[1−11−1]

Step 1 The formula of orthogonal projection of y onto span v1,v2 and this projection is the closest point to y in the subspace W. y^=y⋅v1v1⋅v1v1+y⋅v2v2⋅v2v2 Step 2 because y⋅v1=6,v1⋅v1=12,y⋅v2=6,v2⋅v2=4 a^=612[31−11]+64[1−11−1] Step 3 The required closest point to y in the subspace W. y^=[3−11−1] Result y^=[3−11−1]

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Virginia Caron

Answered question

2021-11-17

Find the closest point to y in the subspace W spanned by $v_{1} v and v_{2}$
$y = [\begin{matrix} 3 \\ 1 \\ 5 \\ 1 \end{matrix}], v_{1} = [\begin{matrix} 3 \\ 1 \\ - 1 \\ 1 \end{matrix}], v_{2} = [\begin{matrix} 1 \\ - 1 \\ 1 \\ - 1 \end{matrix}]$

Answer & Explanation

Blanche McClain

Beginner2021-11-18Added 15 answers

Step 1
$The formula of orthogonal projection of y onto span v_{1}, v_{2} and this projection is the closest point to y in the subspace W.$
$\hat{y} = \frac{y \cdot v_{1}}{v_{1} \cdot v_{1}} v_{1} + \frac{y \cdot v_{2}}{v_{2} \cdot v_{2}} v_{2}$
Step 2
$because y \cdot v_{1} = 6, v_{1} \cdot v_{1} = 12, y \cdot v_{2} = 6, v_{2} \cdot v_{2} = 4$
$\hat{a} = \frac{6}{12} [\begin{matrix} 3 \\ 1 \\ - 1 \\ 1 \end{matrix}] + \frac{6}{4} [\begin{matrix} 1 \\ - 1 \\ 1 \\ - 1 \end{matrix}]$
Step 3
The required closest point to y in the subspace W.
$\hat{y} = [\begin{matrix} 3 \\ - 1 \\ 1 \\ - 1 \end{matrix}]$
Result
$\hat{y} = [\begin{matrix} 3 \\ - 1 \\ 1 \\ - 1 \end{matrix}]$

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