Decompose Vector (2,3,-1): Find Distance to Subspace

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Homework Statement


There is a subspace that contains all the vectors in the form (x, 2y, x). Decompose the vector (2, 3, -1) into a sum of an element from the orthogonal complement of this subspace and an element from the subspace. Find the distance from (2, 3, -1) to this subspace.

The Attempt at a Solution


To find the orthogonal complement of this subspace, I found the kernel, which in this case happens to only contain the zero vector. That means only a particular solution exists, but obviously (2, 3, -1) is not a particular solution, so I'm not sure how to decompose this, much less find the distance.
 
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PhizKid said:

Homework Statement


There is a subspace that contains all the vectors in the form (x, 2y, x). Decompose the vector (2, 3, -1) into a sum of an element from the orthogonal complement of this subspace and an element from the subspace. Find the distance from (2, 3, -1) to this subspace.

The Attempt at a Solution


To find the orthogonal complement of this subspace, I found the kernel, which in this case happens to only contain the zero vector. That means only a particular solution exists, but obviously (2, 3, -1) is not a particular solution, so I'm not sure how to decompose this, much less find the distance.

Why do you say the kernel is only the zero vector? ##(x,2y,x)=x(1,0,1)+y(0,2,0)## is clearly only two dimensional.
 
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Found the kernel of what?
 
The kernel of
##\ \left( \begin{array}{ccc}
1 & 0 \\
0 & 1 \\
1 & 0 \end{array} \right)##
 
Your domain should be R3, not R2.
 
So then it's:
##\ \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)##
?
 
PhizKid said:
So then it's:
##\ \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & \color{red}{2} & 0 \\
1 & 0 & 0 \end{array} \right)##
?

Yes, that matrix represents the transformation ##(x,y,z)\rightarrow (x,2y,x)##.

[Edit]: There would be a 2 in the center.
 
Last edited:
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