8 orthogonal projection innequality

nhrock3
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8)
U=\{x=(x_{1},x_{2},x_{3},x_{4})\in R^{4}|x_{1}+x_{2}+x_{4}=0\}
is a subspace of R^{4}
v=(2,0,0,1)\in R^{4}
find u_{0}\in U so ||u_{0}-v||<||u-v||
how i tried:
U=sp\{(-1,1,0,0),(-1,0,0,1),(0,0,1,0)\}
i know that the only u_{0} for which this innequality will work
is if it will be the orthogonal projection on U parallel to v
i am not sure about the theory of finding it
what to do next?
 
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nhrock3 said:
8)
U=\{x=(x_{1},x_{2},x_{3},x_{4})\in R^{4}|x_{1}+x_{2}+x_{4}=0\}
is a subspace of R^{4}
v=(2,0,0,1)\in R^{4}
find u_{0}\in U so ||u_{0}-v||<||u-v||
how i tried:
U=sp\{(-1,1,0,0),(-1,0,0,1),(0,0,1,0)\}
OK, this is a basis for U.
nhrock3 said:
i know that the only u_{0} for which this innequality will work
is if it will be the orthogonal projection on U parallel to v
i am not sure about the theory of finding it
what to do next?
You can think of U as being a plane through the origin in R4. It doesn't hurt to visualize this as a plane in three dimensions. U is actually a hyperplane, being a space of one dimension less than the space it's in, but it's more convenient to think of this problem as a plane in R3.

Vector v is not in the "plane" (U), so the vector u0 that is closest to v will be the vector that is directly under v, and lying in U. In other words, u0 is the projection of v onto the "plane" U. Surely in what you're studying there are some examples of how to find the projection of one vector onto another vector or a vector onto a plane.
 
actually there no examples
:)
 
Here's a drawing to help you out.
plane.PNG


The drawing shows the subspace U as a plane. Vector v sticks up out of the plane, starting from the origin in R4. Vector u0 lies in the plane, and its tail is also at the origin.

Vector w is perpendicular to the plane.

The three vectors shown form a triangle, so B]u[/B]0 + w = v.
The triangle is a right triangle, so u0 \cdot w = 0.
Any vector in the plane U can be written as a linear combination of the vectors in the basis you found. What must be true of every vector in the plane in relation to a vector that is perpendicular to the plane?

I was able to solve this problem after I drew the picture shown here.
 
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