8 orthogonal projection innequality

[nhrock3]

8)
[itex]U=\{x=(x_{1},x_{2},x_{3},x_{4})\in R^{4}|x_{1}+x_{2}+x_{4}=0\}[/itex]
is a subspace of [itex]R^{4}[/itex]
[itex]v=(2,0,0,1)\in R^{4}[/itex]
find [itex]u_{0}\in U[/itex] so [itex]||u_{0}-v||<||u-v||[/itex]
how i tried:
[itex]U=sp\{(-1,1,0,0),(-1,0,0,1),(0,0,1,0)\}[/itex]
i know that the only [itex]u_{0}[/itex] 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?

 

[Mark44]

8)
[tex]U=\{x=(x_{1},x_{2},x_{3},x_{4})\in R^{4}|x_{1}+x_{2}+x_{4}=0\}[/tex]
is a subspace of [itex]R^{4}[/itex]
[itex]v=(2,0,0,1)\in R^{4}[/itex]
find [itex]u_{0}\in U[/itex] so [itex]||u_{0}-v||<||u-v||[/itex]
how i tried:
[itex]U=sp\{(-1,1,0,0),(-1,0,0,1),(0,0,1,0)\}[/itex]

OK, this is a basis for U.

i know that the only [itex]u_{0}[/itex] 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.

 

[nhrock3]

actually there no examples
:)

 

[Mark44]

Here's a drawing to help you out.

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 [itex]\cdot[/itex] 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.