8 orthogonal projection innequality

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Homework Help Overview

The problem involves finding a vector \( u_0 \) in the subspace \( U \) of \( \mathbb{R}^4 \) defined by the equation \( x_1 + x_2 + x_4 = 0 \), such that the distance from \( u_0 \) to a given vector \( v = (2,0,0,1) \) is minimized. The context is within linear algebra, specifically focusing on orthogonal projections and subspaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of the subspace \( U \) and its basis. There is an attempt to understand the concept of orthogonal projection and how it relates to minimizing the distance between vectors. Questions arise regarding the theory behind finding the orthogonal projection and the geometric interpretation of the problem.

Discussion Status

Some participants have provided visual aids and geometric interpretations to clarify the relationship between the vectors and the subspace. There is acknowledgment of the need for further exploration of the projection theory, but no consensus has been reached on the specific steps to take next.

Contextual Notes

Participants note the absence of examples in their study materials regarding projections, which may limit their ability to apply theoretical concepts to this problem. The discussion also highlights the dimensionality of the subspace and its implications for the projection process.

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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