Show that U intersection W does not equal ((0,0,0))

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Homework Help Overview

The discussion revolves around proving that the intersection of two vector subspaces, U and W, is not equal to the zero vector, specifically in the context of linear algebra. U is defined as the set of vectors of the form (a, 0, a) and W as (c, d, c + 2d), where a, c, and d are real numbers. The goal is to explore the implications of this intersection on the direct sum of the two subspaces.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions under which U and W can be considered a direct sum, questioning the uniqueness of vector representation in the sum. There are inquiries about the existence of non-zero vectors in the intersection and the implications of scalar values a, c, and d being zero or non-zero.

Discussion Status

The conversation is ongoing, with participants seeking clarification on the definitions and properties of vector subspaces. Some guidance has been offered regarding the conditions for a direct sum, but there is no consensus on the specifics of the intersection or the implications of the variables involved.

Contextual Notes

There is a noted misconception regarding the nature of a, c, and d as scalars rather than vectors, which may affect the understanding of the problem. Participants are also grappling with the requirement to find a non-zero vector in the intersection to demonstrate that it is not trivial.

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show that U intersection W does not equal ((0,0,0)) and hence that U+W is not a direct sum

U being (a,0,a) and W being (c,d,c+2d) and we know that c d and a are elements of R, why are there not a direct sum? is there a rule that prevents any of a c or d from being the zero vector? if this is the case obviously its easy to solve?
 
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It is only a direct sum if they have no elements in common, that is: you can only write every element from U+W in the form u+w (with u in U and w in W) in one and only one way.
This follows from the first part, namely: that the intersection is not empty. Because, if x is an element in the intersection it can be written both as x + 0 and 0 + x, so the decomposition is not unique.

To show that it is non-empty, you only need to give a single non-zero element that is in both, which is not hard.
 
sorry that went right over my head, can someone dumb it down a bit please?
 
To show that U\cap W\neq\{(0,0,0)\}, you need to find some other vector (x,y,z) which belongs in both U and W where at least one of x, y, or z is not zero.

In other words, if (a, 0, a) = (c,d,c+2d), what equations can you set up to solve for a, c, and d? Do all three variables have to equal 0?
 
franky2727 said:
is there a rule that prevents any of a c or d from being the zero vector?

Just a note: unless the above is a typo, it seems that you have a pretty severe misconception about vectors. a, c, and d are elements of \mathbb{R}, i.e. scalars. Thus none of them can "be the zero vector." The zero vector in this case (since we are working in \mathbb{R}^3) is the vector (0, 0, 0).
 

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