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Questions regarding vector spaces

9. Let u, x, y, z be elements in a vector space V where

u = 3x - 4y +5z, v = 4x + 5y - 6z, and 2u + 3v = x + y + z

Show that {x,y,z} is a linearly dependent set.

10. Find a basis for the space spanned by the following vectors (see attached file)

11. Let W = {p epsilon P2 | p'(1) = 0}. That is, W are the functions in P2 whose derivative at 1 is 0. Show that W is a subspace of P2.
(a) Make a conjecture about the dimension of W. Explain your choice.
(b) Confirm (or reject) your conjecture by finding a basis for W.

12. Find a basis for the subspace V of R^5 defined by the following equation (see attached file).

13. Find a basis for 2 x 2 matrices that includes [ 0 1; 1 1], [ -1 0: 0 1], and [ 0 0; 1 0] or explain why this is not possible.

14. Without any calculation, explain whether the following statements are true or false.
(a) We can add one vector to the set (3, 5, 2), (-3, 1, 2) to make a bsis for R^3.
(b) A basis for R^3 is the set {(2,3,1), (2,4,1), (4,6,5), (3,4,2)}
(c) A basis for R^3 is the set {(12,-3,10), (0,14,9), (0,0,5)}
(d) A basis for P2 is {37x^2 - 7x + 88,28x - 1,101}

Attachments

Solution Preview

See attached file for solution.

Solutions:

9) Let's write the given equalities in the form:
( 1)
By replacing u and v in the last equation with the first two expressions in x, y and z, we get:
( 2)
After some algebra, one yields:
( 3)
Since x, y and z are linked by a linear relation then, according to the definition of "linear dependency", they are linear dependent.
It can be seen that equation (3) represents a plane in the geometric space E3, which passes through origin. As such, since (0, 0, 0) satisfies (3), the plane can be considered a subspace of the vector space E3.

10) In order to find a basis in the vector space spanned by the given 6 vectors, we will put them in the form of a rectangular matrix, denoted as A:
( 4)
The dimension of the spanned vector space is given by the rank of the above matrix. Thus, we need to find within A a square matrix as large as possible, whose determinant is not null. The matrix defined by first 4 columns has the determinant:
( 5)
We try with columns 1,2,3 and 5:
( 6)
The columns 1,2,3 and 6 yield:

The columns 3,4,5 and 6 yield:
( 7)
We have determined that there is no 4x4 non-singular matrix within matrix A which means that
( 8)
We have to check now the 3x3 matrices. Let's start with the lower matrix formed by the first 4 columns:
( 9)
Since there is a 3x3 non-singular ...

Solution Summary

A basis of a vector space contains only linear independent vectors and the number of vectors of the basis which span a given vector space defines its dimension. Any vector of a given vector space will be represented as an unique linear combination of the vectors of basis. This solution provides answers for questions regarding vectors.

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