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Linear Algebra: Basis and Dimension

10. Find the coordinate vector of p relative to the basis S = {p1, p2, p3}.

(a) p = 4 - 3x + x2; p1 = 1, p2 = x, p3 = x2
(b) p = 2 - x + x2; p1 = 1 + x, p2 = 1 + x2, p3 = x + x2

22. Find the standard basis vectors that can be added to the set {v1, v2} to produce a basis for R4.

v1 = (1, -4, 2, -3), v2 = (-3, 8, -4, 6)

30. Prove: Any subspace of a finite-dimensional vector space is finite-dimensional.
Hint: You can do a proof by contradiction by using 2 or 3 Theorems from section.

See attached file for full problem description.

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Solution This solution is FREE courtesy of BrainMass!

10. Find the coordinate vector of p relative to the basis S = {p1, p2, p3}.

The coordinates of p relative to the basis S are the numbers c1, c2, c3 such that p = c1p1 + c2p2 + c3p3. The coordinate vector of p would be the vector {c1, c2, c3}. (Note how this relates to theorem 5.4.1 - there is only one coordinate vector for each p.)

(a) p = 4 - 3x + x2; p1 = 1, p2 = x, p3 = x2

p = c1(p1) + c2(p2) + c3(p3)
p = c1(1) + c2(x) + c3(x2)

c1 = 4, c2 = -3, c3 = 1

(b) p = 2 - x + x2; p1 = 1 + x, p2 = 1 + x2, p3 = x + x2

p = c1(p1) + c2(p2) + c3(p3)
p = c1(1 + x) + c2(1 + x2) + c3(x + x2)

2 - x + x2 = c1(1 + x) + c2(1 + x2) + c3(x + x2)
2 - x + x2 = c1 + c1x + c2 + c2x2 + c3x + c3x2
2 - x + x2 = (c1 + c2) + (c1+ c3)x + (c2 + c3)x2

2 = c1 + c2
-1 = c1+ c3
1 = c2 + c3

Solve this system of equations. You should get:

c1 = 0, c2 = 2, c3 = -1

22. Find the standard basis vectors that can be added to the set {v1, v2} to produce a basis for R4.

v1 = (1, -4, 2, -3), v2 = (-3, 8, -4, 6)

Standard basis vectors have a 1 in one position and a 0 in all the other positions. So, the standard basis vectors for R4 are:

e1 = (1, 0, 0, 0)
e2 = (0, 1, 0, 0)
e3 = (0, 0, 1, 0)
e4 = (0, 0, 0, 1)

These are the vectors that we have to choose from. However, in this problem, we only need to add two standard basis vectors (a basis for R4 will consist of 4 vectors). We need to find two vectors that are linearly independent of v1 and v2.

You can add e2 and e3, e2 and e4, or e3 and e4 to v1 and v2 to make a basis. (You can check that they are linearly independent of one another by calculating the determinant of the matrix made up of the set of four vectors - if the determinant is 0, then they are linearly dependent; if the determinant is not 0, they are independent.)

For example, the determinant of the following matrix is -4:

1 -3 0 0
-4 8 0 0
2 -4 1 0
-3 6 0 1

Note: e1 is NOT linearly independent of v1 and v2, so it cannot be included in the basis:

e1 = -2v1 - v2
(1, 0, 0, 0) = (-2)(1, -4, 2, -3) + (-1)(-3, 8, -4, 6)

30. Prove: Any subspace of a finite-dimensional vector space is finite-dimensional.
Hint: You can do a proof by contradiction by using 2 or 3 Theorems from section.

We can use a proof by contradiction like the hint says. However, I think we only need to use one theorem.

Assume that we have a finite-dimensional vector space V, and W is an infinite dimensional subspace of V.

By theorem 5.4.7, dim(W) ≤ dim(V). (Keep in mind that we know dim(V) is not infinite since we are assuming that V is a finite-dimensional vector space. By definition, dim(V) will be a finite number.)

Since dim(W) is less than dim(V), dim(W) must also be finite. This contradicts the assumption that V is an infinite-dimensional vector space.

I suppose that you could also do a more round-about proof using 2 or 3 of the theorems, but I think the simpler proof is good enough. A more complicated proof might be like the following:

if V is finite-dimensional (it is n-dimensional), then it has a basis with n elements. However, W would need an infinite number of elements to make up a basis, but since it is a subset of a V, a basis of W would have less than or equal to n elements.

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