• Support PF! Buy your school textbooks, materials and every day products Here!

Linear Algebra

  • Thread starter Ted123
  • Start date
  • #1
446
0

Homework Statement



[PLAIN]http://img219.imageshack.us/img219/2950/linl.jpg [Broken]

Homework Equations



The Attempt at a Solution



Is this how I do part (iii)?

From (ii) I get:

[itex]M^{\mathcal C}_{\mathcal C} (\phi) = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix}[/itex]

3(iii)

[itex]M^{\mathcal C}_{\mathcal C} (\phi) {\bf v} = {\bf w}[/itex] where [itex]{\bf w}[/itex] is the coordinate vector of [itex]\phi (v)[/itex] with respect to [itex]\mathcal C[/itex] and [itex]{\bf v}[/itex] is the coordinate vector of [itex]v\in V[/itex] .

[itex]{\bf v} = \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix}[/itex]

So [itex]{\bf w} = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix} = \begin{bmatrix} 2\pi -8 \\ 10 \\ 2\pi \end{bmatrix}[/itex]
 
Last edited by a moderator:

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
Yes, that's the correct method.
 
  • #3
14
0
Is the matrix M^C_C(phi) supposed to be the matrix of phi in the ordered basis C? If so, your matrix is incorrect; 1+x, 1-x, and 1+x^2 are eigenvectors of phi, with eigenvalues 1, 3, and 2, respectively. The matrix you are looking for is a diagonal matrix.

Your approach for (iii) is correct; simply multiply the matrix of phi in the basis C by the coordinate matrix v; that will give you the coordinate matrix of phi(v).
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
I concur. I get a different matrix for part (ii).
 
  • #5
446
0
Is the matrix M^C_C(phi) supposed to be the matrix of phi in the ordered basis C? If so, your matrix is incorrect; 1+x, 1-x, and 1+x^2 are eigenvectors of phi, with eigenvalues 1, 3, and 2, respectively. The matrix you are looking for is a diagonal matrix.

Your approach for (iii) is correct; simply multiply the matrix of phi in the basis C by the coordinate matrix v; that will give you the coordinate matrix of phi(v).
Well this is my approach to the whole question:

Part (i)

[itex]\phi (1) = 2-x = 2(1) -1(x) + 0(x^2)[/itex]

[itex]\phi (x) = -1 + 2x = -1(1) + 2(x) + 0(x^2)[/itex]

[itex]\phi (x^2) = x + 2x^2 = 0(1) + 1(x) + 2(x^2)[/itex]

[itex]M^{\mathcal B}_{\mathcal B}(\phi) = \begin{bmatrix} 2 & -1 & 0 \\ -1 & 2 & 1 \\ 0 & 0 & 2 \\ \end{bmatrix}[/itex]

Part (ii)

Since [itex]{\mathcal C}[/itex] has size 3 and [itex]\text{dim}(\mathbb{R}_2[x])=3[/itex] it is enough to show that [itex]\mathcal C}[/itex] is either linearly independent or spans [itex]V[/itex] . For completeness let's do both.

[itex]{\mathcal C}[/itex] is linearly independent if

[itex]\alpha (1+x) + \beta (1-x) + \gamma (1+x^2) = 0 \Rightarrow \alpha = \beta = \gamma = 0[/itex]

[itex]\alpha + \alpha x +\beta - \beta x + \gamma + \gamma x^2 = 0[/itex]

[itex]\alpha + \beta + \gamma + (\alpha - \beta)x + \gamma x^2 = 0[/itex]

Equate coefficients.

[itex]\alpha + \beta + \gamma = 0[/itex] (1)

[itex]\alpha - \beta = 0 \Rightarrow \alpha = \beta[/itex] (2)

[itex]\gamma = 0[/itex] (3)

Subbing (2) and (3) in (1) gives [itex]2\alpha = 0 \Rightarrow \alpha = 0 \Rightarrow \beta = 0[/itex] from (2).

Hence [itex]\alpha = \beta = \gamma = 0[/itex] and [itex]\mathcal C[/itex] is linearly independent.

[itex]\mathcal C[/itex] spans [itex]V[/itex] if

[itex]a_1 (1+x) + a_2 (1-x) + a_3 (1+x^2) = b_1 + b_2 x + b_3 x^2[/itex]

has at least one solution for every set of coefficients [itex]b_1,b_2,b_3[/itex]

[itex]a_1 + a_2 + a_3 + (a_1 - a_2)x + a_3x^2 = b_1 + b_2 x + b_3 x^2[/itex]

Equating coefficients yields the following system of equations (*):

[itex]a_1 + a_2 + a_3 = b_1[/itex]

[itex]a_1 - a_2 = b_2[/itex]

[itex]a_3 = b_3[/itex]

Transforming the augmented matrix

[itex]\begin{bmatrix} 1 & 1 & 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{bmatrix}[/itex]

into row echelon form we get

[itex]\begin{bmatrix} 1 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 & 0 & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & 1 & 0 & 0 & 1 \\ \end{bmatrix}[/itex]

which corresponds to the following system of equations:

[itex]a_1 = \frac{1}{2}b_1 + \frac{1}{2}b_2 -\frac{1}{2}b_3[/itex]

[itex]a_2 = \frac{1}{2}b_1 -\frac{1}{2}b_2 -\frac{1}{2}b_3[/itex]

[itex]a_3 = b_3[/itex]

This system {and therefore the system (*)} are consistent for all RHS values, therefore [itex]\mathcal C[/itex] spans [itex]V[/itex] .

We can therefore conclude that [itex]\mathcal C[/itex] is certainly a basis for [itex]V[/itex] .

[itex]\phi (1+x) = (2-1) + (-1+2)x = 1 + x = 1(1) + 1(x) + 0(x^2)[/itex]

[itex]\phi (1-x) = (2+1) + (-1-2)x = 3-3x = 3(1) - 3(x) + 0(x^2)[/itex]

[itex]\phi (1+x^2) = 2 + (-1+1)x + 2x^2 = 2+2x^2 = 2(1) + 0(x) + 2(x^2)[/itex]

[itex]M^{\mathcal C}_{\mathcal C} (\phi) = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix}[/itex]

Part (iii)

[itex]M^{\mathcal C}_{\mathcal C} (\phi) {\bf v} = {\bf w}[/itex] where [itex]{\bf w}[/itex] is the coordinate vector of [itex]\phi (v)[/itex] with respect to [itex]\mathcal C[/itex] and [itex]{\bf v}[/itex] is the coordinate vector of [itex]v\in V[/itex] .

[itex]{\bf v} = \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix}[/itex]

So [itex]{\bf w} = \begin{bmatrix} 1 & 3 & 2 \\ 1 & -3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ -3 \\ \pi \end{bmatrix} = \begin{bmatrix} 2\pi -8 \\ 10 \\ 2\pi \end{bmatrix}[/itex]
 
  • #6
14
0
You don't need to do all of that work. For (i) (and (ii)) simply find the coordinate matrices of phi(1), phi(x) and phi(x^2) in the ordered basis B (or phi(1+x), phi(1-x), phi(1+x^2) in the ordered basis C). These coordinate matrices are, in order, the columns of the matrix you seek.

Now, for part (ii), to show that those three functions are a basis, it suffices to show that they are linearly independent. This is not difficult to do, and your way works fine.

If M^B_B(phi) is indeed meant to represent the matrix of phi in the ordered basis B, then I think your only mistake is that you are misunderstanding what this matrix is.
 
  • #7
446
0
You don't need to do all of that work. For (i) (and (ii)) simply find the coordinate matrices of phi(1), phi(x) and phi(x^2) in the ordered basis B (or phi(1+x), phi(1-x), phi(1+x^2) in the ordered basis C). These coordinate matrices are, in order, the columns of the matrix you seek.

Now, for part (ii), to show that those three functions are a basis, it suffices to show that they are linearly independent. This is not difficult to do, and your way works fine.

If M^B_B(phi) is indeed meant to represent the matrix of phi in the ordered basis B, then I think your only mistake is that you are misunderstanding what this matrix is.
The matrix for (1) is definitely correct (I was told it was by the lecturer) so the method I am using is correct (and is what we were taught).

I know I only need to show EITHER linear independence OR span for part (ii) but for COMPLETENESS (and to make sure I could do both) I did BOTH!

[PLAIN]http://img832.imageshack.us/img832/1083/matrixtp.jpg [Broken]

I've just noticed that I've wrote the images of the vectors in C as a linear combination of the vectors in B rather than C for part (ii)...
 
Last edited by a moderator:
  • #8
14
0
Sorry, just realized you already wrote that; I was confusing the numbers on your question. Under part (i) I think you wrote something different on the solution from what was written in the problem; it seems as though part (i) of the problem is under part (ii) of your solution. Yes, you are exactly right.
 
Last edited:
  • #9
14
0
Just read your last post; that definition is what I was thinking of.

I didn't mean to reiterate that point about only needing to show independence; I see you already wrote that.

Hope I could help.
 
  • #10
446
0
The coordinate matrices for phi(1), phi(x), phi(x^2) in the ordered basis B be should be obtained as follows.

phi(1) = 2 - x = 2(1) - 1(x) + 0(x^2);
phi(x) = -1 + 2x = -1(1) + 2(x) + 0(x^2);
phi(x^2) = x + 2x^2 = 0(1) + 1(x) + 2(x^2).

(Wish I knew how to use Tex, for your sake!) From the above equations, you can read of the coordinates; those are the columns of your matrix. In fact, your matrix should be the transpose of the coefficient matrix of the above system of equations. Deal similarly with the ordered basis C (it's a bit easier; as I said before, the elements of C are each eigenvectors of phi).
I know - that's how I got the matrix for part (i)!

Sorry I just noticed I posted my working to a completely different question before...
 
  • #11
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
I've just noticed that I've wrote the images of the vectors in C as a linear combination of the vectors in B rather than C for part (ii)...
Yeah, that's your mistake.
 
  • #12
14
0
Sorry about that confusion, I just realized that.
 
  • #13
446
0
Just read your last post; that definition is what I was thinking of.

I didn't mean to reiterate that point about only needing to show independence; I see you already wrote that.

Hope I could help.
Sorry everyone!

Just to clarify I posted my working to a completely different question before and mixed up the numbering of the question parts in my working!

It seems my mistake for part (ii) is writing the images as a linear combination of vectors in B rather than C.
 
  • #14
446
0
Yeah, that's your mistake.
So is this correct:

[itex]M^{\mathcal C}_{\mathcal C} (\phi) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix}[/itex]

(And obviously then I've got to correct part (iii) )
 
  • #15
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,631
1,267
Yeah, that's the right matrix.
 

Related Threads on Linear Algebra

Replies
9
Views
715
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
761
  • Last Post
Replies
0
Views
706
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
893
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
968
Top