Linear Algebra Question before my exam

Anakin · 2011-12-16 00:54:32

Hello everyone; it's been quite some time since I last posted. Anyway, I'm going for my Linear Algebra II exam on Saturday and all my studying is nearly done and prepared for.. except one question! I'll post the question and my work and see if someone can guide me:

I'm not sure I even did anything properly there as I somewhat missed that unit because I had a heavy week.

Thank you!

bobbym · 2011-12-16 03:21:27

Hi Anakin;

Have not seen you in a long time. I have checked the computation on the end and at least that is okay. The rest I do not know.

Anakin · 2011-12-16 03:33:35

Hi Bobbym!

Yup, second year of university and work make it nearly impossible to get online nowadays. Can't wait until the holidays start.

The computation is correct but my question was that is this how one would carry out the procedure to find the linear transformation of (x^3 + x^2 - x + 1) with basis B'?

Something leads me to believe that perhaps my final answer isn't wrong because say that we ignore the matrix representation of T relative to B, B'. Then if we do T(x^3 + x^2 - x + 1), we notice that a = 1, b = 2, c = -1, d = 1.

The linear transformation would be the matrix:
|1-1 1+(-1)|
|(-1)+1 1-1|
=
|0 0|
|0 0|

which basis B' would just be [0, 0, 0, 0] as I got as my answer to number 2.

Does that sound reasonable?

bobbym · 2011-12-16 03:40:00

Hi Anakin;

T(x^3 + x^2 - x + 1), we notice that a = 1, b = 2, c = -1, d = 1.

Looks like you will be teaching me. How did you get that?

Anakin · 2011-12-16 04:05:38

Well the linear transformation is T(ax^3 + bx^2 + cx + d) = the matrix
|a-b b+c|
|c+d d-a|

If we try to find T(x^3 + x^2 - x + 1), it appears that the values of a, b, c, d are 1, 1, -1, 1 respectively.

The reason why we try to the find matrix representation relative to B,B' in the first part is so we can just do a matrix multiplication to find the answer without having to compute a-b, b+c, etc. and then convert it to basis B'.

With the matrix, we can just take the representation of the input (x^3 + x^2 - x + 1 in this case) with respect to basis B, which is [1, -1, 1, 1] in this case and do a matrix multiplication, and we get our result in basis B'.

Last edited by Anakin (2011-12-16 04:10:16)

bobbym · 2011-12-16 04:18:42

Hi;

Okay, I see that. T(1)?

Anakin · 2011-12-16 04:54:34

T(1) means that the values of a, b, c = 0 and d = 1. Thus the linear transformation results in a matrix
|0 0|
|1 1|

With respect to basis B', it means we need -1 of the 2nd matrix, 1 of the 3rd and 1 of the 4th to get it because
|0 -1| + |0 1| + |0 0|
|0 0| |1 0| |0 1|

=
|0 0|
|1 1|

Last edited by Anakin (2011-12-16 04:55:23)

bobbym · 2011-12-16 05:04:47

Hi;

Thanks for that. One last thing, in the first line your notation of

Means what?

Anakin · 2011-12-16 05:12:07

No problem.

It means that it is the matrix outputted by the linear transformation but needs to be converted with respect to ordered basis B', that would be = [0, -1, 1, 1]. The matrix itself that you showed has not been changed to B' coordinates until it is listed as [0, -1, 1, 1].

So essentially we want to be able to take any polynomial with degrees less than or equal to 3 (including the zero polynomial), these are in ordered basis B. So 3x^2 - 2x would be [0, -2, 3, 0] in basis B. We want to take their linear transformation, which outputs a matrix. Then we want to represent that output with respect to basis B'.

By creating the matrix [T] (subscript B,B') in part A of my question, we can simply take a polynomial's representation in basis B and multiply it with this matrix [T] (subscript B,B') to get the linear transformation in basis B'.

Last edited by Anakin (2011-12-16 05:13:38)

bobbym · 2011-12-16 05:17:38

Hi;

How did you get the vector

Is it a straight multiplication? Which element of B1 was used?

Anakin · 2011-12-16 05:39:22

I got the vector by inspection; though I suppose if the elements of B' were really complicated, we'd have to solve using a system of equations.

The matrix
|0 0|
|1 1|
has to be represented with the elements of basis B'. So I just used inspection.

The reason why I found T(1), T(x), T(x^2), T(x^3) is because those are the elements of the basis B and any polynomial whose linear transformation we want, that polynomial can be written in terms of these elements of basis B.

I'm gonna get a quick 8 hour nap before I power through all my notes one last time for my exam. I'll be back in a little while.

Last edited by Anakin (2011-12-16 05:41:01)

bobbym · 2011-12-16 05:41:26

Hi Anakin;

Thanks for helping me out on it. I am sorry that I do not know the answer.

Good luck on your exam, do well!

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#1 2011-12-16 00:54:32

Linear Algebra Question before my exam

#2 2011-12-16 03:21:27

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#3 2011-12-16 03:33:35

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#4 2011-12-16 03:40:00

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#5 2011-12-16 04:05:38

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#6 2011-12-16 04:18:42

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#7 2011-12-16 04:54:34

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#8 2011-12-16 05:04:47

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