Linear Algebra - Change of Basis

[raay]

Hi please i need help in number 3 of the tutorial questions. It is not an assignment its just a tutorial (read title in the image). I am currently studying for my final and i need help in (3b). the only way I am thinking of solving this questions is to use the equation given in part (d). But after reading the whole question i think i have to use a different method to solve part (b). Any ideas or hints ?

thanks

 

[vela]

Can you explain in words what $[ S]_{C \to C}$ is?

 

[geoffrey159]

Hi, if I understand, you are looking for the matrix of $S$ in the basis ${\cal C}$ at source and destination, are you ?

I have a review question for you that will probably answer your question too :
In general, what does it mean that $A = (a_{i,j})_{1\le i \le n , 1 \le j \le p}$ is the matrix of a linear map $f$ relative to a source basis ${\cal B}$ and destination basis ${\cal C }$ ?

 

[raay]

Can you explain in words what $[ S]_{C \to C}$ is?

I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

 

[raay]

Hi, if I understand, you are looking for the matrix of $S$ in the basis ${\cal C}$ at source and destination, are you ?

I have a review question for you that will probably answer your question too :
In general, what does it mean that $A = (a_{i,j})_{1\le i \le n , 1 \le j \le p}$ is the matrix of a linear map $f$ relative to a source basis ${\cal B}$ and destination basis ${\cal C }$ ?

an example of what i mean is :
I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

 

[Fredrik]

I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

I didn't check all the details, but it looks like you're using the right method. So you just need to do the same in problem 3. (You seem to think that you need a different method when the two bases are the same, but the method you just used works for arbitrary bases).

 

[raay]

I didn't check all the details, but it looks like you're using the right method. So you just need to do the same in problem 3. (You seem to think that you need a different method when the two bases are the same, but the method you just used works for arbitrary bases).

I don't know how to imply it. working with matrices confuses me :S

 

[geoffrey159]

I don't understand what is the problem. If you have managed to do q1, q2, q3-a , what is different in q3-b ?
What about my question: what is the matrix of a linear map relative to two basis ?

 

[raay]

I know that q3a is just the standard matrix. I am used to solving q3b by using the equation in part (d). but q3b is asking to solve by a method similar to the method in q2 like the example i have written. the thing is I am not used to solving it using matrices i know how to solve polynomials and integrals using the method in q2. for your question i really have no idea at the moment lol because I am studying calculus and my final is tmrw when i finish i will go through algebra again. I just need an explanation so when i reach this topic i can just move on.

 

[Fredrik]

You seem to be imagining a difficulty where there is none. The method that you used to solve problem 2c, also solves problems 3a to 3c.

The FAQ post may be useful:
https://www.physicsforums.com/threads/matrix-representations-of-linear-transformations.694922/

 

[vela]

I can do an example like 2c:
T(1) = 1 + 0 = 1 = 1/2(1+t) + 1/2(1- t) + 0(t^2) = (1/2 1/2 0)
T(t) = t + 1 = 1(1+t) 0(1-t) + 0(t^2) = (1 0 0 )
T(2) = t^2 + 2t = 1(1+t) + (-1)(1-t) + 1(t^2) = (1 -1 1)

so [T](B-->C) is equal to the matrix :
[1/2 1 1]
[1/2 0 -1]
[ 0 0 1]

This isn't really an explanation. It seems, perhaps, that you understand the mechanics of solving some types of problems, but you get stuck because you don't completely understand conceptually what you're doing.

You wrote "T(t) = t + 1 = 1(1+t) + 0(1-t) + 0(t^2) = (1 0 0)." Express in words the answer to the following two questions:

1. Why did you calculate T(t)?
2. Why did you express the result t+1 as 1(1+t) + 0(1-t) + 0(t^2)?

What would be the equation analogous to "T(t) = t + 1 = 1(1+t)+ 0(1-t) + 0(t^2) = (1 0 0)" for problem 3b?

 

[geoffrey159]

The FAQ post may be useful

I would say 'The FAQ post is your last chance'