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Determinant of matrix of linear transformation

  • Thread starter freetonik
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  • #1
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Homework Statement


Linear transformation T: P2->P2
T(f) = -5f + 8f'
Need to find detA (A is a matrix of T)

Homework Equations



T(f) = Af

The Attempt at a Solution


The basis of P2 is B={1, x, x2}. Some polynomial f with respect to B looks like this in general:

(a, b, c)T

right? So, T(f) = -(a, b, c)T + 8 (0, b, 2c)T = (-a, 7b, c)T.

So, A*(a, b, c)T = (-a, 7b, c)T, therefore A =

1 0 0
0 7 0
0 0 1

And detA = -7. This is how I was thinking about it, but the answer is wrong and I am confused :(
 

Answers and Replies

  • #2
gabbagabbahey
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So, T(f) = -(a, b, c)T + 8 (0, b, 2c)T
What happened to the factor of 5 from T(f)=-5f+8f'?
 
  • #3
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Oh!
Thanks!
It always happens to me.

So, fixing that mistake, T(f)=

-5a
3b
11c

And A becomes:

-5 0 0
0 3 0
0 0 11

and detA = -165 and it is still wrong :(
 
  • #4
gabbagabbahey
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Well, if [itex]f(x)=a+bx+cx^2[/itex], then [itex]f'(x)=b+2cx[/itex]

So, if you are representing your basis functions as

[tex]x^0\to\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; x^1\to\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} \;\;\;\text{and}\;\;\;\; x^2\to\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}[/tex]

Then you have [tex]f(x)\to\begin{pmatrix}a \\ b \\ c \end{pmatrix}[/tex] and [tex]f'(x)\to\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}[/tex]
 
  • #5
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Okay, thank you again!
Now I have:

-5(a, b, c)T + 8 (b, 2c, 0)T =

-5a+8b
-5a+16c
-5a

And:

A(a, b, c)T = (-5a+8b, -5a+16c, -5a)T, so A=

-5 8 0
-5 0 16
-5 0 0

detA = -640 and it is still wrong :(((
 
  • #6
gabbagabbahey
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Okay, thank you again!
Now I have:

-5(a, b, c)T + 8 (b, 2c, 0)T =

-5a+8b
-5a+16c
-5a
Errm...

[tex]-5\begin{pmatrix}a \\ b \\ c \end{pmatrix}+8\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}=\begin{pmatrix}-5a+8b \\ -5b+16c \\ -5c \end{pmatrix}[/tex]
 
  • #7
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Oh, I'm dumb! Thank you very much!
 
  • #8
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There is a similar problem I cannot figure out..

T(f(t)) = f(2t) - 3f(t)

Again need to find detA.

In this case I don't even know how to start...
 
  • #9
gabbagabbahey
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Is T again a mapping from P2 to P2?

If so, do it in the same way, just with [itex]t[/itex] as the variable instead of [itex]x[/itex]... if [itex]f(t)=a+bt+ct^2[/itex], what is [itex]f(2t)[/itex]?
 
  • #10
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f(2t) = a + b2t = c2t2?
 
  • #11
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f(2t) = a + b2t = c2t2?
No. If f(t) = a + bt + ct2, then f(2t) = a + b*2t + c(2t)2. This can be simplified.
 
  • #12
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Thank you very much!
 
  • #13
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Sorry, I've faced one more problem...
There is T: R2->R2, s.t. T(V1)=8V2, and T(V2)=-4V1.

Need to find detA.

This is how I started thinking:
AV1=8V2 => A=8V2 / V1
AV2=-4V1 => A=-4V1 / V2

Don't know how to think further...
 
  • #14
gabbagabbahey
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Sorry, I've faced one more problem...
There is T: R2->R2, s.t. T(V1)=8V2, and T(V2)=-4V1.
Okay, and what are [itex]V_1[/itex] and [itex]V_2[/itex]?
 
  • #15
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They are just some vectors in R2, they're not given.
 
  • #16
gabbagabbahey
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They are just some vectors in R2, they're not given.
I don't believe that. If this was true for all [itex]V_1[/itex] and [itex]V_2[/itex] in [itex]\mathbf{R}^2[/itex], then the matrix [itex]A[/itex] would have to be full of zeroes. You'd need only look at some case where [itex]V_2=V_1[/itex] to prove that to yourself.

I suspect that they are supposed to represent the basis vectors of [itex]\mathbf{R}^2[/itex]. In which case, you might as well look at two orthogonal basis vectors

[tex]V_1\to\begin{pmatrix}1 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}0 \\ 1 \end{pmatrix}[/tex]
 
  • #17
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  • #18
gabbagabbahey
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This is how the problem stated:
http://imgur.com/bU4b7.png
Okay, so they are some specific vectors, you just aren't told exactly what they are.

In that case, just say

[tex]V_1\to\begin{pmatrix}a_1 \\ b_1\end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}a_2 \\ b_2\end{pmatrix}[/tex]

and go from there.
 

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