# Determinant of matrix of linear transformation

## Homework Statement

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

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 :(

gabbagabbahey
Homework Helper
Gold Member
So, T(f) = -(a, b, c)T + 8 (0, b, 2c)T

What happened to the factor of 5 from T(f)=-5f+8f'?

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 :(

gabbagabbahey
Homework Helper
Gold Member
Well, if $f(x)=a+bx+cx^2$, then $f'(x)=b+2cx$

So, if you are representing your basis functions as

$$x^0\to\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; x^1\to\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix} \;\;\;\text{and}\;\;\;\; x^2\to\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}$$

Then you have $$f(x)\to\begin{pmatrix}a \\ b \\ c \end{pmatrix}$$ and $$f'(x)\to\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}$$

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 :(((

gabbagabbahey
Homework Helper
Gold Member
Okay, thank you again!
Now I have:

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

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

Errm...

$$-5\begin{pmatrix}a \\ b \\ c \end{pmatrix}+8\begin{pmatrix}b \\ 2c \\ 0 \end{pmatrix}=\begin{pmatrix}-5a+8b \\ -5b+16c \\ -5c \end{pmatrix}$$

Oh, I'm dumb! Thank you very much!

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...

gabbagabbahey
Homework Helper
Gold Member
Is T again a mapping from P2 to P2?

If so, do it in the same way, just with $t$ as the variable instead of $x$... if $f(t)=a+bt+ct^2$, what is $f(2t)$?

f(2t) = a + b2t = c2t2?

Mark44
Mentor
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.

Thank you very much!

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...

gabbagabbahey
Homework Helper
Gold Member
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 $V_1$ and $V_2$?

They are just some vectors in R2, they're not given.

gabbagabbahey
Homework Helper
Gold Member
They are just some vectors in R2, they're not given.

I don't believe that. If this was true for all $V_1$ and $V_2$ in $\mathbf{R}^2$, then the matrix $A$ would have to be full of zeroes. You'd need only look at some case where $V_2=V_1$ to prove that to yourself.

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

$$V_1\to\begin{pmatrix}1 \\ 0 \end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}0 \\ 1 \end{pmatrix}$$

gabbagabbahey
Homework Helper
Gold Member
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

$$V_1\to\begin{pmatrix}a_1 \\ b_1\end{pmatrix} \;\;\;\;\text{and}\;\;\;\; V_2\to\begin{pmatrix}a_2 \\ b_2\end{pmatrix}$$

and go from there.