Question

4. Define a transformation from P2 to R3 by:
p(x)−>(∫01p(x)d(x),∫-10p(x)d(x),∫-11p(x)d(x))
(a) Is this a homomorphism?
(b) Is it onto?
(c) Is it one-to-one?
(d) Is it an isomorphism?
(e) Find a matrix for the transformation between the representation of the standard basis of P2 and the standard basis of R3.

Answer 1

@eliassaab Can you help me please?

Answer 2

First question: what is P2?

Answer 3

space 2

Answer 4

I definitely haven't learned about this yet. @Zarkon @satellite73 might be able to help when they get on.

Answer 5

ok Thsnk you @KingGeorge

Answer 6

Are the integral 0 to 1, then from -1,0 and then from -1, 1

Answer 7

It is linear since the integral is linear.
If we answer the last question first, we will be able to answer the rest.
The matrix is
\[
\left(
\begin{array}{ccc}
1 & \frac{1}{2} & \frac{1}{3}
\\
1 & -\frac{1}{2} &
\frac{1}{3} \\
2 & 0 & \frac{2}{3} \\
\end{array}
\right)
\]
The first column is the image of the polynomial 1
The second column is the image of the polynomial x
The third column is the image of the polynomial x^2

Answer 8

Yes

Answer 9

If you add the first two rows you get the third one. This means the
matrix is not invertible. So the map is not an isomorphism.

Answer 10

How did you get these number?

Answer 11

The firs column is the image of 1
\[
|int_0^1 1 dx =1\\
\int_{-1}^0 1 dx =1\\
\int_{-1}^{1} 1 dx =2\\
\]

Answer 12

You do the same thing for the second row by integrating x instead of one

Answer 13

How did you get these numbers of the matrix?

Answer 14

third column you integrate x^2 instead of 1

Answer 15

I explained it to you. See my previous posts.

Answer 16

I don't understand it? why the second one is = 1?

Answer 17

What is the integral from -1 to 0 of 1?

Answer 18

1

Answer 19

\[
\int_{-1}^0 1 dx = \left [x \right]_{x=-1}^{0}= 0 -(-1)=1
\]

Answer 20

That is why the second element of the first column is 1

Answer 21

is the second element should be x^2?

Answer 22

The second column is the image of x by your map

Answer 23

ok

Answer 24

\[
|\int_0^1x dx =\frac 1 2\\
\int_{-1}^0 x dx =-\frac1 2\\
\int_{-1}^{1}x dx =0\\
\]

Answer 25

ok

Answer 26

For the third column
\[
|\int_0^1x^2 dx =\frac 13\\
\int_{-1}^0 x^2 dx =\frac13\\
\int_{-1}^{1}x^2 dx =\frac 2 3\\
\]

Answer 27

ok got it

Answer 28

How can I check if it is onto or one to one?

Answer 29

should the matrix be like this ?
1 1 1
( 1 1 -1 )
1 2 0

Answer 30

@eliassaab

Answer 31

No the matrix should look like
\[\left(
\begin{array}{ccc}
1 & \frac{1}{2} & \frac{1}{3}
\\
1 & -\frac{1}{2} &
\frac{1}{3} \\
2 & 0 & \frac{2}{3} \\
\end{array}
\right)
\]

Answer 32

Then rank of the matrix is 2, the map is not one to one.

Answer 33

A linear map from a vector space of dimension 3 to another one of dimension 3 is onto if and only if is one to one if and only if it is an isomorphism

Answer 34

Try to find the image of the polynomial 3x^2 -1 by your map,
you will find that its image is the vector (0,0,0)
This by itself shows that your map is not one to one.

Answer 35

I am now going hiking to
http://www.beirutnightlife.com/tourism/kornet-el-sawda/
I will not be available for about 10 hours.

Answer 36

Also you can show that the vector (1,1,1) cannot be the image of any polynomial by your map.