READ THIS FIRST, PLEASE: I\'m not looking so much for the answer to this problem

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READ THIS FIRST, PLEASE: I'm not looking so much for the answer to this problem as I am how to work it (but the correct answer is required.) I'm having a hard time understanding how this problem is worked and looking for help in this regard. PLEASE READ PAST THE QUESTION PASTED BELOW TOO, THANKS.

Here's the question:

For the first part of the problem, I know that

L(x^2 , x, 1) = [1 +x, 1-x] * [ matrix we need to find ].

From the equation above, I know that we get L(x^2) = 4x. From that, somehow we get the vector <2, -2>. I do not understand how to exactly go about finding that vector, I thought it was a simple matter of breaking down the exponent but it doesn't seem to be that way.

Moving on, L(x) = 3. That's fine, but somehow this gives us the vector of <3/2, 3/2>. I know this should be obvious, but I can't see why. If you can break it down Barney style, so much the better.

Finally, L(1) gives us the vector of <0,0> which is pretty obvious.

A = [ 2 3/2 0

-2 3/2 0 ].

Finally, I have no clue how to even approach this final part of the question here with the coordinate vector.

Explanation / Answer

once you know the operation of L on the basis vectors of the P3, you need to express the result as a sum of the basis vectors of P2

For example, since (L(x^2)=4x)

now, express 4x as a sum of (1+x) and (1-x) (the given basis of p2)

the coefficients of (1+x) and (1-x) are the coordinates of 4x in the given basis of P2

we, have (4x = 2(1+x)+(-2)(1-x))

thus, you can see that the coordinates of 4x will be (2,-2)

similarly, (L(x)=3=(3/2)(1+x)+(3/2)(1-x))

thus the coordinates of L(x) will be (3/2,3/2)

and lastly, (L(1)=0=0*(1+x)+0*(1-x))

thus the coordiantes of L(0) are (0,0)

this is how you get these vectors.

Now to get the matrix reperesentation of L, you note that the coordiantes of (x^2) in the given basis of P3 are (1,0,0)

thus is P is the matrix representation of L then,

since P takes a three dimensional vector to a two dimensional one (takes a vector from P3 to P2), so P should be (2 imes 3) dimensional

so, let (p=egin{pmatrix}a&b&c\d&e&fend{pmatrix})

(Pegin{pmatrix}1\0\0end{pmatrix}=egin{pmatrix}2\-2end{pmatrix})

thus (egin{pmatrix}a\dend{pmatrix}=egin{pmatrix}2\-2end{pmatrix})

which gives a=2 and d=-2

similarly, in the basis of P3 the vector x has coordiantes (0,1,0)

doing the algebra as done above (try it I suggest, you will get better insight), we get

b=3/2, e=3/2

similarly, 1 is represented in the given basis of P3 as (0,0,1)

again by doing a similar algebra we will get

c=0 and f=0

You could also have got the matrix representation of L by simply arraging the coordinates of (L(x^2), L(x) and L(1)), which we found earlier in consecutive columns.

But, now I think, after doing the algebra I suggested, you have better understanding of why the arranging of the vectors in subsequent columns gives you the correct matrix representation of the operator L.

I hope this helps you.

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