- anonymous

hello
can anyone help me solve this?
the questions in english are :
a) show that F is linear
b) determine the associated matrix relative to the basis A and B . Is this matrix invertible? yes, give the reverse
c) determine by hand:
the image of the vector
the polynomial Px
d)determine the kernel ker(f) and image Im(f)

- jamiebookeater

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- anonymous

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- Loser66

To show F is linear, we need show F(aP + Q)(x) = aF(P(x) +F(Q(x)
YOu can break it out by doing 2 parts : F(P + Q)(x) = F(P)+F(Q) and F(aP(x))= aF(P(x))
I do the first one
Let P, Q in R^2[x], a in R
then aP (x) + Q(x) = (aP+Q)(x)
Now, apply F on
\(F(aP + Q)(x) = ((aP+Q)(0), (aP+Q)'(1),\int_0^1 (aP+Q)(x))\)

- Loser66

All you need is "translate" the right hand side to \(F(aP(x)) +F(Q(x)\)

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## More answers

- Loser66

for the first term: \((aP + Q)(x)\) , the property of function give us \((aP+Q)(x) = aP(x) +Q(x)\)
It likes (f+g)(x) = f(x) +g(x)

- Loser66

the second term is derivative, and we know that derivative of the sum = sum of derivative, right? like (f+g)' = f'+g'
hence we have \((aP+Q)'(1) = (aP)'(1) +Q'(1)\)

- Loser66

the last term is integral, exactly the same, we still have
\[\int_0^1 (aP+Q)(x)dx=\int_0^1aP(x) dx +\int_0^1 Q(x) dx\]

- Loser66

Now, combine all

- Loser66

You know that if I have vectors
\[\left[\begin{matrix}a +b \\c +d\end{matrix}\right] =\left[\begin{matrix}a \\c \end{matrix}\right]+\left[\begin{matrix}b \\c \end{matrix}\right]\], right?

- Loser66

hence, your result can be broken down like that

- Loser66

\[\left[\begin{matrix}aP(x) + Q(x)\\aP'(1) +Q'(1)\\\int_0^1aP(x)dx +\int_0^1Q(x)\end{matrix}\right]=\left[\begin{matrix}aP(x) \\aP'(1) \\\int_0^1aP(x)dx \end{matrix}\right]+\left[\begin{matrix} Q(x)\\Q'(1)\\\int_0^1Q(x)\end{matrix}\right]\]

- Loser66

From the right, the first matrix is F(aP(x), and the second one is F(Q(x)
Done, right?

- anonymous

ok that seems quite logical , i really got confused by the writing

- Loser66

Of course, you must jot down in neat. I inserted the explanation when solving it. It makes the proof has many irrelevant terms. hehehe...

- anonymous

okay thank you , and do you know b, c and d? ( I'm very grateful that you help me)

- Loser66

For b) matrix A is invertible since its determinant =-12 \(\neq 0\)

- Loser66

So does B, det(B) = 2

- anonymous

oh but the question is to combine those two bases to get the associated matrix

- Loser66

And you know how to find their inverse, right?

- anonymous

yes the inverse and determinant aren't the problem , that's just math but i need the matrix that is formed by those two bases given F

- Loser66

I don't get the question.

- anonymous

it's called the associated matrix in dutch. But i can't find a decent translation for it

- Loser66

You meant the transformation that changes base A to base B??

- anonymous

yes i think so

- Loser66

so, just put [B|A] and try to change to [Identity| required matrix] by rref. Dat sit

- anonymous

this is an example of what we did on another exercise

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- Loser66

|dw:1437695240946:dw|

- Loser66

wait, my master here, let me tag him
@oldrin.bataku

- anonymous

with base of starting vector space [(1,0,0),(1,1,0),(1,1,1) ] to the ending vector space [(0,1,0,0),(0,0,1,0),(0,0,0,1),(1,0,0,0)]

- anonymous

b) determine the associated matrix relative to the basis A and B . Is this matrix invertible? yes, give the reverse
what are your bases?

- anonymous

those are on the first picture

- anonymous

okay, so the columns of the matrix representing a given linear map or transformation correspond to the way the matrix transforms the basis vectors (in order)

- anonymous

so here our ordered basis for the first space is \((3x^2-2,4x,1)\) -- now look at how our map \(F\) transforms these

- anonymous

consider that if \(P(x)=3x^2-2\) then $$P(0)=3(0)^2-2=-2\\P'(x)=6x\implies P'(0)=0\\\int_0^1 P\, dx=\left[\frac13 x^3-2x\right]_0^1=\frac13-2=-\frac53$$in other words, \(F\) maps \(3x^2-2\to (-2,0,-5/3)\)

- anonymous

now take \(P(x)=4x\), so: $$P(0)=0\\P'(x)=4\implies P'(0)=4\\\int_0^1 P\, dx=\left[2x^2\right]_0^1=2$$ so \(4x\mapsto (0,4,2)\)
and lastly \(P(x)=1\) gives $$P(0)=1\\P'(x)=0\implies P'(0)=0\\\int_0^1 P\, dx=1$$ so \(1\mapsto (1,0,1)\)

- anonymous

now, we want to express these output vectors in terms of the given basis on \(\mathbb{R}^3\), which is \((1,0,0),(0,2,1),(1,0,1)\)

- anonymous

so we want to figure out \(c_1,c_2,c_3\): $$(-2,0,-5/3)=c_1 (1,0,0)+c_2 (0,2,1)+c_3(1,0,1)$$

- anonymous

so we can do this by considering \(c_1+c_3=-2,c_3=-5/3\) which gives: $$c_1=-1/3\\c_3=-5/3\\c_2=0$$

- anonymous

so in terms of coefficients of our basis vectors we have that \(3x^2-2\mapsto (-2,0,-5/3)\) translates to \(\begin{bmatrix}1\\0\\0\end{bmatrix}\to\begin{bmatrix}-1/3\\0\\-5/3\end{bmatrix}\) so our first column is that

- anonymous

|dw:1437696322479:dw| this is the solution

- Loser66

Hold on, the second term is P'(1) , not P'(0)

- Loser66

for 3x^2-2

- anonymous

oops, then our results are actually: $$P(x)=3x^2-2\implies P'(x)=6x\implies P'(1)=6\\P(x)=4x\implies P'(x)=4\implies P'(1)=4\\P(x)=1\implies P'(x)=0\implies P'(1)=0$$

- anonymous

so we have: $$3x^2-2\mapsto (-2,6,-5/3)$$

- anonymous

which isn't the solution i'm afraid

- anonymous

lol well obviously that's not the solution

- anonymous

now express that in terms of the basis for \(\mathbb{R}^3\), and that's your first column

- anonymous

then express the results for the next two in terms of those bases and those are your second nad third columns, and then you have your matrix

- anonymous

i think you maybe understand my question wrong, and that's why you don't get to the solution

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- anonymous

and we solved it like this

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- anonymous

and i also think that px represents ax²+bx+c because its part of R2(x)

- Loser66

I am with @oldrin.bataku
He made mistake at \(F(3x^2 -2) \)
\(P(x) = 3x^2 -2--> P(0) =-2\\P'(x) = 6x \rightarrow P'(1) =6\\\int_0^1 3x^2-2 dx= (\dfrac{3x^3}{3}-2x)|^1_0= -1\)
Hence \(F(3x^2-2)= (-2,6,-1)~~or~~\left(\begin{matrix}-2\\6\\-1\end{matrix}\right)\). However, it is in standard base, not \(\beta\) base. We need transfer it to \(\beta\) base by doing:
\[\left(\begin{matrix}-2\\6\\-1\end{matrix}\right)= a_1\left(\begin{matrix}-1\\0\\0\end{matrix}\right)+a_2\left(\begin{matrix}0\\2\\1\end{matrix}\right)+a_3\left(\begin{matrix}1\\0\\1\end{matrix}\right)\]

- Loser66

sorry, the first entry of a_1 is 1, not -1
Solving it, we get \(a_1+a_3=-2\\2a_2=6\\a_2+a_3=-1\)
Hence \(a_1= 2:a_2=3:a_3=-4\) .This is the first row of the matrix.

- Loser66

Do the same with F(4x) and F(1), you will get other 2 rows. Combine all, that is the required matrix.

- anonymous

okay i found the solutions for that and the inversible matrix but still got stuck for the vector 6x²-8x-1 . the solution should be (-1,4,-3) but i really can't find it . This exercise drives me nuts :(

- Loser66

what is your matrix?

- anonymous

the one we just calculated

- anonymous

|dw:1437751753106:dw|

- Loser66

(-1,4,-3) is the answer for standard base, not for \(\beta\) base
Let's apply F to (6x^2-8x-1)
P(0) =-1
P'(x) = 12x-8--> P'(1) = 4
\(\int_0^1 (6x^2-8x-1)dx = (2x^3-4x^2-x)|^1_0=-3\)
hence, F(6x^2-8x-1)= (-1,4,-3)

- anonymous

so u dont need the matrix to solve this? im so confused .Do you know any decent theory and exercises with the same level of difficulty for this? because in my text book it's just so badly explained

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