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anonymous
 one year ago
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)
anonymous
 one year ago
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)

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9To 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
 one year ago
Best ResponseYou've already chosen the best response.9All you need is "translate" the right hand side to \(F(aP(x)) +F(Q(x)\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9for 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
 one year ago
Best ResponseYou've already chosen the best response.9the 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
 one year ago
Best ResponseYou've already chosen the best response.9the 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
 one year ago
Best ResponseYou've already chosen the best response.9You 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
 one year ago
Best ResponseYou've already chosen the best response.9hence, your result can be broken down like that

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9\[\left[\begin{matrix}aP(x) + Q(x)\\aP'(1) +Q'(1)\\\int_0^1aP(x)dx +\int_0^1Q(x)\end{matrix}\right]=\left[\begin{matrix}aP(x) \\aP'(1) \\\int_0^1aP(x)dx \end{matrix}\right]+\left[\begin{matrix} Q(x)\\Q'(1)\\\int_0^1Q(x)\end{matrix}\right]\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9From the right, the first matrix is F(aP(x), and the second one is F(Q(x) Done, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok that seems quite logical , i really got confused by the writing

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9Of course, you must jot down in neat. I inserted the explanation when solving it. It makes the proof has many irrelevant terms. hehehe...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you , and do you know b, c and d? ( I'm very grateful that you help me)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9For b) matrix A is invertible since its determinant =12 \(\neq 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh but the question is to combine those two bases to get the associated matrix

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9And you know how to find their inverse, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 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
 one year ago
Best ResponseYou've already chosen the best response.9I don't get the question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's called the associated matrix in dutch. But i can't find a decent translation for it

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9You meant the transformation that changes base A to base B??

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9so, just put [BA] and try to change to [Identity required matrix] by rref. Dat sit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is an example of what we did on another exercise

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9wait, my master here, let me tag him @oldrin.bataku

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with 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
 one year ago
Best ResponseYou've already chosen the best response.0b) determine the associated matrix relative to the basis A and B . Is this matrix invertible? yes, give the reverse what are your bases?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0those are on the first picture

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0so here our ordered basis for the first space is \((3x^22,4x,1)\)  now look at how our map \(F\) transforms these

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that if \(P(x)=3x^22\) then $$P(0)=3(0)^22=2\\P'(x)=6x\implies P'(0)=0\\\int_0^1 P\, dx=\left[\frac13 x^32x\right]_0^1=\frac132=\frac53$$in other words, \(F\) maps \(3x^22\to (2,0,5/3)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0now, 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0so in terms of coefficients of our basis vectors we have that \(3x^22\mapsto (2,0,5/3)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437696322479:dw this is the solution

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9Hold on, the second term is P'(1) , not P'(0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, then our results are actually: $$P(x)=3x^22\implies P'(x)=6x\implies P'(1)=6\\P(x)=4x\implies P'(x)=4\implies P'(1)=4\\P(x)=1\implies P'(x)=0\implies P'(1)=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we have: $$3x^22\mapsto (2,6,5/3)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which isn't the solution i'm afraid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol well obviously that's not the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now express that in terms of the basis for \(\mathbb{R}^3\), and that's your first column

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then 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
 one year ago
Best ResponseYou've already chosen the best response.0i think you maybe understand my question wrong, and that's why you don't get to the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we solved it like this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i also think that px represents ax²+bx+c because its part of R2(x)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9I am with @oldrin.bataku He made mistake at \(F(3x^2 2) \) \(P(x) = 3x^2 2> P(0) =2\\P'(x) = 6x \rightarrow P'(1) =6\\\int_0^1 3x^22 dx= (\dfrac{3x^3}{3}2x)^1_0= 1\) Hence \(F(3x^22)= (2,6,1)~~or~~\left(\begin{matrix}2\\6\\1\end{matrix}\right)\). 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
 one year ago
Best ResponseYou've already chosen the best response.9sorry, 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
 one year ago
Best ResponseYou've already chosen the best response.9Do the same with F(4x) and F(1), you will get other 2 rows. Combine all, that is the required matrix.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i found the solutions for that and the inversible matrix but still got stuck for the vector 6x²8x1 . the solution should be (1,4,3) but i really can't find it . This exercise drives me nuts :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the one we just calculated

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437751753106:dw

Loser66
 one year ago
Best ResponseYou've already chosen the best response.9(1,4,3) is the answer for standard base, not for \(\beta\) base Let's apply F to (6x^28x1) P(0) =1 P'(x) = 12x8> P'(1) = 4 \(\int_0^1 (6x^28x1)dx = (2x^34x^2x)^1_0=3\) hence, F(6x^28x1)= (1,4,3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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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