anonymous
  • 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)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Loser66
  • 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
  • Loser66
All you need is "translate" the right hand side to \(F(aP(x)) +F(Q(x)\)

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Loser66
  • 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
  • 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
  • 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
  • Loser66
Now, combine all
Loser66
  • 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
  • Loser66
hence, your result can be broken down like that
Loser66
  • 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
  • Loser66
From the right, the first matrix is F(aP(x), and the second one is F(Q(x) Done, right?
anonymous
  • anonymous
ok that seems quite logical , i really got confused by the writing
Loser66
  • 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
  • anonymous
okay thank you , and do you know b, c and d? ( I'm very grateful that you help me)
Loser66
  • Loser66
For b) matrix A is invertible since its determinant =-12 \(\neq 0\)
Loser66
  • Loser66
So does B, det(B) = 2
anonymous
  • anonymous
oh but the question is to combine those two bases to get the associated matrix
Loser66
  • Loser66
And you know how to find their inverse, right?
anonymous
  • 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
  • Loser66
I don't get the question.
anonymous
  • anonymous
it's called the associated matrix in dutch. But i can't find a decent translation for it
Loser66
  • Loser66
You meant the transformation that changes base A to base B??
anonymous
  • anonymous
yes i think so
Loser66
  • Loser66
so, just put [B|A] and try to change to [Identity| required matrix] by rref. Dat sit
anonymous
  • anonymous
this is an example of what we did on another exercise
Loser66
  • Loser66
|dw:1437695240946:dw|
Loser66
  • Loser66
wait, my master here, let me tag him @oldrin.bataku
anonymous
  • 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
  • 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
  • anonymous
those are on the first picture
anonymous
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
|dw:1437696322479:dw| this is the solution
Loser66
  • Loser66
Hold on, the second term is P'(1) , not P'(0)
Loser66
  • Loser66
for 3x^2-2
anonymous
  • 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
  • anonymous
so we have: $$3x^2-2\mapsto (-2,6,-5/3)$$
anonymous
  • anonymous
which isn't the solution i'm afraid
anonymous
  • anonymous
lol well obviously that's not the solution
anonymous
  • anonymous
now express that in terms of the basis for \(\mathbb{R}^3\), and that's your first column
anonymous
  • 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
  • anonymous
i think you maybe understand my question wrong, and that's why you don't get to the solution
anonymous
  • anonymous
and we solved it like this
anonymous
  • anonymous
and i also think that px represents ax²+bx+c because its part of R2(x)
Loser66
  • 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
  • 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
  • Loser66
Do the same with F(4x) and F(1), you will get other 2 rows. Combine all, that is the required matrix.
anonymous
  • 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
  • Loser66
what is your matrix?
anonymous
  • anonymous
the one we just calculated
anonymous
  • anonymous
|dw:1437751753106:dw|
Loser66
  • 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
  • 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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