Recent content by victoranderson

That is my question. How do we determine the basis for the image and the dimensions? If you say the image of F is four-dimensional, then basis for the image of F can only be span {(1/2,-1,0,0),(5/6,-2,1,-2/3),(17/12,-4,7/2,-7/3)} = span {(1/2,-1,0,0),(5/6,-2,1,-2/3)} Is it correct now?Last time...

so the basis for the image is {##\frac{1}{2},\frac{5}{6},\frac{17}{12}##} and {##-1,-2,-4##}? Another question. Since F: P2(R) -> P3(R), the dim(F) is 3 or 4? After doing this question I think dim(F) = 3, since dim(ker(F)) = 1 and dim(Im(F))=2

Hello I have a question about part (b) Now I know that F(f)(x) = [##\frac{1}{2}a+\frac{5}{6}b+\frac{17}{12}c]+[-a-2b-4c]x+[b+\frac{7}{2}c]x^{2}+[\frac{-2}{3}b-\frac{7}{3}c]x^{3}## so Kernel of F : F(f)(x) =0 => ##\frac{1}{2}a+\frac{5}{6}b+\frac{17}{12}c=0## ##-a-2b-4c=0##...

I have a question about diagonalizable matrix So now I have showed M is a diagonalizable and I am asked to find a matrix N^3=M Obviously if M=PDP^(-1) then N=PD^(1/3)P^(-1) but I am wondering how to show it formally.

Thanks a lot. I think I am done in this question but I really want to know more... If I prove f(0)=1 by contradiction which theorem can we use? I think extreme values theorem , IVT are unrelated..

Please see attached. The question is to determine f(0). In my opinion the question wants me to find f(x).. If f(x)=1, then 1. f:[0,1] -> R is continuous (do I need to prove?) 2. Obviously, f(1) = 1 3. f(x) is a rational number for all xε[0,1] so f(0) = 1 Am I correct?

I regret studying maths-related degree in university.. The reason for the eigenvalue is zero implies dim (Ker D) = n+1 is because n+1-fold composition is the zero map? I think I stuck in here so I cannot give a correct explanation

OK. This is my explanation. Please have a lookFor n=0 Po(R) = a_0 = ker D , where a_0 is in R Basis of Po(R) = {1}, which consists of an eigenvector with eigenvalue 0 for DFor n≥1 Pn(R) = {a_0+...+a_n*x^n} Dim of Pn(R)=n+1=dim(Ker(D))+dim(Im(D)) The basis of Pn(R) should consist of n+1...

just guessing...i m a beginner in this topic do you mean the eigenvectors must be elements of either Im(D) or ker(D), but not both?

Thanks for your reply Yes, I know the basis of Ker (D) = {1} For part (e), we already know D is diagonalizable for the case n=0, because a 1x1 matrix is obviously diagonal So we let the eigenvector of D be v_1, say For n>0 Assume D is diagonalizable So D must have at least two eigenvectors...

Please see attached. From part (a), we know that kernel (D) is a constant function, i.e f(x)=c, say From part (d), we know that eigenvalue of D is zero My question: For part (e), Is it correct to say that D is diagonalizable if and only if P_n (R) = ker (D) ?? so the only solution is...

finally, x^3 satisfy all conditions, right? f(x)=x^3 f'(x)=3x^2 inverse of f(x) = x^(1/3) derivative of the inverse = 1/3 * x^(-2/3) , so not differentiable at x=0

In my opinion trig function is the easiest way to satisfy [-1,1] to [-1,1]

i m almost crazy..may be i m an idiot i cannot think of a equation which satisfy these conditions 1. 1 to 1 2. f:[-1,1] to [-1,1] 3. derivative is 0 in the middle 4. has an inverse i now think of e^x - x but this kind of equation seems i m overthinking

Yes, I have been thinking this part for 3 hours... Is the function f(x)=cosx satisfy the requirements? f'(x)=-sinx so at x=0, f'(x)=0 derivative of the inverse = 1/sin(arcosx), so x=1 or -1, the denominator is 0 correct?