SOLUTION: The space of polynomials in one variable, x, of degree at most 4, is mapped into itself by the linear map, L, that takes a typical polynomial P to the polynomial x(dP/dx)-4P, where

Algebra ->  Rate-of-work-word-problems -> SOLUTION: The space of polynomials in one variable, x, of degree at most 4, is mapped into itself by the linear map, L, that takes a typical polynomial P to the polynomial x(dP/dx)-4P, where      Log On


   

Question 363408: The space of polynomials in one variable, x, of degree at most 4, is mapped into itself by the linear map, L, that takes a typical polynomial P to the polynomial x(dP/dx)-4P, where dP/dx denotes the standard derivative of P.
Thus: L(P)= x(dP/dx) - 4P
1. Prove (show) that L is a linear map.
2. Wat is the kernal of this map? (i.e. describe the polynomials that it contains)
3. What is the dimension of the kernal?
4. What is the dimension of the range of this map?
5. Is there a polynomial of degree 2 that is not in the range of the map? give reasons for your answers to questions 2-5
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
1. L%28p%2Bq%29+=+xd%28p%2Bq%29%2Fdx-4%28p%2Bq%29+=+x%28dp%2Fdx+%2B+dq%2Fdx%29+-+4p-4q. Hence
.
Also , where alpha is a scalar. Therefore L is a linear map.
2. Let L%28p%29+=+O, the zero polynomial. This is the same as L%28p%29+=+xdp%2Fdx-4p+=+O. Solving,
xdp%2Fdx+=+4p,
dp%2Fp+=+4dx%2Fx,
lnp+=+4lnx+%2B+lnk, where k is a constant.
lnp+=+ln%28k%2Ax%5E4%29, or p+=+k%2Ax%5E4. Thus all polynomials of this form comprise kerL.
3. Consequently from part 2 above, dim ker L = 1.
4. Since dim kerL+ dim rangeL = 5 in this case, dim range L = 4. This also becomes obvious if we put in p+=+a+%2Bbx%2B+cx%5E2+%2B+dx%5E3+%2B+ex%5E4into L(p)and find out that L%28p%29+=+-4a-3bx-2cx%5E2-dx%5E3.
5. If we let d = e = 0, then L%28a+%2Bbx%2B+cx%5E2+%2B+dx%5E3+%2B+ex%5E4%29=+L%28a+%2Bbx%2B+cx%5E2+%29+=+-4a-3bx-2cx%5E2. Thus by inspection any polynomial of degree 2 will always be in the range of the map.