Find a basis for teh corresponding cyclic subspace

For each of the following find a basis for the corresponding cyclic subspace:

a) T = d/dx : R_{<=3}[x] --> R_{<=3}[x], v = x^{2}

b) T = d/dx: R_{<=3}[x] --> R_{<=3}[x], v = x^{3}

c) T is rotation of R^{2} by 180 degrees, v = [3,4]

d) T is rotation by 30 degrees, v = [3,4]

e) T: R^{3} --> R^{3} given by T(x,y,z) = (x+2z, 2x-y, z), and v = [1,0,0]

My thoughts on what bases are:

a) Would a basis be (x^2, 2x, 2)?

b) Would a basis be (x^3, 3x^2, 6x)?

c) Would a basis be {(3,4),(-3,-4)}?

d) Would a basis be {(3,4),(-3,-4)}? *Not too sure on this one*

e) I have no idea ??

Re: Find a basis for teh corresponding cyclic subspace

By $\displaystyle R_{<=3}[x]$ do you mean the space of deg(3) or less polynomials with real coefficiants?

Re: Find a basis for teh corresponding cyclic subspace

Yes you are correct jakncoke.

Re: Find a basis for teh corresponding cyclic subspace

b and d seem to be incorrect. recall for d, the transformation matrix of rotation is given by $\displaystyle \begin{bmatrix} cos \theta & - sin \theta \\ sin \theta & cos \theta \end{bmatrix} $ For the last one, transformation matrix is given by $\displaystyle \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $

Re: Find a basis for teh corresponding cyclic subspace

So for b) would I have to go a step further and say the basis is {3x^2, 6x, 6}? or am I missing something between a and b that makes a correct and b incorrect?

Re: Find a basis for teh corresponding cyclic subspace

the basis i got for b) was {$\displaystyle x^3.3x^2. 6x.6$}

Re: Find a basis for teh corresponding cyclic subspace

For e) I am having trouble coming up with a basis. Do I multiply the given vector by the transformation matrix, or is the basis just {(1,0,2),(2,-1,0),(0,0,1)}?

Re: Find a basis for teh corresponding cyclic subspace

the basis is $\displaystyle (1,0,0) , (1,2,0) $

so you have $\displaystyle \{(1,0,0), Tv, T^2v\} $ we see that $\displaystyle Tv = (1,2,0) $ and $\displaystyle T^2v = (1,0,0) $

so $\displaystyle \{(1,0,0), (1,2,0), (1,0,0) $ since the third vector makes this linearly dependent, throw it away

and thus resulting $\displaystyle \{(1,0,0), (1,2,0) $ is the basis