# Find a basis for teh corresponding cyclic subspace

• Feb 20th 2013, 02:59 PM
TimsBobby2
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 = x2

b) T = d/dx: R<=3[x] --> R<=3[x], v = x3

c) T is rotation of R2 by 180 degrees, v = [3,4]

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

e) T: R3 --> R3 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 ??
• Feb 20th 2013, 03:37 PM
jakncoke
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?
• Feb 20th 2013, 03:49 PM
TimsBobby2
Re: Find a basis for teh corresponding cyclic subspace
Yes you are correct jakncoke.
• Feb 20th 2013, 04:58 PM
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 $\displaystyle \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
• Feb 20th 2013, 05:09 PM
TimsBobby2
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?
• Feb 20th 2013, 05:24 PM
jakncoke
Re: Find a basis for teh corresponding cyclic subspace
the basis i got for b) was {$\displaystyle x^3.3x^2. 6x.6$}
• Feb 20th 2013, 06:09 PM
TimsBobby2
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)}?
• Feb 20th 2013, 06:15 PM
jakncoke
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