# Finding an invariant subspace of a linear operator.

The question is:

Let $T: \mathbb{R}^3\rightarrow \mathbb{R}^3$ defined by a rotation of 30 degrees about the vector $a = (1,2,3)$; Find an invariant subspace of $T$. Is $T$ a bijection?

What I first did was find the matrix representation of T by using the equation about an unit vector.

$a = (1,2,3)$ and the corresponding unit vector is $\frac1{\sqrt14} \begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix}$

The formula for the rotation matrix around an unit vector is:

$$Q = (cos\Theta)I + (1-cos\theta) \begin{bmatrix} a_1^2 & a_1a_2 & a_1a_3\\ a_1a_2&a_2^2&a_2a_3\\ a_1a_3&a_2a_3&a_3^2 \end{bmatrix} - sin\theta \begin{bmatrix} 0 & a_3 & -a_2\\ -a_3&0&a_1\\ a_2&-a_1&0 \end{bmatrix}$$

Using $\theta = 30, a_1=\frac1{\sqrt14}, a_2=\frac2{\sqrt14}, a_3=\frac3{\sqrt14}$

I got $$Q = \begin{bmatrix} \frac{\sqrt3}2 & 0&0 \\ 0&\frac{\sqrt3}2&0\\ 0&0&\frac{\sqrt3}2 \end{bmatrix} + \frac{2-\sqrt3}{2\sqrt14} \begin{bmatrix} 1&2&3\\2&4&6\\3&6&9\end{bmatrix} - \frac1{2\sqrt14} \begin{bmatrix} 0&3&-2\\-3&0&1\\2&-1&0\end{bmatrix}$$

From here, how would I find all the invariant subspaces of $T$?

If $T$ is a rotation around a vector in $\mathbb{R^3}$, then the subspace generated by that vector is invariant. There are no other one dimensional invariant subspaces, can you find a two dimensional one? I suggest thinking about this geometrically first, and then diving into computations.
• My professor does want us to find an invariant subspace other than the 0 subspace and the $\mathbb{R}^3$. So would the invariant subspace be the subspace generated by that vector? Would that mean all the linear combinations of that vector with a scalar? Dec 16 '15 at 5:35
• The subspace $W$generated by the vector $(1,2,3)$ is an invariant subspace, yes. And by the definition of a subspace generated by a vector, $W = \{(x,y,z) \in \mathbb{R^3} : (x,y,z) = \alpha (1,2,3), \alpha \in \mathbb{R} \}$, so yes, scalar multiples of the vector. Dec 16 '15 at 5:39
• What would be an explanation of this? Does the degree of rotation matter in this case or did he just put it there for no reason? He did hint that $T(v)$ = (the vector you get by rotating $v$ 30 degrees counterclockwise about $a = (1,2,3)$. Dec 16 '15 at 5:40
• Perhaps he wanted you to find all the invariant subspaces of $T$? In that case, the degree of rotation becomes relevant. Or perhaps he wants you to prove things using a matrix representation of your transformation, for which you would need this information. As for the explanation, what happens to a vector if you rotate it about itself? Dec 16 '15 at 5:42