How to prove this is a basis for $\mathbb{R}^3$ even after being rotated I have $\{\vec{v_1},\vec{v_2},\vec{v_3}\}=\{(1,0,0)^T,(0,1,0)^T,(0,0,1)^T\}$ as a basis for $\mathbb{R}^3$. I know intuitively that $\{A\vec{v_1},A\vec{v_2},A\vec{v_3}\}$ is also a basis for $\mathbb{R}^3$ where $A$ is the transformation represented by rotating 30 degrees counterclockwise about the axis spanned by $(1,1,1)^T$ because the second basis simply represents a rotated version of the original coordinate system of $\mathbb{R}^3$. 
How would I go about proving this though? I'm not sure how to rigorously show that this second basis is still a basis for $\mathbb{R}^3$ even though I know this is true geometrically.
 A: The key thing here is that $A$ is invertible; as such, linear independence of $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$ implies linear independence of $\{A\vec{v}_1,A\vec{v}_2,A\vec{v}_3\}$.
Why?  Suppose that $\{A\vec{v}_1,A\vec{v}_2,A\vec{v}_3\}$ is not a linearly independent set.  Then we can write
$$
\vec{0}=\alpha_1A\vec{v}_1+\alpha_2A\vec{v}_2+\alpha_3A\vec{v}_3=A(\alpha_1\vec{v}_1+\alpha_2\vec{v}_2+\alpha_3\vec{v}_3),
$$
for some $\alpha_1,\alpha_2,\alpha_3$ which are not all zero. Invertibility of $A$ then implies 
$$
\alpha_1\vec{v}_1+\alpha_2\vec{v}_2+\alpha_3\vec{v}_3=\vec{0},
$$
contradicting linear independence of $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$.
So, as long as you can argue that $A$ is invertible (which shouldn't be too hard), you can also argue that this new set is linearly independent. The fact that $\mathbb{R}^3$ is 3-dimensional gets you the rest of the way there.
A: You could get the matrix $A$ for the rotation and show that it's nonsingular. (This may be a little tricky; I remember doing a problem like this.  First move $(1,1,1)$ to the $z$-axis,  say, by a couple rotations.  Then rotate about the $z$-axis.  Then move back.)
Alternatively,  note that the kernel of such a rotation is necessarily $0$.  Therefore it's nonsingular and takes a basis to a basis. 
