How does T rotate points in \mathbb{R}^2?

[flyingpig]

Homework Statement

Find the standard matrix of T

T: \mathbb{R}^2 \to \mathbb{R}^2 rotates points (about the origin) through \frac{3 \pi}{2} radians counterclockwise

The Attempt at a Solution

I just substitute \frac{3 \pi}{2} into the rotation matrix and I got \begin{bmatrix}<br /> 0 &amp; 1 \\ <br /> -1 &amp; 0 <br /> \end{bmatrix}

The book got this answer too, but they did something weird

They did T(\vec{e_1}) = -\vec{e_1} and T(\vec{e_2}) = \vec{e_1}

I don't understand how they got T(\vec{e_1}) = -\vec{e_1}

 

[fzero]

It looks like a typo. What did you find for T(\vec{e_1})?

 

[flyingpig]

I meant T(\vec{e_1}) = -\vec{e_2}

 

[fzero]

I meant T(\vec{e_1}) = -\vec{e_2}

I'm confused whether that's what you obtained or what the book obtained. In either case, that's correct, since

\vec{e}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, ~~ \vec{e}_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.

You can explicitly rotate these by +3\pi/2 to find the action of T. The matrix form can then be deduced straightforwardly.

[STRIKE]Incidentally, you should be using \theta = -3\pi/2, since it's a counterclockwise rotation. You got the signs correct somehow anyway, but if the angle were different, you'd have gotten the wrong answer.[/STRIKE]

Edit: Sorry, counterclockwise corresponds to a positive rotation, so ignore what I said here.

 

[flyingpig]

No That's what the book meant

I don't understand what they are doing. The way I did it is plug them into

\begin{bmatrix}<br /> cos\theta &amp; -sin\theta \\ <br /> sin\theta &amp; cos\theta <br /> \end{bmatrix}

 

[fzero]

OK, it's easy if you draw a picture that I'll explain in words. \vec{e}_1 is the unit vector in the direction of the +x-axis, while \vec{e}_2 is that for the +y-axis. We put the bases of both vectors at the origin for convenience. Now, if we rotate \vec{e}_1 by 3\pi/2 radians counterclockwise, that corresponds to 270^\circ, which corresponds to the -y-axis or -\vec{e}_2. We can do the same thing for \vec{e}_2 to find T(\vec{e}_2).

Now if you know the action of T on the basis vectors, that's enough information to solve for all of the entries of the matrix by writing the system of equations

T(\vec{e}_i) = \sum_j T_{ij} \vec{e_j}.

This is just a decomposition of whatever we compute for T(\vec{e}_i) in terms of a linear combination of the basis vectors. The coefficients T_{ij} are precisely the matrix elements of T. Because we are using the Cartesian orthonormal basis for \mathbb{R}^2 this is the standard form.