# What are they doing ?

## 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 $$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$

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

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

fzero
Homework Helper
Gold Member
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.

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No That's what the book meant

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

$$\begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}$$

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.