What are they doing ?

  • Thread starter flyingpig
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  • #1
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Homework Statement



Find the standard matrix of T

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

The Attempt at a Solution



I just substitute [tex]\frac{3 \pi}{2}[/tex] into the rotation matrix and I got [tex]\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}[/tex]

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

They did [tex]T(\vec{e_1}) = -\vec{e_1}[/tex] and [tex]T(\vec{e_2}) = \vec{e_1}[/tex]

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

Answers and Replies

  • #2
fzero
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It looks like a typo. What did you find for [tex]T(\vec{e_1})[/tex]?
 
  • #3
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I meant [tex]T(\vec{e_1}) = -\vec{e_2}[/tex]
 
  • #4
fzero
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I meant [tex]T(\vec{e_1}) = -\vec{e_2}[/tex]

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

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

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

[STRIKE]Incidentally, you should be using [tex]\theta = -3\pi/2[/tex], 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.
 
Last edited:
  • #5
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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

[tex]\begin{bmatrix}
cos\theta & -sin\theta \\
sin\theta & cos\theta
\end{bmatrix}[/tex]
 
  • #6
fzero
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OK, it's easy if you draw a picture that I'll explain in words. [tex]\vec{e}_1[/tex] is the unit vector in the direction of the [tex]+x[/tex]-axis, while [tex]\vec{e}_2[/tex] is that for the [tex]+y[/tex]-axis. We put the bases of both vectors at the origin for convenience. Now, if we rotate [tex]\vec{e}_1[/tex] by [tex]3\pi/2[/tex] radians counterclockwise, that corresponds to [tex]270^\circ[/tex], which corresponds to the [tex]-y[/tex]-axis or [tex]-\vec{e}_2[/tex]. We can do the same thing for [tex]\vec{e}_2[/tex] to find [tex]T(\vec{e}_2)[/tex].

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

[tex] T(\vec{e}_i) = \sum_j T_{ij} \vec{e_j}.[/tex]

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

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