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how do I do the following:

Consider the matrix $\begin{bmatrix} \cos x & -\sin x\\ \sin x & \cos x \end{bmatrix}$ and the vector $\begin{bmatrix} \cos y\\ \sin y \end{bmatrix}$. Show that the product of the $2\times 2$ matrix with $2\times 1$ matrix $\begin{bmatrix} \cos(x+y)\\ \sin(x+y) \end{bmatrix}$ graphically, by sketching.

I can show this using trig identities, but I am not sure how to graph the $2\times 1$ trig vector in first place, let alone explain the rest of the question graphically. All the help very much appreciated!

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2 Answers 2

up vote 7 down vote accepted

[cos(y),sin(y)] is the unique point on the unit circle with angle $y$ from the origin. Your $2x2$ matrix rotates the plane about the origin by angle $x$, which is why the result is a point on the same circle with angle $x+y$.

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Thank you very much! –  LinAlgStudent Jan 9 '11 at 0:51

Every point $(x,y)$ can be thought of as a $ 2 \times 1 $ matrix.

If $(x,y)$ is a point in the 2D plane, then for a $2\times 2$ matrix $A$, the product of $A$ and $[x,y]^{T}$ gives another 2D point.

In this case, you $A$ is a rotation matrix, and the $2 \times 1$ matrix you have corresponds to a point on the unit circle.

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Thanks a lot, so I am visualizing it now, it makes sense that it's a unit circle. How do I go about showing that the 2x2 matrix rotates all of the points by angle x though? It doesn't seem apparent to me, even after I multiply the matrices out. –  LinAlgStudent Jan 9 '11 at 0:52
    
@user5157: The point (cos y, sin y) corresponds to an angle of y. By your trigonometry, the new point is (cos (x+y), sin (x+y)), which corresponds to an angle x+y... –  Aryabhata Jan 9 '11 at 0:55
    
Sorry, I wasn't clear. I am supposed to show that without using trig identities at first, just by sketching. My question is, how do I show it via sketching the unit circle and having that 2x2 matrix, that 2x2 matrix just shifts all of the points by angle x. –  LinAlgStudent Jan 9 '11 at 0:57
    
@User: Consider what happens to the point (1,0). That is enough to make the claim that all points on the circle get rotated (why?). –  Aryabhata Jan 9 '11 at 1:03

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