Teacher: Let's start simple, and consider a vector \(u_o = (1,1)\), with the usual standard basis vectors \(e_1 = (1,0)\) and \(e_2 = (0,1)\) in 2D.
Student: Are we going to stay in 2D thoughout?
Teacher: For now, at least. It makes visualization much easier. To begin with, we are going to consider rotation of the vector \(u_o\) by an angle \(\theta\) to the new vector \(u_n\).
Student: I guess the subscripts "o" and "n" stand for "old" and "new". So you mean I should conjure a picture like this?
Teacher: That's excellent. What do we really mean by the vector \(u_o = (1,1)\) in terms of the bases?
Student: I know. I know. We mean \(u_0 = 1 e_1 + 1 e_2\). The "co-ordinates" of a vector tell me what linear combination of the basis I should take.
Teacher: Very impressive! You can also write the vector in "matrix notation" as \[u_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.\] Remember that the matrix notation implicitly assumes a certain basis. We will revisit this subtle but confusing point later, when we talk about non-standard bases.
Student: Okay. So as long as I use the same basis or co-ordinate system throughout a problem, I don't have to spend too much time thinking about it?
Teacher: Right. Now let us consider the 2 by 2 rotation matrix \(Q\), which performs the rotation. One can derive it from pure geometrical considerations. This magic matrix looks like:
\[Q = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}\]
Teacher: Very simple. To rotate \(u_0\) counter-clockwise by an angle \(\theta\) just multiply
\[u_n = Q(\theta) u_0.\]
Student: Cool. Let me try. If I set \(\theta = \pi/2\). I get:
\[u_n =\begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} .\]
Teacher: Try \(\theta = \pi/4\), and tell me what you see?
Student: I get \(u_n = (0, \sqrt{2})\). Hmmm.
I see. I see. The length of the vector is unchanged. Both \(u_0\) and \(u_n\) are of length \(\sqrt{2}\).
Teacher: Pure rotation, my friend. The matrix \(Q\) is special. A general matrix \(A\) would have been less gentle. In general, a 2 by 2 matrix rotates and stretches the vector on which it acts.
Student: Cool! Is there anything else that is special about \(Q\)
Teacher: Yes! I am glad you asked. \(Q\) belongs to a special class of matrices called orthogonal matrices. These matrices have the interesting property that the transpose is the inverse.
\[Q^{-1} = Q^T.\]
Student: Does that have any relevance to our discussion?
Teacher: Sure it does! Given \(u_n = Q u_0\) we can use the property of orthogonal matrices to write \(u_0 = Q^T u_n\). That is, we can rotate a vector clockwise by an angle by multiplying by \(Q^T\).
Student: Wait a minute. Shouldn't I have to multiply with \(Q(-\theta)\), instead of \(Q^T\)?
Teacher: A rose by any other name is still a rose!
Student: Aha! They are the same matrix.
Teacher: Excellent. Next time, we'll consider the problem we alluded to earlier. What happens to a vector when we change the basis.
Student: Are we going to stay in 2D thoughout?
Teacher: For now, at least. It makes visualization much easier. To begin with, we are going to consider rotation of the vector \(u_o\) by an angle \(\theta\) to the new vector \(u_n\).
Student: I guess the subscripts "o" and "n" stand for "old" and "new". So you mean I should conjure a picture like this?
Teacher: That's excellent. What do we really mean by the vector \(u_o = (1,1)\) in terms of the bases?
Student: I know. I know. We mean \(u_0 = 1 e_1 + 1 e_2\). The "co-ordinates" of a vector tell me what linear combination of the basis I should take.
Teacher: Very impressive! You can also write the vector in "matrix notation" as \[u_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}.\] Remember that the matrix notation implicitly assumes a certain basis. We will revisit this subtle but confusing point later, when we talk about non-standard bases.
Student: Okay. So as long as I use the same basis or co-ordinate system throughout a problem, I don't have to spend too much time thinking about it?
Teacher: Right. Now let us consider the 2 by 2 rotation matrix \(Q\), which performs the rotation. One can derive it from pure geometrical considerations. This magic matrix looks like:
\[Q = \begin{bmatrix}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{bmatrix}\]
Student: How do I use it?
Teacher: Very simple. To rotate \(u_0\) counter-clockwise by an angle \(\theta\) just multiply
\[u_n = Q(\theta) u_0.\]
Student: Cool. Let me try. If I set \(\theta = \pi/2\). I get:
\[u_n =\begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} .\]
Teacher: Try \(\theta = \pi/4\), and tell me what you see?
Student: I get \(u_n = (0, \sqrt{2})\). Hmmm.
Teacher: Pure rotation, my friend. The matrix \(Q\) is special. A general matrix \(A\) would have been less gentle. In general, a 2 by 2 matrix rotates and stretches the vector on which it acts.
Student: Cool! Is there anything else that is special about \(Q\)
Teacher: Yes! I am glad you asked. \(Q\) belongs to a special class of matrices called orthogonal matrices. These matrices have the interesting property that the transpose is the inverse.
\[Q^{-1} = Q^T.\]
Student: Does that have any relevance to our discussion?
Teacher: Sure it does! Given \(u_n = Q u_0\) we can use the property of orthogonal matrices to write \(u_0 = Q^T u_n\). That is, we can rotate a vector clockwise by an angle by multiplying by \(Q^T\).
Student: Wait a minute. Shouldn't I have to multiply with \(Q(-\theta)\), instead of \(Q^T\)?
Teacher: A rose by any other name is still a rose!
Student: Aha! They are the same matrix.
Teacher: Excellent. Next time, we'll consider the problem we alluded to earlier. What happens to a vector when we change the basis.
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