**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}\]

**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.

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.

**: Excellent. Next time, we'll consider the problem we alluded to earlier. What happens to a vector when we change the basis.**

Teacher

Teacher

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