Teacher: We now know how to rotate a vector u counter-clockwise by an angle \theta using a rotation matrix Q. In this lesson, we are not going to transform the vector u - instead we are going to investigate how the matrix representation changes when we move from the standard basis vectors to some other basis.
Student: Right, that last time around you did remark that in the matrix representation of a vector u = (1,1) the basis was tacitly assumed. So I guess, we have to first talk about a new basis.
Teacher: Yep. Let's again consider a vector u = (u_1, u_2) and the standard basis vectors e_1 = (1,0) and e_2 = (0,1) in 2D. Thus,
u = u_1 e_1 + u_2 e_2 = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}.
Student: I notice you dropped the subscript "o" from last time, because we are not going to touch the vector per se. Also I notice that we using a general representation instead of something specific u = (u_1, u_2).
Teacher: Good observation. We will toggle between a general and a specific u, depending on the situation. Now let's pick two new basis vector E_1 and E_2. Just to reiterate lessons from last time, let us generate this new basis by rotating the standard basis by 90 degrees.
Student: Okay. Let me figure this part out. I set \theta = \pi/2. I can compute Q(\pi/2) and get:
E_1 =Q(\pi/2) e_1 = \begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} .
Similarly, E_2 = Q(\pi/2) e_2 = (-1, 0). I guess that means I can draw a picture such as:
Teacher: Great. Now let's consider the point u = (1,1) as before, and ask ourselves how its representation in the new basis U = (U_1, U_2) looks like.
Student: Hang on. I thought we were not going to do anything to the vector u.
Teacher: We aren't! We are simply looking at the same geometrical object u with a different lens (basis). It similar to saying: "Texas is to the west", when you are in Florida, and "Texas is to the East", when you are in California. Texas hasn't moved. You have.
Student: Aha! I see what you mean. We are trying to represent the same geometrical object u_1 e_1 + u_2 e_2 = U_1 E_1 + U_2 E_2.
Teacher: Great. Since we know the relationship between the old and new basis, we should be able to figure out the co-ordinates in the new basis.
U_1 E_1 + U_2 E_2 = U_1 Q e_1 + U_2 Q e_2 = Q U_1 e_1 + Q U_2 e_2
That is: QU = u, or U = Q^T u.
Student: Okay let me try to keep things straight. Q codifies the relationship between the old and new axis. You are telling me that a point u in the old basis is the same as the point Q^T u in the new basis.
Teacher: Yes! Why don't you try out the example?
Student: Okay. I already know what Q is. I can get:
U = Q^T u = \begin{bmatrix} 0 & 1\\ -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} .
Teacher: Does that look like the right answer?
Student: U = 1 E_1 + (-1) E_2 . That looks about right from this figure!
Teacher: Right. If you want to rotate a vector you multiply it by Q. If you want to represent the same vector in a different basis you multiply it by Q^T.
Student: Question. Here the new basis was a simple rotation of the old basis given by Q. What about a general new basis.
Teacher: Good question. We can figure this out using linear algebra for a n-dimensional space. We know what the standard basis for such a space is. e_i = [0~... 0, 1, 0, ..., 0]^T, where the 1 is in the i-th row. Consider an alternative basis packed together as columns in the matrix C:
C = (v_1~v_2~...v_n)
Student: Each of the v_i is a n-dimensional vector, and since they form a basis for R^n, they are linearly independent. And C is really a n by n matrix, that is invertible.
Teacher: Someone here remembers their linear algebra! Okay now we can write a point u \in R^n in terms of the new basis
u = U_1 v_1 + ... + U_n v_n = C \begin{bmatrix} U_1 \\ \vdots \\ U_n \end{bmatrix} = C U
And hence, U = C^{-1} u.
Student: I get it! In our case, the the basis transformation was simply a rotation C = Q. Therefore
U = C^{-1} u = Q^T u. That really does tie it all in together.
Teacher: I am so glad that it does. You know what else. Khan Academy has a bunch of nice videos on this topic.
Student: I sure will check them out.
Student: Right, that last time around you did remark that in the matrix representation of a vector u = (1,1) the basis was tacitly assumed. So I guess, we have to first talk about a new basis.
Teacher: Yep. Let's again consider a vector u = (u_1, u_2) and the standard basis vectors e_1 = (1,0) and e_2 = (0,1) in 2D. Thus,
u = u_1 e_1 + u_2 e_2 = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix}.
Student: I notice you dropped the subscript "o" from last time, because we are not going to touch the vector per se. Also I notice that we using a general representation instead of something specific u = (u_1, u_2).
Teacher: Good observation. We will toggle between a general and a specific u, depending on the situation. Now let's pick two new basis vector E_1 and E_2. Just to reiterate lessons from last time, let us generate this new basis by rotating the standard basis by 90 degrees.
Student: Okay. Let me figure this part out. I set \theta = \pi/2. I can compute Q(\pi/2) and get:
E_1 =Q(\pi/2) e_1 = \begin{bmatrix} 0 & -1\\ 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} .
Similarly, E_2 = Q(\pi/2) e_2 = (-1, 0). I guess that means I can draw a picture such as:
Teacher: Great. Now let's consider the point u = (1,1) as before, and ask ourselves how its representation in the new basis U = (U_1, U_2) looks like.
Student: Hang on. I thought we were not going to do anything to the vector u.
Teacher: We aren't! We are simply looking at the same geometrical object u with a different lens (basis). It similar to saying: "Texas is to the west", when you are in Florida, and "Texas is to the East", when you are in California. Texas hasn't moved. You have.
Student: Aha! I see what you mean. We are trying to represent the same geometrical object u_1 e_1 + u_2 e_2 = U_1 E_1 + U_2 E_2.
Teacher: Great. Since we know the relationship between the old and new basis, we should be able to figure out the co-ordinates in the new basis.
U_1 E_1 + U_2 E_2 = U_1 Q e_1 + U_2 Q e_2 = Q U_1 e_1 + Q U_2 e_2
That is: QU = u, or U = Q^T u.
Student: Okay let me try to keep things straight. Q codifies the relationship between the old and new axis. You are telling me that a point u in the old basis is the same as the point Q^T u in the new basis.
Teacher: Yes! Why don't you try out the example?
Student: Okay. I already know what Q is. I can get:
U = Q^T u = \begin{bmatrix} 0 & 1\\ -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} .
Teacher: Does that look like the right answer?
Student: U = 1 E_1 + (-1) E_2 . That looks about right from this figure!
Teacher: Right. If you want to rotate a vector you multiply it by Q. If you want to represent the same vector in a different basis you multiply it by Q^T.
Student: Question. Here the new basis was a simple rotation of the old basis given by Q. What about a general new basis.
Teacher: Good question. We can figure this out using linear algebra for a n-dimensional space. We know what the standard basis for such a space is. e_i = [0~... 0, 1, 0, ..., 0]^T, where the 1 is in the i-th row. Consider an alternative basis packed together as columns in the matrix C:
C = (v_1~v_2~...v_n)
Student: Each of the v_i is a n-dimensional vector, and since they form a basis for R^n, they are linearly independent. And C is really a n by n matrix, that is invertible.
Teacher: Someone here remembers their linear algebra! Okay now we can write a point u \in R^n in terms of the new basis
u = U_1 v_1 + ... + U_n v_n = C \begin{bmatrix} U_1 \\ \vdots \\ U_n \end{bmatrix} = C U
And hence, U = C^{-1} u.
Student: I get it! In our case, the the basis transformation was simply a rotation C = Q. Therefore
U = C^{-1} u = Q^T u. That really does tie it all in together.
Teacher: I am so glad that it does. You know what else. Khan Academy has a bunch of nice videos on this topic.
Student: I sure will check them out.
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