Wednesday, June 26, 2013

Vector Representation in Different Basis

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.

: 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 \(u = (u_1, u_2)\).

: 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 \(u\).

: 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 \(u_1 e_1 + u_2 e_2 = U_1 E_1 + U_2 E_2\). 

: 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.

: Yes! Why don't you try out the example?

Student: Okay. I already know what \(Q\) is. I can get:
\[U = Q^T u =  \begin{bmatrix} 0 & 1\\ -1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} .\]

: Does that look like the right answer?

Student: \(U = 1 E_1 + (-1) E_2 \). That looks about right from this figure!

: 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.

: 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.

: 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. 

: 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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