Homework Two

A 2×2 matrix changes a vector to a different vector except for the Trivial case where the 2×2 matrix is the Identity Matrix.

We start with two equations:

L^1_1 v^1 + L^2_1 v^2 = \tilde v^1

L^1_2 v^1 + L^2_2 v^2 = \tilde v^2

At this point all four L symbols are unknown and we assume they are all different (see Appendix A — Notice of Payment).

We are following a convention (Index Notation) that the subscript of L must match the superscript of v since L and v are involved in a summation.

L^2_1 v^1 correct

L^1_2 v^1 incorrect

Notice that for the two equations, a new component is a Linear Combination of the set of all old components.

We now go back to those two equations for \tilde v^1 and \tilde v^2. We can make those equations using a matrix and a vector:

\begin{bmatrix} L^1_1 && L^2_1 \\ && \\ L^1_2 && L^2_2 \end{bmatrix} \begin{bmatrix} b_1 \\ \\ b_2 \end{bmatrix} = \begin{bmatrix} \tilde{b}_1 \\ \\ \tilde{b}_2 \end{bmatrix}

And we can also get the two equations with the following:

\begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} L^1_1 && L^1_2 \\ && \\ L^2_1 && L^2_2 \end{bmatrix} = \begin{bmatrix} \tilde{b}_1 & \tilde{b}_2 \end{bmatrix}

Scrutinize the two matrices until you see that they are not identical. One is the transpose of the other.

Recall that vectors can be represented by the following: here

v = \begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix}

\tilde v = \begin{bmatrix} \tilde{b}_1 & \tilde{b}_2 \end{bmatrix} \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

v = \tilde v

\begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} \tilde{b}_1 & \tilde{b}_2 \end{bmatrix} \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

Work higher up gave us a calculation for \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix} and we put it in:

\begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} b_1 & b_2 \end{bmatrix} \begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix} \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

If Operator O on something equals Operator O on elsething then something and elsething are equal. In the above equation, the \begin{bmatrix} b_1 & b_2 \end{bmatrix} is an operator operating on everything else. We can remove it from both sides.

\begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix} \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

We are close, but this is not what we want. We want something that operates on v to give \tilde v.

It is legal to take a matrix and put a -1 superscript on it to make the inverse matrix:

\begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix}^{-1} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix}^{-1} \begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix} \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

A matrix and its inverse together go to the identity matrix and then that disappears:

\begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix}^{-1}

\begin{bmatrix} L^1_1 && L^1_2 \\ \\ L^2_1 && L^2_2 \end{bmatrix}^{-1} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

The matrix that changes old basis vectors to new basis vectors is related to the matrix that changes old components to new components of tne othet matrix. One is the inverse of the transpost of the othet.

This homework utilizes Rabbits Two.

Appendix A – Notice of Payment

If we start out with a set of things and we say that we are assuming at the beginning that they are all different, if you do work the shows we can relate one of those to another of those, document it and turn in your documentation and we will pay you. Work done showing things assumed to be different are actually related is work we deem valuable.