Rabbits Two

The equation below is true if the new vector is the same as the old vector:

v^1 b_1 + v^2b_2 = \tilde{v}^1 \tilde{b}_1 + \tilde{v}^2\tilde{b}_2

The two equations below are true if each component of a new basis vector is a Linear Combination of all the components of the old basis vector.

  • \tilde{b}_1 = L^1_1 b_1 + L^2_1 b_2
  • \tilde{b}_2 = L^1_2 b_1 + L^2_2 b_2

At this point we know nothing of what numbers for L^1_1, L^1_2, L^2_1 and L^2_2. We have made a choice that constrains all solutions we might consider to the assumption we’ve made.

The equation below takes the above two linear equations and puts them into a single equation that uses a matrix:

\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}

Carefully scrutinize the equation below and notice that the matrix in this one is the transpose of the matrix in the equation above. Notice also that the equation below is true if the two linear equations we postulated are true:

\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}

The result above will soon be needed. We now return to an equation that must be true if the new vector is the same as the old vector. The equation below is the equation at the start of this document, recast to express it as an equality of matrix multiplications:

\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}

In the next equation \begin{bmatrix} \tilde{b}_1 & \tilde{b}_2 \end{bmatrix} is replaced using an above equation (matrix operating on old basis vectors produces new basis vectors):

\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}

\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}

Look closely at the matrices with a capital l. It changes. The change is taking the Transpose of the first matrix.

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

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

  • Feed your head
  • Feed the head
  • Feed the hare
  • Be the hare!

Notice carefully:

  • L original on old b gives new b
  • L transpose on new v gives old v

Notice both the transposition and the switch?

Now for convenience, let M = L^T so that for the next few equations we won’t have to keep writing the T superscript.

\begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = M \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

M^{-1} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = M^{-1} M \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

M^{-1} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = (1) \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

M^{-1} \begin{bmatrix} v^1 \\ v^2 \end{bmatrix} = \begin{bmatrix} \tilde{v}^1 \\ \tilde{v}^2 \end{bmatrix}

Appendix A

Now to put it all together using a more compact notation:

  • b L = \tilde b
  • L^T \tilde v = v
  • (L^T)^{-1} v = \tilde v

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

Working through all this math, we see that we need the inverse of a transposition. If we merely considered what we need for a change from inches to centimeters, we would not have noticed the transposition part because we would have been working with diagonal matrices such as \begin{bmatrix}2.54 && 0 && 0 \\ 0 && 2.54 && 0 \\ 0 && 0 && 2.54 \end{bmatrix} then we would not have noticed the transposition because a diagonal matrix is its own transpose.

Appendix B

We need to know when we can move things around.

Scalars (such as ‘a’ and ‘b’ below) can be moved around while still maintaining the equality.

  • aAbB = C
  • abAB = C
  • dAB = C (where d=ab)

Here is one way of explaining it. For objects A and B, that scalar on the outside can be moved into the object and when it moves in, every component in that object gets the scalar.

If the math AB is always a product of one component from A and one component from B to make a component in the new object, then every component in the new object will have an ‘ab’ on it and we could turn around and pull them all out, putting an ‘ab’ in front of the new object.

Also, if there is summation, that ‘ab’ will be on every term, and we can pull it out and put in in front.