White Rabbit One

Our White Rabbit friend is in a hurry. He spent a lot of time learning things that were deemed necessary prior to actually doing anything with tensors and he wants to actually DO SOMETHING with a real tensor. Twice, we will give you a formula that is needed without any explanation, justification, proof, etc.

I am indebted to Brian Keng who offered several examples on his tensor page and I took one and made small modifications to create what we have below where $L = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$ and yes, L is a tensor.

$v = \begin{bmatrix} x \\ y \end{bmatrix}$

$L = \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix}$

For this example, the calculation of Lv is extremely simple. You might even be able to look at it and see that the answer is $\begin{bmatrix} ax \\ by \end{bmatrix}$ .

We need for you to imagine that we have one or more reasons to say we don’t want to do the Lv calculation and that some knight came along and told us to try a calculation with modifications of the two things, $\tilde L \tilde v$ . The magic is–the math thing that we use to get these modifications is related to the math thing that takes $\tilde L \tilde v$ back to Lv.

Quick Example to Show the Trick

“Help! I need to know (3)(5)! I can’t multiply two odd numbers. All my work has been with both even or at least one even.”

We do something that changes 3 to 6 [2(3)=6]
We do two things that change 5 to 5 [(1/2)5(2) = 5]
(6)(5) is 30
We do the inverse of the something and it changes 30 to 15 [(1/2)30 = 15

Read Appendix B if it bothers you that we did two things to 5 just to get 5.

You might recognize that for the above our thing was “double” and the inverse was “half”. Below the thing will be rotation by an angle and the inverse will be a rotation by the negative of the angle.

R will be a Rotation Matrix that does a rotation of 30 degrees.

$R = \begin{bmatrix} cos(\theta) & - sin(\theta) \\ sin(\theta) & cos(\theta) \end{bmatrix}$

$cos(30\deg) = 0.8$
$sin(30\deg) = 0.6$

$R = \begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix}$

We have a formula we can use to calculate the inverse matrix based on the above four numbers. However, we can take an easier route and do a second rotation matrix where the angle is -30 degrees.

$cos(-30\deg) = 0.8$
$sin(-30\deg) = -0.6$

$R^{-1} = \begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix}$

We can confirm that one matrix is the inverse of the other by proving that when we multiply them together we get the Identity Matrix:

$\begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix} \begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

We will use $R$ to make a new vector from our old vector:

$Rv = \tilde v$

$\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0.8x & 0.6y \\ -0.6x & 0.8y \end{bmatrix}$

Another purpose of R is so that we can create a new tensor from the old tensor, or, $\tilde L$ from $L$ . The formula for that will be given here.

$\tilde L = R^{-1} L R$

Below (here) we calculate $R^{-1} L R$ :

$\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix}$

$\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} \begin{bmatrix} 0.8a & -0.6a \\ 0.6b & 0.8b \end{bmatrix}$

$\begin{bmatrix} 0.64a + 0.36b & -0.48a + 0.48b \\ -0.48a + 0.48b & 0.36a + 0.64b \end{bmatrix}$

(one more matrix calculation)

$R \tilde L \tilde v$

$\begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix} \begin{bmatrix} 0.64a + 0.36b & -0.48a + 0.48b \\ -0.48a + 0.48b & 0.36a + 0.64b \end{bmatrix} \begin{bmatrix} 0.8x & 0.6y \\ -0.6x & 0.8y \end{bmatrix}$

$\begin{bmatrix} 0.8 & -0.6 \\ 0.6 & 0.8 \end{bmatrix} \begin{bmatrix} 0.8ax & 0.6by \\ -0.6ax & 0.8by \end{bmatrix}$

$\begin{bmatrix} ax & 0 \\ 0 & by \end{bmatrix}$

For the next work, we will be confirming something that we’ve been given. For that, we need $R^{-1}Lv$ and we need $\tilde L \tilde v$ . Our goal is to show they are equal. (No)

$R^{-1}Lv$

$\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

$\begin{bmatrix} 0.8 & 0.6 \\ -0.6 & 0.8 \end{bmatrix} \begin{bmatrix} ax \\ by \end{bmatrix}$

$\begin{bmatrix} 0.8ax & 0.6by \\ -0.6ax & 0.8by \end{bmatrix}$

$R^{-1}Lv = \begin{bmatrix} 0.8ax & 0.6by \\ -0.6ax & 0.8by \end{bmatrix}$

$\tilde L \tilde v$

$\begin{bmatrix} 0.64a + 0.36b & -0.48a + 0 48b \\ -0.48a + 0.48b & 0.36a + 0.64b \end{bmatrix} \begin{bmatrix} 0.8x + 0.6y \\ -0.6x + 0 8y \end{bmatrix}$

$\begin{bmatrix} (0.64a + 0.36b)(0.8x + 0.6y) + (-0.48a + 0.48b)(-0.6x + 0.8y) \\ (-0.48a + 0.48b)(0.8x + 0.6y) + (0. 36a + 0.64b)(-0.6x + 0.8y) \end{bmatrix}$

$\begin{bmatrix} I + II \\ III + IV \end{bmatrix}$

0.64a +0.36b
0.8x + 0.6y

0.512ax + 0.384ay + 0.288bx + 0.216by

-0.48a + 0.48b
-0.6x + 0.8y

0.288ax – 0.384ay – 0.288bx + 0.384by

I + II = 0.8ax + 0.6by

III

-0.48 + 0.48b
0.8x + 0.6y

-0.384ax – 0.288ay + 0.384bx + 0.288by

0.36a + 0.64b
-0.6x + 0.8y

-0.216ax + 0.288ay – 0.384bx + 0.512by

III + IV = -0.6ax + 0.8by

$\tilde L \tilde v = \begin{bmatrix} 0.8ax + 0.6by \\ -0.6ax + 0.8by \end{bmatrix}$

From above (this is a copy):

$R^{-1}Lv = \begin{bmatrix} 0.8ax & 0.6by \\ -0.6ax & 0.8by \end{bmatrix}$

$\tilde L \tilde v = R^{-1}Lv = \tilde L \tilde v$

Note carefully th

Now to explain why we might want all this. Maybe we can’t do the calculation Lv, or, maybe we don’t want to do it if a simplification is possible.

Notice the following:

$R^{-1}LR$ lets us calculate a new L, which we write as $\tilde L$ .
Rv lets us calculate a new v, which we write as $\tilde v$ .
As our last step we do the calculation for $R \tilde L \tilde v$ and that gives us the answer for Lv.

We’re assuming that you want to do this because there is something cool like $L = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $\tilde L = \begin{bmatrix} e & 0 \\ 0 & g \end{bmatrix}$ and we love those zeroes in the new L.

Appendix A

The first question might be why we used the reverse matrix to make a new vector from the old vector. We are using the rotation for the coordinate system. The vector moves opposite so we want the “reverse” rotation, which is done by the inverse matrix $R^{-1}$ .

One thing to notice, the transformation matrix (R) and the tensor (L) are two different things. This might not be clear to a student who is just starting to learn the topic (there is so much to cover when starting).

Appendix B

When we work with very simple things in a simple coordinate system (the kind that Euclid used) some features necessary for the high end stuff don’t have the opportunity to show us why they are needed.

Appendix C

It’s fair to ask “from whence cometh those formulas?”

We will be working on it. Part of the thing with the White Rabbit is recognition of his stress for getting bogged down in all the studies required before you can start working with tensors.

We were given $\tilde L = R^{-1} L R$ .

One rather obvious trick is to hit that R negative one with the opposite to cancel it out and whatever we use on the right side to do that needs to be put on the left side.

$R \tilde L = R R^{-1} L R$

$R \tilde L = L R$

There is another strategy we can use:

We start with Lv = w which is some physics law.

$\tilde L \tilde v = \tilde w$

It’s the same physics law expressed in new coordinates.

At this point we know nothing about any possible relation between $L$ and $\tilde L$ .

We use R and its inverse to move between the two coordinate systems:

$R v = \tilde v$
$R^[-1} \tilde v= v$
$R w = \tilde w$
$R^{-1} \tilde w = w$

We already know that $R$ and $R^{-1}$ are such that one is the inverse of the other. Any way we multiply the two together we will get the Identity Matrix.

Our trick is to start with $\tilde L \tilde v = \tilde w$ and then do several changes to [latex] \tilde w[/latex]:

[latex]\tilde w[/latex]

$Rw[/latex] [latex]RLv[/latex] [latex]RL R^{-1} \tilde v[/latex] Now that we have [latex]\tilde L \tilde v = RL R^{-1} \tilde v[/latex] we can take the last piece off each expression and we have the following: [latex]\tilde L = R L R^{-1}$

We've noticed authors having R operate on a variable in the new coordinate system to make the variable in the old coordinate system. This is just the opposite to how we do it and when they do it that way R inverse is the first operator.

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