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I've just learned how to go from one basis to another. After a few examples, I wrote myself the following method. Method: which ingredients of the output basis do I need to express the input basis?

For example, in 2 dimensions, say I need -1 of the first basis vector and 1 of the second basis vector (both of the output basis) to get the first basis vector of the input basis. For the the second basis vector of the input basis, say I can take 2 of the first basis vector (of the output basis) and -1 of the second basis vector (of the output basis). Then my transition matrix is $$\begin{pmatrix}-1 & 2\\1&-1\end{pmatrix}.$$

This feels reversed to me. If I'm going from the input basis to the output basis, I expected that I'd ask myself which ingredients of the input basis I need in order to get to the output vectors. Do I make sense here?

How can I turn this procedure more natural to me?

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  • $\begingroup$ This is a thing that gets everybody confused. You are not alone. I did not make sense of this until I learned about tensors, with covariant and contravariant indices. I suggest not to worry too much, for the moment. $\endgroup$ – Giuseppe Negro Jan 18 '18 at 11:28
  • $\begingroup$ Not the news I hoped to get :-), but I surely appreciate the blatant truth. But check @Stella's answer below. It's quite good! $\endgroup$ – R. Chopin Jan 18 '18 at 20:41
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You’re right that this is counter-intuitive. However, I find the following analogy insightful:

Suppose you have a parallelepiped that is $2\times 3\times 1$, with all the side lengths measured in inches. What are the side lengths, as measured in centimeters?

To answer this question, we have to calculate how many centimeters comprise $2,3,$ and $1$ inches respectively, just like in the case of changes of basis. The reason for this is that the object you’re measuring is staying the same, but the language with which you’re describing it is changes. So you need to work backwards, and ask yourself “how do I say the same thing I just said, but in my new language.”

This is very different from the case of if the object was to change size, and then need to be expressed in the original language. You might express that as a problem as follows:

Suppose you have a uniform density parallelepiped of ice that is $2\times 3\times 1$, with all the side lengths measured in inches. After being left out in the sun, it melts so that it loses half of its total mass, and shrinks equally in every dimension. What are the new dimensions of the cube?

Here the universe didn’t change, just the object in question, and so you have to find the size of the object in terms of the old basis. Of course, we could combine these two types of problems and make you do both simultaneously.

In my answer to your other question, I pointed out that your textbook didn’t name the linear function, only it’s matrix representation with respect to the standard basis. The key insight here is that the metric representation isn’t the base thing that exists in the world, rather it’s a description of the thing that exists in a particular language, just like $2in\times 3in\times 1in$ is. The true underlying object, the parallelepiped, is the linear transformation. The units we measure it in, the language we use to talk about the properties of the thing, is the matrix representation.

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  • $\begingroup$ That is a fantastic answer. I had not much hope I'd understand this, but I think you just turned it all much brighter and I even think the unnatural way might actually be quite natural from now on! Thank you so much for giving me such excellent education! $\endgroup$ – R. Chopin Jan 18 '18 at 20:46
  • $\begingroup$ Is this the idea? $$ \begin{pmatrix}2\\3\\1\end{pmatrix}_{\text{inch}} = 5.08\begin{pmatrix}1\\0\\0\end{pmatrix}_{\text{cm}} + 7.62\begin{pmatrix}0\\1\\0\end{pmatrix}_{\text{cm}} + 1\begin{pmatrix}0\\0\\1\end{pmatrix}_{\text{cm}} $$ $\endgroup$ – R. Chopin Jan 18 '18 at 21:13
  • $\begingroup$ @R.Chopin Precisely. $\endgroup$ – Stella Biderman Jan 18 '18 at 21:15
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There are two opposite directions you can think of this. One textbook calls them "alias" and "alibi". For "alias" you keep your point where it is but give it a new name (express it in a different basis). For "alibi" you keep the coordinate system the same, but move your point instead.

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    $\begingroup$ I have seen this as "passive" and "active" point of view respectively, especially in physics books (Sakurai's "Modern quantum mechanics", for example). $\endgroup$ – Giuseppe Negro Jan 18 '18 at 11:58

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