intuition.
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18
  19. 19
  20. 20
  21. 21
  22. 22
Linear Algebra
Step 18 of 22

Change of basis

Understand first: inverse · span and basis

Back in Act I you saw that a basis is a choice, and that {(1,1),(1,−1)}{(1,1),(1,-1)} works as well as the standard one. Time to collect on that.

The same arrow, two descriptions

An arrow sits in the plane. Nothing about it will move during this entire step.

In the standard basis you call it (3,1)(3, 1): three steps along ı^\hat{\imath}, one along ȷ^\hat{\jmath}.

Now describe it using b1=(1,1)b_1 = (1,1) and b2=(1,−1)b_2 = (1,-1). How many of each?

2(1,1)+1(1,−1)=(2,2)+(1,−1)=(3,1)2(1,1) + 1(1,-1) = (2,2) + (1,-1) = (3,1)

So in that basis the same arrow is called (2,1)(2, 1).

Two different addresses, one unmoved arrow. Coordinates were never properties of the vector. They are properties of the vector plus the basis you measured it with — the point deferred back in step three, now cashed in.

Translating between languages

Build a matrix BB whose columns are the new basis vectors:

B=[111−1]B = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix}

Read it with the column picture. Feed it (2,1)(2,1) — the coordinates in the new language — and it returns 2⋅(1,1)+1⋅(1,−1)=(3,1)2 \cdot (1,1) + 1 \cdot (1,-1) = (3,1), the standard coordinates. So BB translates into standard coordinates, and B−1B^{-1} translates out of them.

Translating a transformation

Here is the move that makes Act IV work. Suppose MM performs some motion in standard coordinates, and you want the same motion expressed in the new basis:

B−1MBB^{-1} M B

Read it right to left as a sentence. BB takes your new-language vector into standard language; MM does the motion there, where it is already understood; B−1B^{-1} translates the result back. It is not a formula to memorise — it is a round trip through a language everyone speaks.

Which sets up the obvious question. If changing basis changes how a transformation looks, is there a basis that makes it look as simple as possible? For that, you need to find the directions the transformation itself prefers.