Comments on: Rotating a matrix cannot be done with matrix multiplication

By: Vincent

Vincent — Thu, 07 Jan 2010 12:59:33 +0000

Hi Roie, your analysis is sound.

And the Onion link’s fine… ooh Egyptians fail terribly at constructing monumental cubes…

By: Roie

Roie — Wed, 06 Jan 2010 17:36:23 +0000

Sorry, I don’t know how that Onion link got accidentally pasted instead of my name…

By: Rhttp://www.theonion.com/content/news_in_photos/a_long_elaborate_history

Rhttp://www.theonion.com/content/news_in_photos/a_long_elaborate_history — Wed, 06 Jan 2010 17:35:26 +0000

Raymond Chen revived this thread on his blog, so I just got here, over a year too late.

In any case, look at the second equation you wrote: A(0,0)b + A(0,1)d = a. This means that our matrix A generates a from b and d, “withoug even knowing” what a was in the first place. In other words, even if we change a (the top-left element of our original matrix), the top-right element of the rotated matrix will not change.

Here’s a more formal way of looking at it, that generalizes to N*N matrices: Suppose we have our original matrix A, and the “operation matrix” O (which we’d like to be a “rotation matrix”). Write OA=B. Then, assuming O is constant, B[N,1] doesn’t depend on A[1,1] at all: B[N,1] is the result of multiplying the first row of O by the LAST COLUMN of A.

By: John

John — Wed, 25 Mar 2009 14:26:46 +0000

I do think (but haven’t proved to myself) that you should be able to do this with something like a similarity transform with “permutation-like” matrices (despite the fact that neither will be square and it may not follow all the rules of a permutation)

By: Messy indices on the right please | Polymath Programmer

Messy indices on the right please | Polymath Programmer — Fri, 26 Sep 2008 09:03:46 +0000

[...] do I mean by messy indices on the right? Variable assignment. I mentioned something of this in the matrix rotation article, and I want to talk more on it here. And I will tell you why I prefer them on the right side of the [...]

By: Vincent Tan

Vincent Tan — Sun, 14 Sep 2008 15:41:34 +0000

Raymond – I was never much of a student of empirical proofs… Fondness of elaborate math arguments have blinded me!

Matthew – Thank you for sharing your proof. Much more refined than mine. Can’t believe I went into simultaneous equations and such…

By: Matthew van Eerde

Matthew van Eerde — Fri, 12 Sep 2008 17:40:12 +0000

Here’s the proof I came up with:
http://blogs.msdn.com/matthew_van_eerde/archive/2008/09/12/rotating-a-matrix-redux.aspx

By: Raymond

Raymond — Mon, 08 Sep 2008 14:25:38 +0000

Shorter proof:

[ A(0,0) A(0,1) ] [ 1 0 ] = [ 0 1 ]
[ A(1,0) A(1,1) ] [ 0 1 ] [ 1 0 ]

Therefore, A =

[ 0 1 ]
[ 1 0 ]

but this doesn’t work for the matrix

[ 0 1 ] [ 2 0 ] != [ 0 2 ]
[ 1 0 ] [ 0 1 ] != [ 1 0 ]