Topics: Mathematics, Matrices, Non-commutativity
Last edit 2023-07-17 -JH
Assumes: Basic Familiarity with Matrices
Notation: We represent the [multiplicative] inverse of A as A-1.
Many many years ago (about 1970 or so), in High School, I was introduced to Matrices, and they've been my friends ever since.
At that time the instructor pointed out [correctly] that Matrix multiplication is not commutative.
This this means that for two matrices A and B it is NOT true, in general, that the product AB has the same value as the product BA. (However it is commutative for some special cases, such as identity matrices.)
I asked the instructor "If BA is not equal to AB, what is it equal to?". The instructor answered "It is equal to whatever you get when you multiply B by A, in that order".
While this *is* true, it is not very enlightening. However, this has apparently been the "standard"
answer instructors have been handing to students for that question since matrices were first taught.
But, sometime in the last 50 years, I did find a truthful answer that actually says something
marginally interesting.
For two square matrices, A and B,
if AB = C, then BA = A-1CA = BCB-1
(Assuming that A has an inverse in the first case, and that B has an inverse in the second case.)
Trivial proof: Substitute AB for C, and cancel all multiplications of anything by its inverse.
It also follows, near trivially, that if E = A-1FA then F = AEA-1 assuming that A has an inverse, for any conformable E or F what so ever.
Proof: E = A-1FA
AE = FA # Premultiply both sides by A, and cancel.
AEA-1 = F # Postmultiply both sides by A-1, and cancel.
And this observation holds for things other than Matrices... as long as the system has a multiplication, and that multiplication has inverses. Although, I admit, it is only interesting when the multiplication is not commutative.
Technical Aside:
In the general case of more that two factors, the technique above rotates a list of factors, which in the case of only two factors happens to be the same as exchanging them.
This can be made more obvious by considering, for A, B, C, and D non-singular:
D = ABC.
A-1DA = A-1ABCA = BCA
CDC-1 = CABCC-1 = CAB
As you can see, these are just rotations of the factors. The first rotates them circularly left, the second rotates them circularly right.
Epilog:
Given the number of places in Matrix Algebra that expressions of the form "A-1CA" occur, I have to wonder where there is any place where the observations of this post could be of any value what so ever. I personally doubt it, but I welcome comments on the topic.
*sigh* I begin to suspect that any time I publicly express doubt that something has any value,
I will be shortly have a counter example drop into my lap. (See below)
Addendum: (Jul 17th 2023) -JH Added addendum.
Mike Brookes (Imperial College London) has pointed out to me that, for a non-singular X,
if B=X-1AX, then A and B are
Similar.
He then makes the observation that my identities "translate to" (quoting him)
"two square matrices,
P and Q, are similar if and only if
there exist matrices X and Y (not both singular)
such that P = XY and Q = YX".
And I now make the trivial observation that for square A and B both non-singular, AB is therefore similar to BA, and the non-trivial observation that for square P, Q and non-singular X:
XP = QX implies that P and Q are similar.
The proof is trivial... pre multiply both sides by X-1, and cancel.
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