| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
A question from a linear algebra problem set I've got. Any ideas.
a matrix B in M nxn(C) is unitary. Prove |det(B)| = 1
_ ______ _ ___
Hint Prove that det(B) = det(B) where (B)ij = Bij
Dennis
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1205.1 | AITG::DERAMO | Dan D'Eramo, nice person | Wed Mar 07 1990 12:29 | 3 | |
What is the definition of "unitary"? Dan | |||||
| 1205.2 | define unitary | ARCANA::ESTRELLA | Wed Mar 07 1990 13:09 | 9 | |
* * ___
a B matrix is unitary when BB = I , where (B )ij = Bij
the complex conjugate of Bij.
If this is not clear let me know.
Dennis
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| 1205.3 | 4GL::GILBERT | Ownership Obligates | Wed Mar 07 1990 15:59 | 39 | |
_ ______
First, let's prove that det(B) = det(B). This is trivially true for a 1x1
matrix. Using induction, we assume it's true for any nxn matrix. Then
_ __ __ ___ __ ___ __
det(B) = B det(S ) - B det(S ) + ... � B det(S ), where the S are the
11 1 12 2 1n n i
appropriate submatrices (from the usual definition of det). Then
_ __ _______ __ _______ __ _______
det(B) = B det(S ) - B det(S ) + ... � B det(S )
11 1 12 2 1n n
____________________________________________
= (B det(S ) - B det(S ) + ... � B det(S ))
11 1 12 2 1n n
because the product of complex conjugates is the complex conjugate of
the product, and the sum of complex conjugates is the complex conjugate
of the sum. We see that this last equation is simply
_ ______
det(B) = det(B)
Which, by the inductive hypothesis, is true for all square matrices.
QED.
(This result could've been reached more succinctly -- it's true of any
polynomial having real coefficients).
_
Suppose now that B is unitary. That is, B B = I. We have that
_ _ _ ______
det(B B) = det(B) det(B) = det(I) = 1. But det(B) = det(B).
_
Let det(B) = a + ib. We have det(B) det(B) = (a + ib)(a - ib) = a� + b� = 1.
That is, |det(B)| = 1.
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