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 |
Proposed by G. Trenkler, University of Dortmund, Dortmund, Germany Let A be a square matrix with real entries satisfying A^2 = A^T. (i) Find its Moore-Penrose inverse A^+ in terms of A. (ii) Assume A is a 2x2 matrix. Find all solutions to A^2 = A^T that are not symmetric. [The "^" character in the above represents superscripting -- "A^T" is the transpose of A, and "A^+" is A with a superscript "+".]
T.R | Title | User | Personal Name | Date | Lines |
---|---|---|---|---|---|
1944.1 | Solution to (ii) | FLOYD::YODER | MFY | Thu Feb 23 1995 17:30 | 19 |
We want 2 /a b\ /a^2+bc ab+bd\ /a c\ ! ! = ! ! = ! ! \c d/ \ac+cd bd+d^2/ \b d/ where b != c. Looking at the off-diagonal elements, we see that any of b=0, c=0, or a+d=0 will imply b=c=0, so they are all false, and we can take the ratio of the off-diagonal elements to get b/c = c/b, so b^2 = c^2, so b = -c (since b != c). Now b(a+d) = c = -b, so a+d = -1. Taking the difference of the diagonal elements, a^2 - d^2 = a-d, (a-d)(a+d) = a-d, so -(a-d) = a-d, a-d = 0, and so a=d. From a+d = -1 we now get a=d=-1/2; and from a^2+bc = a we get b^2 = -bc = a^2 - a = 3/4. So b is +- 1/2 sqrt(3), and c = -b. It is easy to check these work; the only two solutions are Q and Q^T where /-1/2 -sqrt(3)/2\ Q = ! ! \sqrt(3)/2 1/2 / | |||||
1944.2 | AUSSIE::GARSON | achtentachtig kacheltjes | Thu Feb 23 1995 19:18 | 3 | |
re .0 Anyone know what the Moore-Penrose inverse is? | |||||
1944.3 | typo | FLOYD::YODER | MFY | Thu Mar 02 1995 10:27 | 2 |
The last part of .1 is missing a minus sign before the lower right-hand entry; it should be -1/2, not 1/2. | |||||
1944.4 | WRKSYS::ROTH | Geometry is the real life! | Mon Mar 13 1995 12:41 | 10 | |
> Anyone know what the Moore-Penrose inverse is? It's also called the pseudo-inverse and is the inverse mapping from the column space of a matrix to the complement of the null space. It's matrix representation is unique, though it's better to think of what it actually does. - Jim |