Build a Relation Between Matrix Elements
1 Ansicht (letzte 30 Tage)
Ältere Kommentare anzeigen
I have a complex valued matrix for example:
a = complex(1.1,-1.3); b = complex(1.3,1.2);c = complex(1.5,-1.4); d = complex(1.8,1.2); % random numbers
A = [a b; c d];
How can I build a relation between the elements of the matrix as, and then get the sume for each element. All the relations would be calculated as:
A1_1 = A(1) + A(2) + A(3) + A(4);
A1_2 = A(1) + A(2) + A(3) - A(4);
A1_3 = A(1) + A(2) - A(3) + A(4);
A1_4 = A(1) - A(2) + A(3) - A(4);
A1_5 = A(1) + A(2) - A(3) - A(4);
A1_6 = A(1) - A(2) - A(3) + A(4);
A1_7 = A(1) - A(2) + A(3) + A(4);
A1_8 = A(1) - A(2) - A(3) - A(4);
A2_1 = A(2) + A(3) + A(4) + A(1);
A2_2 = A(2) + A(3) + A(4) - A(1);
A2_3 = A(2) + A(3) - A(4) + A(1);
A2_4 = A(2) - A(3) + A(4) - A(1);
A2_5 = A(2) + A(3) - A(4) - A(1);
A2_6 = A(2) - A(3) - A(4) + A(1);
A2_7 = A(2) - A(3) + A(4) + A(1);
A2_8 = A(2) - A(3) - A(4) - A(1);
A3_1 = A(3) + A(4) + A(1) + A(2);
A3_2 = A(3) + A(4) + A(1) - A(2);
A3_3 = A(3) + A(4) - A(1) + A(2);
A3_4 = A(3) - A(4) + A(1) - A(2);
A3_5 = A(3) + A(4) - A(1) - A(2);
A3_6 = A(3) - A(4) - A(1) + A(2);
A3_7 = A(3) - A(4) + A(1) + A(2);
A3_8 = A(3) - A(4) - A(1) - A(2);
A4_1 = A(4) + A(3) + A(2) + A(1);
A4_2 = A(4) + A(3) + A(2) - A(1);
A4_3 = A(4) + A(3) - A(2) + A(1);
A4_4 = A(4) - A(3) + A(2) - A(1);
A4_5 = A(4) + A(3) - A(2) - A(1);
A4_6 = A(4) - A(3) - A(2) + A(1);
A4_7 = A(4) - A(3) + A(2) + A(1);
A4_8 = A(4) - A(3) - A(2) - A(1);
A1 = A1_1 + A1_2 + A1_3 + A1_4 + A1_5 + A1_6 + A1_7 + A1_8
A2 = A2_1 + A2_2 + A2_3 + A2_4 + A2_5 + A2_6 + A2_7 + A2_8
A3 = A3_1 + A3_2 + A3_3 + A3_4 + A3_5 + A3_6 + A3_7 + A3_8
A4 = A4_1 + A4_2 + A4_3 + A4_4 + A4_5 + A4_6 + A4_7 + A4_8
ANew = [A1 A2; A3 A4]
4 Kommentare
Akzeptierte Antwort
Voss
am 2 Jun. 2022
Bearbeitet: Voss
am 2 Jun. 2022
If that's really what you want to do, notice that if you sum these 8 equations:
A1_1 = A(1) + A(2) + A(3) + A(4);
A1_2 = A(1) + A(2) + A(3) - A(4);
A1_3 = A(1) + A(2) - A(3) + A(4);
A1_4 = A(1) - A(2) + A(3) - A(4);
A1_5 = A(1) + A(2) - A(3) - A(4);
A1_6 = A(1) - A(2) - A(3) + A(4);
A1_7 = A(1) - A(2) + A(3) + A(4);
A1_8 = A(1) - A(2) - A(3) - A(4);
You get A1_1+A1_2+...+A1_8 = 8*A(1)
That's because all the A(2), A(3), and A(4) terms on the right-hand side add to zero. That is, there are 4 positive copies and 4 negative copies of each of A(2), A(3), A(4), so their sum is 0.
Therefore, the end result you're after is:
ANew = 8*A
(I think you have it transposed in the question, i.e., it should be ANew = [A1 A3; A2 A4]; that is, A2 comes from A(2), which is c, not b.)
2 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!