COMPLEX SQUARE MATRIX ( 2x2)
(p,q) (a+b
, c+d) (a,c)
(b,d) (a,c) (b,d)
(p,q) = = + + + where p=a+b, q=c+d, r=e+f, s=g+h where etc. are the determinants of respective matrices. (r,s) (e+f , g+h) (e,g) (f,h) (f,h) (e,g) (r,s) |
(p,q) (a+ib ,
c+id) (a,c)
(b,d) (a,c) (b,d)
(p,q) = = - + i +i where p=a+ib, q=c+id, r=e+if, s=g+ih where etc. are the determinants of respective matrices. (r, s) (e+if , g+ih) (e,g) (f,h) (f,h) (e,g) (r,s)
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(p,q) [(a+ib)+(a'+ib')]
, [(c+id) +(c'+id')] (a+ib,
c+id) (a'+ib', c'+id')
(a+ib, c+id) (a'+ib', c'+id')
= = + + + (r, s) [(e+if )+(e'+if')], [(g+ih) +(g'+ih')] (e+if, g+ih) (e'+if', g'+ih') (e'+if', g'+ih') (e+if, g+ih)
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* If A is a square, non-singular matrix, then it can be
diagonalized using a transformation known as Similarity
Transformation. If T is a similarity matrix, then T-1AT=D,
a diagonal
matrix, the diagonal elements being the eigen values of A. If A is a
2x2 matrix, and
| p q| | a b | |D11 D12| |1 0| T= and A= then D= = | r s| | c d | |D21 D22| |0 1| where D11=[s(ap+br)-q(cp+dr)] /(ps-qr) ; D22= [p(ds+cq)-r(aq+bs)]/(ps-rq) ; D12=-[bs2 +q(a-d)s -cq2 ] / (ps-rq) ; D21= -[br2 +p(a-d)r -cp2 ] / (ps-rq) ; Since in D12=0, denominator is zero and s=q*((d-a) ± √[(d-a)2 + 4bc])/2b Since in D21=0, denominator is zero and r=p*((d-a) ∓ √[(d-a)2 + 4bc])/2b
then for determining the elements of T, we can arbitrarily assign values to either p or r and either q or s. Thus choosing the arbitrary values of any pairs will do--(p,q),(r,s),(p,s),(q,r) . In our calculation ,we have chosen the arbitrary values of (p,q). Then other two values can be found out. |
* If A= | p q| | r s| is a real matrix, then its inverse is |s -q| |-r p| divided by the determinant of A. Similarly, if A is |p+ip1 q+iq1| |r+ir1 s+is1| then its inverse is |s+is1 -(q+iq1)| |-(r+ir1) p+ip1| divided by the determinant of complex A. If Determinant is a +i aa where a =(ps-p1s1+q1r1-qr) and aa=(p1s+ps1-q1r-qr1) Now the inverse is given by |[(sa+aa*s1) +i (s1a-s*aa)]/z [(aa*q1-qa) +i (q*aa-aq1)]/z | |[(aa*r1-ra) +i (r*aa-ar1)]/z [(pa+aa*p1) +i (p1a-p*aa)]/z | where z=a2 +aa2
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* For the Complex Matrix, to find out the eigenvectors
corresponding to eigenvalue λ1+iλ1a,
If one eigen vector is x +ix1 where x,x1 are any arbitrary real
numbers, then the other eigenvector is y +iy1 where |K -K1| |x| |y| = |K1 K | |x1| |y1| where K=a12(λ1-a11)+a12a(λ1a-a11a) / (a12*a12 + a12a*a12a) ; K1=a12(λ1a-a11a)-a12a(λ1-a11) / (a12*a12 + a12a*a12a) ; Similarly for the other eigen value. * Any hermitian matrix H can be diagonalized by a suitable Unitary matrix U such that U†HU =D. Above, we have enumerated how to find U(normalized). First for First eigenvalue, find pairs of normalized eigenvectors.These pairs constitute the first column of U. for second eigenvalue, find the next pair which constitute the second column. Thus U is constructed. Among eigenvalue pairs, we have put arbitrary value of first eigenvector and with the relationship, 2nd vector is found. There can be different choices on the matter. |