Lecture21f.pdf

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Department
Mathematics
Course
MATB24H3
Professor
Sophie Chrysostomou
Semester
Winter

Description
University of Toronto at Scarborough Department of Computer & Mathematical Sciences MAT B24S Fall 2011 MAT B24 Lecture 21 We have proven that all Hermitian matrices are unitarily diagonalizable. Questions arise now as to which other questions have this property: Are all matrices unitarily diagonalizable? Are the Hermitian matrices the only ones that are unitarily diagonalizable?   1 0 i EXAMPLE: Determine if A = 0 1 0  is unitarily diagonalizable 0 0 i solution.   1 − λ 0 i det(A − λI) = det 0 1 − λ 0  = (1 − λ) (i − λ) 0 0 i − λ There are two eigenvalu1s: λ = 1 of algebraic multipli2ity 2 and λ = i of algebraic multiplicity 1.       0 0 i 1 0 Eλ1= nullspace 0 0 0  = sp 0 , 1 0 0 i − 1 0 0 so, 1 has geometric multiplicity 2=alg. multiplicity.  1 − i 0 i  1 + i 0 −1  1  Eλ2= nullspace  0 1 − i 0 = nullspace  0 1 0 = sp 0  0 0 0 0 0 0 1 + i 1 and λ2has geom mult. =1=alg. mult. Thus A is diagonalizable and     1 0 1 1 0 0 if C =  0 1 0  then C −1AC = 0 1 0  = D a diagonal. 0 0 1 + i 0 0 i Notice however that Eλ1and E λ2are not orthogonal. In fact, ▯     ▯ 1 1 0 , 0  = 1 ▯= 0 0 1 + i thus this non Hermitian matrix is not unitarily diagonalizable. So: not all diagonalizable matrices are unitarily diagonalizable. Next question we face then is: are the unitarily diagonalizable matrices the Hermitian matrices only, or could we find a non Hermitian, but yet a unitarily diagonalizable matrix? The answer is in the theorem that follows. However, we get a definition first. DEFINITION: A square complex matrix A is normal if AA = A A. ∗ THEOREM: A square complex matrix A is unitarily diagonalizable if and only if it is normal ie. AA = A A proof. (=⇒) We assume that A is unitarily diagonalizable. We want to prove that A is normal. Since A is unitarily diagonalizable, then there is a unitary matrix U such ∗ that U AU = D where D is the diagonal matrix ▯ 0 if i ▯= j [D]ij . dii if i = j Since D is a diagonal, then D = D = D. ∗ Note: DD = DD which has in its diagonal entries the form d jj,jj 2 ∗ and D D = DD which has in its diagonal entries the form d d jj jj since d d = d d then we have that DD = D D so D is normal. jj jj jj jj ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Now A = UDU so A = (U ) D U =
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