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**Unformatted text preview: **CANONICAL FORMS OF 2 2 MATRICES AND THEIR APPLICATIONS KIAM HEONG KWA 1. Canonical Forms of 2 2 Matrices The characteristic equation of a 2 2 complex matrix, say A = a 11 a 12 a 21 a 22 , is the quadratic equation det( A- I 2 ) = 2- (tr A ) + det A = 0 , where I 2 = 1 0 0 1 is the 2 2 identity matrix and tr A = a 11 + a 22 and det A = a 11 a 22- a 12 a 21 are respectively the trace and the determinant of A . Thus A has two distinct eigenvalues given by 1 , 2 = tr A p (tr A ) 2- 4 det A 2 , except when tr A = 2 det A so that 1 = 2 . The goals of this section are (1) to show that there is an invertible 2 2 complex matrix S such that A can be Jordan decomposed into one of the following forms: S 1 2 S- 1 , S 1 1 S- 1 = 1 I 2 , S 1 1 1 S- 1 ; (2) to identify such a matrix S ; (3) to give a sufficient and necessary condition on the eigenvalues of A so that A has a particular Jordan decomposition. In either of the Jordan decomposition A = SJS- 1 , the matrix J is called a Jordan canonical form of A . Date : August 17, 2010. 1 2 KIAM HEONG KWA To begin with, let us consider the case 1 6 = 2 . Let s i = s 1 i s 2 i be an eigenvector associated with i , i = 1 , 2. Since 1 6 = 2 , the vectors s 1 and s 2 are linearly independent. Thus the matrix S = ( s 1 s 2 ) = s 11 s 12 s 21 s 22 is invertible. Now AS = a 11 a 12 a 21 a 22 s 11 s 12 s 21 s 22 = a 11 s 11 + a 12 s 21 a 11 s 12 + a 12 s 22 a 21 s 11 + a 22 s 21 a 21 s 12 + a 22 s 22 = ( A s 1 A s 2 ) = ( 1 s 1 2 s 2 ) = 1 s 11 2 s 12 1 s 21 2 s 22 = s 11 s 12 s 21 s 22 1 2 = S 1 2 , from which it follows that A = S 1 2 S- 1 . Next, let us consider the case 1 = 2 and there are two linearly independent eigenvectors s 1 and s 2 associated with 1 . By setting S = ( s 1 s 2 ) , it can be shown as in the previous case that A = S 1 1 S- 1 = I 2 thanks to the linear independence of s 1 and s 2 . This implies that A has only one eigenvalue with two linearly inde- pendent eigenvectors if and only if A is a scalar matrix, i.e., a multiple of I 2 . Finally, let us consider the case 1 = 2 and there is only one eigenvector t 1 = t 11 t 21 associated with 1 up to linear independence. From the preceding paragraph, we know that A is not a scalar ma- trix. We first show that there is an invertible matrix T such that A = T 1 1 T- 1 for some scalar 6 = 0. Replacing t 1 by t 1 k t 1 k if necessary, we may assume that t 1 is a unit vector. Let t 2 = t 12 t 22 be CANONICAL FORMS OF 2 2 MATRICES AND THEIR APPLICATIONS 3 a unit vector in C 2 that is orthogonal to t 1 so that t 11 t 12 + t 21 t 22 = 0....

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