<p>Diagonalization of a Matrix:</p><p>Definition of diagonalization of a matrix: A matrix A is diagonalizable if there exists a nonsingular matrix P and a diagonal matrix D such that D  P1 AP .</p><p>Example:</p><p>Let  4  6 A    .  3 5  Then,</p><p>1 2 0  1  2  4  61  2 1 D           P AP, 0 1  1 1   3 5  1 1  where 2 0  1  2 D   , P    . 0 1  1 1 </p><p>Theorem: An n  n matrix A is diagonalizable if and only if it has n linearly independent eigenvector.</p><p>[proof:]</p><p>:</p><p>A is diagonalizable. Then, there exists a nonsingular matrix P and a diagonal matrix </p><p>1 1 0 ⋯ 0   0  ⋯ 0  D   2  ,  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  such that </p><p>D  P 1 AP  AP  PD </p><p>1 0 ⋯ 0   0  ⋯ 0  .  Acol (P) col (P) ⋯ col (P) col (P) col (P) ⋯ col (P) 2  1 2 n 1 2 n  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  Then,</p><p>Acol j (P)   j col j (P), j  1,2,, n. That is, </p><p> col1 (P),col2 (P),,coln (P) are eigenvectors associated with the eigenvalues 1 ,2 ,,n .</p><p>Since P is nonsingular, thus col1 (P),col2 (P),,coln (P) are linearly independent. </p><p>:</p><p>Let x1 , x2 ,, xn be n linearly independent eigenvectors of A associated with the eigenvalues 1 ,2 ,,n . That is, </p><p>Ax j   j x j , j 1,2,, n. Thus, let </p><p>P  x1 x2 ⋯ xn  i.e., col j (P)  x j  and </p><p>2 1 0 ⋯ 0   0  ⋯ 0  D   2  .  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n </p><p>Since Ax j   j x j ,</p><p>1 0 ⋯ 0   0  ⋯ 0  AP  Ax x ⋯ x   x x ⋯ x  2   PD . 1 2 n 1 2 n  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  Thus, P 1 AP  P 1PD  D ,</p><p>1 P exists because x1 , x2 ,, xn are linearly independent and thus P is nonsingular.</p><p>Important result: An n  n matrix A is diagonalizable if all the roots of its characteristic equation are real and distinct.</p><p>Example:</p><p>Let  4  6 A    .  3 5  Find the nonsingular matrix P and the diagonal matrix D such that D  P1 AP and find An , n is any positive integer.</p><p>[solution:]</p><p>3 We need to find the eigenvalues and eigenvectors of A first. The characteristic equation of A is   4 6 det(I  A)    1  2  0 .  3   5    1 or 2 . By the above important result, A is diagonalizable. Then, </p><p>1. As   2 , 1 Ax  2x  2I  Ax  0  x  r , r  R.  1  2. As   1,  2 Ax  x   I  Ax  0  x  t , t  R.  1  Thus, 1  2   and    1   1  are two linearly independent eigenvectors of A.</p><p>Let 1  2 2 0  P    and D    .  1 1  0 1 Then, by the above theorem, D  P1 AP .</p><p>To find An , </p><p> n n 2 0  1 1 1 1 n D   n   P APP AP⋯P AP P A P  0 1  n times Multiplied by P and P 1 on the both sides, </p><p>4 n 1 n 1 1 n 1 n 1  22 0 1  2 PD P  PP A PP  A    n    1 1  0 1  1 1   2 n  2  1n1  2 n1  2  1n1        n n1 n1 n1   2  1 2  1 </p><p>Note (very important): If A is an n  n diagonalizable matrix, then there exists an nonsingular matrix P such that D  P1 AP , where col1 (P),col2 (P),,coln (P) are n linearly independent eigenvectors of A and the diagonal elements of the diagonal matrix D are the eigenvalues of A associated with these eigenvectors. </p><p>Note: For any n  n diagonalizable matrix A, D  P1 AP, then Ak  PDk P 1, k 1,2, where </p><p> k 1 0 ⋯ 0    0 k ⋯ 0 Dk   2  .  ⋮ ⋮ ⋱ ⋮  k   0 0 ⋯ n </p><p>Example:</p><p>5  3 Is A    diagonalizable? 3 1</p><p>[solution:]</p><p>5   5 3 2 det(I  A)     2  0 .  3  1</p><p>Then,   2, 2 .</p><p>As   2,</p><p>1 2I  Ax  0  x  t , t  R. 1 1 Therefore, all the eigenvectors are spanned by   . There does not exist two linearly 1 independent eigenvectors. By the previous theorem, A is not diagonalizable. </p><p>Note: An n  n matrix may fail to be diagonalizable since  Not all roots of its characteristic equation are real numbers.  It does not have n linearly independent eigenvectors.</p><p>Note:</p><p>The set S j consisting of both all eigenvectors of an n  n matrix A</p><p> n associated with eigenvalue  j and zero vector 0 is a subspace of R . S j is</p><p> called the eigenspace associated with  j .</p><p>6</p>