# We found p ap p 1 ap d for the matrices p 2 i 6 2 i

• 147
• 100% (8) 8 out of 8 people found this document helpful

This preview shows page 127 - 130 out of 147 pages.

We found P * AP = P - 1 AP = D for the matrices P = (2 + i ) / 6 (2 + i ) / 30 - 1 / 6 5 / 30 and D = 0 0 0 6 . 121
7 7. Complex matrices and vector spaces Therefore the spectral decomposition of A is A = 0 E 1 + 6 E 2 . Clearly, in this case, we only need to find the matrix E 2 = x 2 x * 2 . We have A = 6 1 30 2 + i 5 ( 2 - i 5 ) = 1 5 5 5(2 + i ) 5(2 - i ) 25 = 1 2 + i 2 - i 5 , as expected. So the spectral decomposition of this matrix is just the matrix A itself. Example 7.23 In Section 4.1 we orthogonally diagonalised the matrix B = 2 1 1 1 2 1 1 1 2 . We found that P T BP = D with the matrices P = - 1 6 - 1 2 1 3 - 1 6 1 2 1 3 2 6 0 1 3 and D = 1 0 0 0 1 0 0 0 4 . The spectral decomposition of B is B = 1 E 1 + 1 E 2 + 4 E 3 , where the matrices E i = x i x * i are obtained from the orthonormal eigenvectors of B (which are the columns of P ). We have, E 1 = 1 6 2 - 1 - 1 2 ( - 1 - 1 2 ) = 1 6 1 1 - 2 1 1 - 2 - 2 - 2 4 and similarly, E 2 = 1 2 1 - 1 0 - 1 1 0 0 0 0 , E 3 = 1 3 1 1 1 1 1 1 1 1 1 . The spectral decomposition of B is B = 1 6 1 1 - 2 1 1 - 2 - 2 - 2 4 + 1 2 1 - 1 0 - 1 1 0 0 0 0 + 4 3 1 1 1 1 1 1 1 1 1 . Activity 7.22 Check all the calculations for this example. Let us take a closer look at the matrices E i . As the following theorem shows, we will find that they are Hermitian, they are idempotent, and, for i 6 = j , they satisfy E i E j = 0 where 0 is the zero matrix. For a proof, see the Anthony and Harvey book. Theorem 7.16 If { x 1 , x 2 , . . . , x n } is an orthonormal basis of C n , then the matrices E i = x i x * i have the following properties: (i) E * i = E i . 122
7 7.7. Spectral decomposition (ii) E 2 i = E i . (iii) E i E j = 0 if i 6 = j . The fact that each E i is an idempotent matrix means that it represents a projection. To see what this projection is, look at its action on the orthonormal basis vectors: E i x i = ( x i x * i ) x i = x i ( x * i x i ) = x i h x i , x i i = x i (1) = x i and, for i 6 = j , E i x j = ( x i x * i ) x j = x i ( x * i x j ) = x i h x j , x i i = x i · 0 = 0 . If v is any vector in C n , v can be written as a unique linear combination v = a 1 x 1 + a 2 x 2 + · · · + a n x n of these basis vectors, so that E i v = E i ( a 1 x 1 + a 2 x 2 + · · · + a n x n ) = a 1 E i x 1 + a 2 E i x 2 + · · · + a i - 1 E i x i - 1 + a i E i x i + a i +1 E i x i +1 + · · · + a n E i x n = a i x i . Therefore E i is the orthogonal projection of C n onto the subspace spanned by the vector x i . Activity 7.23 Look at the previous example and write down the orthogonal projection of C n onto the subspace Lin { (1 , 1 , 1) T } . Matrices which satisfy properties (ii) and (iii) of Theorem 7.16 have an interesting application. Suppose E 1 , E 2 , E 3 are three such matrices, that is E i E j = E i for i = j 0 for i 6 = j for i = 1 , 2 , 3. Then for any real numbers α 1 , α 2 , α 3 , and any positive integer n , we will show that ( α 1 E 1 + α 2 E 2 + α 3 E 3 ) n = α n 1 E 1 + α n 2 E 2 + α n 3 E 3 .