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# lecture_22-23 - 92 Lecture 22-23 The concept of eigen...

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Unformatted text preview: 92 Lecture 22-23 The concept of eigen values and eigen vectors is applicable to any Hilbert space and for any linear transformation. The topic that deals with this concept is called linear operator theory. Procedure for finding eigen values and eigen vectors: First find the eigen values and then find the eigen vectors. Let L : S → S be a linear transformation defined by y = Ax where y is the output and A is n × n matrix. Let λ be an eigen value of this transformation. Then there must exist a vector x different from the all zero vector such that. Ax = λx ⇒ ( A − λI n ) x = 0 where I n → identity matrix of size n × n and is the all-zero vector ⇒ the null space of the linear transformation given by ˜ L : S → S , where ˜ L ( x ) = ( A − λI n ) x , contains a vector that is different from the all-zero vector ⇒ det ( A − λI n ) = 0 This follows from the following observation. Suppose det ( A − λI n ) negationslash = 0 , then the matrix ( A − λI n ) has the inverse. ⇒ x = ( A − λI n )- 1 = 0 which is not true because we assumed in the beginning that x negationslash = 0 . Hence by solving det ( A − λI n ) = 0 , we get the eigen values of the linear transformation A. To find the corresponding eigen vectors, go to the null space of ( A − λI n ) , find a collection of vectors which are linearly independent (or orthogonal) and which span the subspace given by this null space of ( A − λI n ) . Let us look at an example: Let S = C n , A = 2 0 0 3 2 1 2 1 2 3 2 det ( A − λI n ) = 0 ⇒ (2 − λ )(( 3 2 − λ ) 2 − 1 4 ) = 0 ⇒ λ = 2 , λ = 2 , λ = 1 Hence there are 2 eigen values λ = 2 and λ = 1 . Let us check the Null space of ( A − λI ) for λ = 1 . The Null space is given by all those vectors x = x 1 x 2 x 3 that satisfy 93 ( A − λI n ) x = 0 1 0 0 1 2 1 2 1 2 1 2 x 1 x 2 x 3 = ⇒ x 1 = 0 , x 2 + x 3 = 0 . This implies that the Null space is a line (see Fig. 66). So we will pick v 1 = 1 √ 2 − 1 √ 2 section of x +x =0 2 section of x =0 1 x 3 3 x 2 x 1 Figure 66: Null Space as our first eigen vector corresponding to λ = 1 . Now let us check the Null space of ( A − 2 I ) − 1 2 1 2 1 2 − 1 2 x 1 x 2 x 3 = ⇒ − x 2 + x 3 = 0 . This implies that the Null space is a plane. So we will pick 2 vectors from this null space that are linearly independent v 2 = 1 , v 3 = 1 √ 2 1 √ 2 Hence for the given linear transformation A, we have found 2 distinct eigen values and 3 distinct (linearly independent) eigen vectors....
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lecture_22-23 - 92 Lecture 22-23 The concept of eigen...

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