Sol2 - cep 2008/09 MT3802 Numerical Analysis SOLUTIONS Tutorial Sheet 2 1 For a sub-ordinate matrix norm and an invertible matrix A where Ax = b

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Unformatted text preview: cep 2008/09 MT3802 - Numerical Analysis SOLUTIONS - Tutorial Sheet 2 1. For a sub-ordinate matrix norm and an invertible matrix A , where Ax = b , show that ( i ) k A s k ≤ k A k s s ∈ N Using Rule 5 (1.9) of sub-ordinate matrix norms we have k A s k ≤ k A kk A s- 1 k ≤ k A k 2 k A s- 2 k ≤ ··· ≤ k A k s- 1 k A k ≤ k A k s ( ii ) 1 k A k ≤ k x k k b k ≤ k A- 1 k Consider the linear system Ax = b ⇒ k Ax k = k b k ≤ k A kk x k ⇒ 1 k A k ≤ k x k k b k (1) and similarly, Ax = b x = A- 1 b ⇒ k x k = k A- 1 b k ≤ k A- 1 kk b k (2) Combining (1) & (2) we get: 1 k A k ≤ k x k k b k ≤ k A- 1 k . 2. Assume that ( A + δ A )- 1 is computed as an approximation to A- 1 . Show that ( A + δ A )- 1- A- 1 =- ( A + δ A )- 1 .δ A . A- 1 =- ( I + A- 1 δ A )- 1 . A- 1 δ A . A- 1 . (3) To find the first expression we can write the LHS as ( A + δ A )- 1- A- 1 = ( A + δ A )- 1 [ I- ( A + δ A ) A- 1 ] = ( A + δ A )- 1 [ I- I- δ AA- 1 ] ....
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This note was uploaded on 09/28/2009 for the course MATH MATH427 taught by Professor Dr.sharpey during the Spring '09 term at Monmouth IL.

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Sol2 - cep 2008/09 MT3802 Numerical Analysis SOLUTIONS Tutorial Sheet 2 1 For a sub-ordinate matrix norm and an invertible matrix A where Ax = b

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