Then the orthonormal vectoru j s are vectoru 1

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always associated with the largest singular value.] Then the orthonormal vectoru j ’s are vectoru 1 = Avectorv 1 3 = 1 3 1 0 1 1 0 1 parenleftbigg 1 / 2 1 / 2 parenrightbigg = 1 6 1 2 1 , vectoru 2 = Avectorv 2 1 = 1 0 1 1 0 1 parenleftbigg 1 / 2 1 / 2 parenrightbigg = 1 2 1 0 1 . 6
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The singular value decomposition of A is then A = σ 1 vectoru 1 vectorv T 1 + σ 2 vectoru 2 vectorv T 2 = 3 6 1 2 1 1 2 ( 1 1 ) + 1 2 1 0 1 1 2 ( 1 1 ) = 1 2 1 1 2 2 1 1 + 1 2 1 1 0 0 1 1 (3) Equivalently, A = U Σ V T , where U is an orthogonal 3 × 3 matrix whose columns are vectoru 1 , vectoru 2 , and vectoru 3 with vectoru 3 a unit vector orthogonal to vectoru 1 and vectoru 2 (we never need to compute vectoru 3 explicitly), V an orthogonal 2 × 2 matrix whose columns are vectorv 1 and vectorv 2 , and Σ a 3 × 2 matrix containing the singular values of A . Thus U = 1 / 6 1 / 2 . . . 2 / 6 0 vectoru 3 1 / 6 1 / 2 . . . , Σ := 3 0 0 1 0 0 , V := parenleftbigg 1 / 2 1 / 2 1 / 2 1 / 2 parenrightbigg . b) Find the best rank 1 approximation to A . Solution: From equation (3), this is σ 1 vectoru 1 vectorv T 1 = 1 2 1 1 2 2 1 1 7
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