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which implies +(α)T =eTR(α)−1V−1, (2.39) 1 = e1T V−1 − w(α)T V−1 . (2.40) The only dependence on α is in w(α). This fact appears unfortunate because w(α) solves the system w(α)T (I − αJ) = (1 − α)(eT − vT V), but J has 1 structure that makes the solution obvious. An additional concern is that +(α)T includes the first row of V−1, a vector that may be arbitrary! Let’s resolve these concerns. We can decompose12 ⎡⎢I ⎤⎥ ⎢⎥ J= ⎢D⎥ , (2.41) ⎢1⎥ ⎢ J ⎥ ⎣ 2⎦ where D1 is a diagonal matrix of all eigenvalues on the unit circle not equal to 1 and J2 is a Jordan matrix with all eigenvalues on the interior of the unit circle. If we conformally partition w(α)T = (w (α)T w (α)T w (α)T ) 012 12 You might think this next step is incorrect because it asserts a form on J that possibly requires reordering V. The algebra, however, still works if we assert this form on step 1. 13 Remember that X and V are going to slightly different because they are Jordan forms of transposed matrices. andV=(V V V )13 then 012 w0(α)T=e1−vTV0, (2.42) w (α)T =−(1−α)vTV (I−αD )−1, and (2.43) w (α)T =−(1−α)vTV (I−αJ )−1. (2.44) 222 in +(α)T . After partitioning we have ⎡⎢ Z ⎤⎥ ⎢ 0⎥ V−1 = ⎢Z ⎥ ⎢ 1⎥ ⎢Z ⎥ ⎣ 2⎦ (2.45) 111 Both of the linear systems for w1 (α) and w2 (α) are non-singular for 0 ≤ α ≤ 1 and w0(α) is a constant function! The solution of w(α)T also resolves our second concern, the eT V−1 factor 1 +(α)T = vTV Z +(1−α)vTV (I−αD )−1Z +(1−α)vTV (I−αJ )−1Z , 00111222 (2.46) 2.7 ⋅ the limit case 39PDF Image | Instagram Cheat Sheet
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