Math 113 Final Exam Solutions
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چکیده
(This can be shown using a basis argument, which would amount to what most of you wrote anyway.) Thus, given S ∈ L(V ) such that imS ⊆ imT , we can define R = (T |W )−1S, from which it follows TR = S. Conversely, if S = TR, then since linear operators preserve inclusion and R(V ) ⊆ V , we have S(V ) = (TR)(V ) ⊆ T (V ); that is, imS ⊆ imT . 2. Let T be a linear operator on V with distinct eigenvalues λ1, . . . , λn and corresponding eigenvectors v1, . . . , vn. For each 1 ≤ i ≤ n, define the normalized Lagrange interpolating polynomial:
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