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**Unformatted text preview: **Homework 2 Problem. (5-2. # 20) Prove that V = k summationdisplay i =1 W k is the direct sum of W 1 , . . . , W k if and only if dim ( V ) = k summationdisplay i =1 dim ( W k ) . Solution. ( ) This is trivial. ( ) Now suppose that dim ( V ) = k summationdisplay i =1 dim ( W k ) . Let i be a basis for W i for i = 1 , . . ., k . And let = 1 k . Then it is obvious that spans V . Therefore, | | dim ( V ) = k summationdisplay i =1 dim ( W i ) = k summationdisplay i =1 | i | | | . Since spans V with | | = dim ( V ) , is a basis for V , i.e., is a linearly independent set. Hence, for each i , span ( i ) span ( k uniondisplay j =1 ,j negationslash = i j ) = { } , that is, W i parenleftBig k summationdisplay j =1 ,j negationslash = i W j parenrightBig = { } , which means that V is the direct sum of W 1 , . . ., W k . Problem. (5-2. # 22) Suppose that T is a linear operator on a finite dimensional vector space V and suppose that the distinct eigenvalues of T are 1 , 2 , . . ., k . Prove that span ( { x V : x is an eigenvector of T } ) = E 1 E 2 E k . 1 Solution. We denote by W the subspace spanned by all the eigenvalues of T , i.e., W = span ( { x V : x is an eigenvector of T } ) ....

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