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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 β i be a basis for W i for i = 1 , . . ., k . And let β = β 1 ∪ ···∪ β k . Then it is obvious that β spans V . Therefore, | β | ≥ dim ( V ) = k summationdisplay i =1 dim ( W i ) = k summationdisplay i =1 | β i | ≥ | β | . Since β spans V with | β | = dim ( V ) , β is a basis for V , i.e., β is a linearly independent set. Hence, for each i , span ( β i ) ∩ span ( k uniondisplay j =1 ,j negationslash = i β 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 λ 1 , λ 2 , . . ., λ k . Prove that span ( { x ∈ V : x is an eigenvector of T } ) = E λ 1 ⊕ E λ 2 ⊕ ··· ⊕ E λ 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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