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Unformatted text preview: (Partial) Solutions to Homework 6 Problem 15.2 Let M = 2 1 0 2 . Find all eigenvalues of M . Does M have two independent eigenvectors? Can M be diagonalized? Solution: The characteristic polynomial of M is (2 ) 2 , so the only eigen value is = 2. Any eigenvector of M is a solution to ( M 2 I ) X = 0. If we try to solve this system, we get 0 1 0 0 x 1 x 2 = . Therefore, the solution set to this system is all vectors of the form t , or in other words, the solu tion set is span { 1 } . This is only onedimensional, so we can only have one linearly independent eigenvector. Therefore, M cannot be diagonalized! Why not? Suppose we had a basis where M was diagonal, say T = { t 1 ,t 2 } was such a basis, and the diagonalized version of M is D = a d . This would mean that Dt 1 = at 1 and Dt 2 = dt 2 . In other words, this would mean that t 1 and t 2 were both eigenvectors. But we only have one eigenvector. The big idea diagonalizable if and only if there is a basis of eigenvectors. Problem 19.2 Show that the set of all linear transformations mapping R 3 R is itself a vector space. Find a basis for this vector space. Do you think your proof could be modified to work for linear transformations R n R ?...
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 Spring '10
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 Eigenvectors, Vectors

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