HW6sol - (Partial Solutions to Homework 6 21 Find all...

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(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 - λ ) 2 , so the only eigen- value is λ = 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 = 0 0 . Therefore, the solution set to this system is all vectors of the form t 0 , or in other words, the solu- tion set is span { 1 0 } . This is only one-dimensional, 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 0 0 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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