MATH 115 SOLUTION SET 7ANSWERS TO SELECTED PROBLEMS§5.17.The point of this problem is to extend the definition of determinant frommatrices to linear transformations. The result is always that a property satisfiedby matrices is always satisfied by linear transformations as well:a. IfQis the change of basis matrix fromβtoγ, then [T]β=Q[T]γQ-1, so that[T]βand [T]γare similar and therefore have the same determinant.b.Tis invertible if and only if the matrix [T]βis invertible, which is so if andonly if det[T]β6= 0. But det[T]βis the same as detT.c. We have det(T-1) = det[T-1]β= det[T]-1β= (detT)-1.d.We have det(TU) = det([TU]β) = det([T]β[U]β) = det([T]β) det([U]β) =det(TU).e. We have det(T-λIV) = det([T-λIV]β) = det([T]-λI). The second equalitycomes from Thm. 2.8./8. a.Tis invertible if and only if detT6= 0, which is true if and only if 0 is nota root of the characteristic polynomial det(T-λIV), which is true if and only if 0is not an eigenvalue ofT.b. We have thatλis an eigenvalue ofTif and only if for somev∈V, we haveTv=λv. But sinceTis invertible, every eigenvalue is nonzero and we may rewrite
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