Answer yes consider an eigenvector associated with ?

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Answer: Yes. Consider an eigenvector associated with λ = 1 / 2. Set x 0 = (1 , - 2) T . Then A t x 0 = (1 / 2) t x 0 . Since (1 / 2) t 0, A t x 0 0 . 3. Two embedding problems: a ) Consider A = (0 , 1) R , and B = { ( x, 0) R 2 : 0 < x < 1 } ⊂ R 2 . Is A open? Is B open? If the two differ, explain why embedding A in R 2 changed the result. Answer: The set A is open. Let x A and set = min( x, 1 - x ). Then B ( x ) A . Since A contains an open ball about each of its points, it is an open set. The set B is not open. Let ( x, 0) B and consider B ( x, 0). The point ( x, / 2) B ( x, 0), but is not in B . Thus there is a point in B so that no open ball about that point is contained in B . (In fact, this is true of all points in B ). Therefore B is not open. The difference occurs because of the extra dimension. Balls in R 2 contain points not on the horizontal axis, and so cannot be contained in sets restricted to the horizontal axis. With balls in R 1 , this is not an issue. b ) Now suppose C is a closed set in R 2 . Show that D = { ( x, y, 0) : ( x, y ) C } is closed in R 3 . Answer: Let ( x n , y n , z n ) D with ( x n , y n , z n ) ( x, y, z ). Since z n = 0, z = 0. Because C is closed and ( x n , y n ) C and ( x n , y n ) ( x, y ), we find ( x, y ) C , so ( x, y, z ) = ( x, y, 0) D . Thus D is closed because it contains all of its limit points.
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