Page 338 - 35Linear Algebra
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338                                                                    Sample Second Midterm


                                  Thus we may conclude P A (A) = P  −1 P A (D)P.
                                                                    
                                                     λ 1  0   · · ·  0
                                                      0            0
                                                         λ 2        
                                                                    
                                  Now suppose D =  .         .    . . Then
                                                     . .      .  .  . . 
                                                      0       · · · λ n
                                  P A (λ) = det(λI − A) = det(λP −1 IP − P  −1 DP) = detP.det(λI − D).detP

                                                                                            
                                                                  λ − λ 1    0    · · ·   0
                                                                     0    λ − λ 2         0
                                                                                            
                                                                                            
                                                                     .                    .
                                            = det(λI − D) = det     .             . .    .  
                                                                    .              .     .  
                                                                     0       0    · · · λ − λ n
                                                      = (λ − λ 1 )(λ − λ 2 ) . . . (λ − λ n ) .
                                  Thus we see that λ 1 , λ 2 , . . . , λ n are the eigenvalues of M. Finally we compute

                                                   P A (D) = (D − λ 1 )(D − λ 2 ) . . . (D − λ n )
                                                                                          
                                         0   0   · · ·  0   λ 1 0 · · ·  0       λ 1  0  · · · 0
                                                     0     0  0      0       0           0 
                                         0 λ 2                                       λ 2
                                                                          
                                     = .        .    .   .      .    .  . . .  .     .   . = 0 .
                                                                                               
                                                                               
                                       
                                                         
                                        . .      . .  .   . .    . .  . .    . .     . .  . . 
                                                      .
                                         0   0   · · · λ n  0   0 · · · λ n       0   0  · · · 0
                                  We conclude the P M (M) = 0.
                               9. A subset of a vector space is called a subspace if it itself is a vector space,
                                  using the rules for vector addition and scalar multiplication inherited from
                                  the original vector space.
                                   (a) So long as U 6= U ∪ W 6= W the answer is no. Take, for example, U
                                                           2

                                       to be the x-axis in R and W to be the y-axis. Then 1, 0 ∈ U and

                                        0, 1 ∈ W, but 1, 0 + 0, 1 = 1, 1 /∈ U ∪ W. So U ∪ W is not
                                       additively closed and is not a vector space (and thus not a subspace).
                                       It is easy to draw the example described.
                                   (b) Here the answer is always yes. The proof is not difficult. Take a vector
                                       u and w such that u ∈ U ∩ W 3 w. This means that both u and w
                                       are in both U and W. But, since U is a vector space, αu + βw is also
                                       in U. Similarly, αu + βw ∈ W. Hence αu + βw ∈ U ∩ W. So closure
                                       holds in U ∩ W and this set is a subspace by the subspace theorem.
                                                                                                         3
                                       Here, a good picture to draw is two planes through the origin in R
                                       intersecting at a line (also through the origin).


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