Page 338 - 35Linear Algebra

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
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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.
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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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