MATH 304 Test 1 Solutions
January 30 , 2007 Name
Put your answers in the space provided. Show your reasoning. The maximum score on the test is 30 points.
Calculators may be used unless specifically restricted.
1. 6 points Diagonalize the matrix A="0.4−0.3
0.4 1.2#.Circle your answer. Use this diagonalization to
show that Akapproaches "−0.5−0.75
1.0 1.50 #as k→ ∞.
0.4−λ−.03
0.4 1.2−λ
= 0.48 −1.6λ+λ2+ 0.12 = λ2−1.6λ+ 0.6 = (λ−0.6)(λ−1)
λ= 1."−0.6−0.3
0.4 0.2#⇒"2 1
0 0 #so the eigenvector is u="1
−2#
λ= 0.6."−0.2−0.3
0.4 0.6#⇒"2 3
0 0 #so the eigenvector is u="3
−2#
Thus A="0.4−0.3
0.4 1.2#=P"1 0
0 0.6#P−1where P="1 3
−2−2#and P−1=1
4"−2−3
2 1 #.
Ak=P DkP−1.Dk="1 0
0 0.6k#and 0.6k→0as k→ ∞. So Dk→"1 0
0 0 #
Finally, Ak→P"1 0
0 0 #P−1=1
4"1 3
−2−2#" 1 0
0 0 #" −2−3
2 1 #=1
4"−2−3
4 6 #
2. 4 points Suppose that xis an eigenvector of Acorresponding to the eigenvalue λ. Show that xis an
eigenvector of 5I−A. What is the corresponding eigenvalue?
(5I−A)x= 5x−Ax= 5x−λx= (5 −λ)x
So xis an eigenvector for 5I−Aand the eigenvalue is 5−λ
3. 4 points For what value(s) of wis the matrix "3 0
1w#diagonalizable?
If w6= 3, then the eigenvalues are 3and w; each with a linearly independent eigenvector and so A
will be diagonalizable.
If w= 3 then 3is an eigenvalue of algebraic multiplicity 2. We find its eigenvectors.
A−3I="0 0
1 0 #⇒"1 0
0 0 #. All eigenvectors will be multiples of "0
1#. We do not have
enough linearly
independent eigenvectors with which to diagonalize A. So Ais diagonalizable for all wexcept 3.
