21-241: Matrix Algebra – Summer I, 2006

Practice Exam 4 Solutions

 
1 −3 −7
1. Let A = 0 −2 −6.
0 2 5
(a) Find the characteristic polynomial of A.
Solution.
 
1−λ −3 −7 µ ¶
−2 − λ −6
det(A − λI) = det  0 −2 − λ −6  = (1 − λ) det
2 5−λ
0 2 5−λ
= (1 − λ)[(−2 − λ)(5 − λ) − (−6)2]
= −λ3 + 4λ2 − 5λ + 2.

(b) Find all eigenvalues and their multiplicities of A.

Solution. Since det(A − λI) = −λ3 + 4λ2 − 5λ + 2 = −(λ − 1)2 (λ − 2), A has two distinct
eigenvalues: λ1 = 1 with multiplicity 2, λ2 = 2 with multiplicity 1.
(c) For each eigenvalue, find a basis for the corresponding eigenspace. Determine which eigenvalues
are complete.
Solution. To find a basis for Vλ1 , we are to solve the system (A − λ1 I)x = 0.
     
0 −3 −7 0 −3 −7 R + 2 R 0 -3 −7
R2 −R1 3 2
A − λ1 I = 0 −3 −6 −−− −−→ 0 0 1  −−−−3−→ 0 0 1
2
R3 + 3 R1 2
0 2 4 0 0 −3 0 0 0

Thus the general solution is

     
x1 x1 1
x = x2  =  0  = x1 0 .
x3 0 0

So (1, 0, 0)T is a basis for Vλ1 . Since dim Vλ1 = 1, less than the multiplicity, λ1 is not complete.
Similarly, we solve the system (A − λ2 I)x = 0.
   
−1 −3 −7 R + 1 R -1 −3 −7
3 2
A − λ2 I =  0 −4 −6 −−−−2−→  0 -4 −6 .
0 2 3 0 0 0

Thus the general solution is

   5   
x1 − 2 x3 −5
x1 
x = x2  = − 32 x3  = −3 .
2
x3 x3 2

So (−5, −3, 2)T is a basis for Vλ2 . Since dim Vλ2 = 1, equal to the multiplicity, λ2 is complete.
(d) Is A complete? diagonalizable?
Solution. Since A has an incomplete eigenvalue, it’s not complete, thus not diagonalizable.

1
Matrix Algebra Practice Exam 4 Solutions

2. Write down a real matrix that has

µ ¶ µ ¶
−1 1
(a) eigenvalues −1, 3 and corresponding eigenvectors , .
2 1
µ ¶ µ ¶
−1 1
Solution. Since , are linearly independent (why?), they form an eigenvector basis
2 1
µ ¶ µ ¶
2 −1 1 −1 0
for R . Thus the matrix, denoted by A, is diagonalizable. Let S = , Λ= .
2 1 0 3
Then, µ ¶µ ¶ µ 1 1¶ µ5 4¶
−1 −1 1 −1 0 −3 3
A = SΛS = 2 1 = 83 31 .
2 1 0 3 3 3 3 3
µ ¶
1+i
(b) an eigenvalue −1 + 2i and corresponding eigenvector .
3i
Solution. Since the wanted matrix, denoted by A, is real, the complex conjugate of µ −1 + 2i,
¶
1+i
namely −1 − 2i, must also be an eigenvalue of A. Moreover, the complex conjugate of ,
3i
µ ¶ µ ¶ µ ¶
1−i 1+i 1−i
namely , must be an eigenvector corresponding to −1 − 2i. Since , are
−3i 3i −3i
linearly independent, A allows an eigenvector basis for C2 .
µ ¶ µ ¶
1+i 1−i −1 + 2i 0
Let S = ,Λ= . Then,
3i −3i 0 −1 − 2i
µ ¶µ ¶ µ1 ¶ µ ¶
−1 1+i 1−i −1 + 2i 0 2 − 16 − 16 i −3 43
A = SΛS = 1 = .
3i −3i 0 −1 − 2i 2 − 16 + 16 i −6 1

3. Show that λ is an eigenvalue of A if and only if λ is an eigenvalue of AT .

Proof. If λ is an eigenvalue of A, then det(A − λI) = 0. Thus,

det(AT − λI) = det(AT − λI T ) = det(A − λI)T = det(A − λI) = 0.

This implies that λ is an eigenvalue of AT .

2x2 + xy + 3xz + 2y 2 + 2z 2
4. Find the minimum and maximum value of the rational function .
x2 + y 2 + z 2
Solution. Let v = (x, y, z)T , then

2x2 + xy + 3xz + 2y 2 + 2z 2 vT Kv
= ,
x2 + y 2 + z 2 kvk2
 1 3

2 2 2
where K =  12 2 0 . By the optimization principles for eigenvalues,
3
2 0 2
½ ¯ ¾ ½ ¯ ¾
vT Kv ¯¯ vT Kv ¯¯
max v =
6 0 = λ1 , min v =
6 0 = λ3 ,
kvk2 ¯ kvk2 ¯

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Matrix Algebra Practice Exam 4 Solutions

where λ1 and λ3 are respectively the largest and smallest eigenvalues of K. The characteristic equation
of K is
 1 3

2−λ 2 2
det(K − λI) = det  12 2−λ 0 
3
0 2−λ
µ2 1 3
¶ µ 1
¶
3 2 2 + (2 − λ) det 2 − λ 2
= det 1
2 2−λ 0 2 2−λ
· ¸ " µ ¶2 #
3 3 1
= 0 − (2 − λ) + (2 − λ) (2 − λ)2 −
2 2 2
· ¸
5
= −(λ − 2) (λ − 2)2 −
2
à r !à r !
5 5
= −(λ − 2) λ − 2 − λ−2+ = 0.
2 2
q q q
5 5 5
Therefore, λ1 = 2 + 2, λ3 = 2 − 2. So, the maximum value of the function is 2 + 2 and the
q
5
minimum value of the function is 2 − 2.