Math 2270 Spring 2004
Homework 24 Solutions
Section 8.1) 2, 6, 14, 16, 22
(2) First, we find the eigenvalues of the matrix A. For a 2 x 2 matrix the characteristic
polynomial is fA (λ) = λ2 − tr(A)λ + det(A) = λ2 − 2λ = λ(λ − 2). The eigenvalues are
λ1 = 0 and λ2 = 2. Next, we must find the corresponding eigenspaces.
E0 = ker
"
#
−1 −1
−1 −1
= ker
"
1 1
0 0
#
= sp
"
1
−1
#
From our work in chapter 8.1 we know that the two eigenspaces must be orthogonal to one
another. Since we are working in ℜ2 , we know that E0 and E2 must"be #the orthogonal
1
complements of one another. We can save some work by noticing that
is orthogonal
1
"
#
1
to E0 , so E2 = sp
. We already have an eigenbasis, so all that remains is to normalize
1
√
the vectors. Both vectors have a length of 2 so the orthonormal eigenbasis is
("
√ # " √ #)
1/√2
1/ √2
,
1/ 2
−1/ 2
(6) First we find the eigenvalues.


λ −2
−2

0 
det(λI − A) = det  −2 λ − 1
 = −2 (2(λ − 1)) + (λ + 1) (λ(λ − 1) − 4)
−2
0
λ+1
= −4λ + 4 + λ3 + λ2 − λ2 − λ − 4λ − 4 = λ3 − 9λ = λ(λ2 − 9) = λ(λ − 3)(λ + 3)
Eigenvalues are λ1 = 0, λ2 = −3 and λ3 = 3. We find the eigenspaces.






1
0 1
1
0 −2 −2





0  = sp  −2 
E0 = ker  −2 −1 0  = ker  0 0

2
1 0 −1/2
−2 0 1
E−3






−2
1 0
1
−3 −2 −2






= ker  −2 −4 0  = ker  0 1 −1/2  = sp  1 
2
0 0
0
−2 0 −2






2
1 0 −2
3 −2 −2





0 
E3 = ker  −2 2
 = ker  0 1 −2  = sp  2 
1
0 0 0
−2 0 4
1
Math 2270 Spring 2004
Homework 24 Solutions
The three eigenspaces are orthogonal to one another, so all we have to do is normalize the
three vectors. All three have length equal to 3, so the basis is








 
 
2/3 
−2/3
1/3


 

2/3
1/3
−2/3 

,
,

1/3 
2/3
2/3
(14) Let S and D be the matrices found in Example 3 on pages 367-369. We know that
S −1 AS = D or A = SDS −1 .
(a) In this case we are interested in the matrix 2A, so we have 2A = 2 (SDS −1) = S(2D)S −1.
The S matrix is identical and the diagonal matrix of eigenvalues is 2D.
(b) In this case we are interested in the matrix A − 3I, so we have A − 3I = SDS −1 −
S(3I)S −1 = S (D − 3I) S −1 . The S matrix is again identical, but the diagonal matrix of
eigenvalues is D − 3I.
(c) In this case we are interested in the matrix (A−I)/2, so we have A/2−I/2 = S(D/2)S −1 −
S(I/2)S −1 = S ((D − I)/2) S −1 . The S matrix is again identical, but the diagonal matrix
of eigenvalues is (D − I)/2.
(16) We use a method similar to that used in class for Example 3 (pages 367-369). First, we
notice that the given matrix A is definitely not invertible since all of the rows are identical.
Therefore, we know that ker(A) contains more than the zero vector and λ = 0 must be an
eigenvalue. We must determine the algebraic and geometric multiplicities of λ = 0.

E0 =



ker 



−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1
−1









=



ker 



1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0









=




sp 



1
−1
0
0
0
 
 
 
 
,
 
 
 
1
0
−1
0
0
 
 
 
 
,
 
 
 
1
0
0
−1
0
 
 
 
 
,
 
 
 
We see that dim(E0 )=4, so the geometric multiplicity of λ = 0 is also 4. Now we have two
possibilities to consider. There are either no other eigenvalues, implying that λ = 0 has an
algebraic multiplicity equal to 5 (greater than its geometric multiplicity), or there is another
eigenvalue to find. We know that if there is another eigenspace Eλ , it must be orthogonal to
E0 . Let’s look for a vector orthogonal to all four vectors in the basis for E0 . By inspection
2
1
0
0
0
−1








Math 2270 Spring 2004

1

 1

we see that 
 1

 1
1
is its associated








Homework 24 Solutions
should work. We must see if this is really an eigenvector, and if so, what
eigenvalue.








1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1








1
1
1
1
1









=







5
5
5
5
5









=



5



1
1
1
1
1








We see that the chosen vector is in fact an eigenvector with eigenvalue λ = 5. Therefore, the
algebraic and geometric multiplicities of λ = 5 are both one and the algebraic and geometric
multiplicities of λ = 0 are both four.
Notice that the given matrix B = A + 2I. We can use a method similar to problem (14) to
find the eigenvalues and their multiplicities. Let S and D be the orthogonal and diagonal
matrices (respectively) such that A = SDS −1 . Then, A + 2I = SDS −1 + S(2I)S −1 =
S (D + 2I) S −1 . Therefore, the matrix B has the same eigenvectors, but they correspond to
different eigenvalues. The new eigenvalues are λ = 0 + 2 = 2 with algebraic and geometric
multiplicities of four and λ = 5 + 2 = 7 with algebraic and geometric multiplicities of one.
The det(B) is equal to the product of its eigenvalues so det(B)= 2 · 2 · 2 · 2 · 7 = 24 · 7 = 112.
(22) We know that all symmetric matrices are diagonalizable. If k = 2, the given matrix is
symmetric and therefore diagonalizable. We know that upper and lower triangular matrices
are easy to work with since their eigenvalues are sitting on the diagonal. If we let k = 0,
the matrix is upper triangular and its only eigenvalue is λ = 0 with algebraic multiplicity of
four. However, we see that the matrix A has a rank of three and therefore a nullity of one.
The dimension of the kernel of A, or of the eigenspace E0 is therefore one. So, λ = 0 has a
geometric multiplicity of one. Thus, the matrix A is not diagonalizable when k = 0.
3