EXERCISES AND SOLUTIONS

IN LINEAR ALGEBRA

Mahmut Kuzucuoğlu
Middle East Technical University
matmah@metu.edu.tr
Ankara, TURKEY
March 14, 2015
ii

TABLE OF CONTENTS
CHAPTERS
0. PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1. LINEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ??
2. MAP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ??
3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ??
4. ?? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ??
5. ??? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ??
 Preface

I have given some linear algebra courses in various years. These

problems are given to students from the books which I have followed
that year. I have kept the solutions of exercises which I solved for the
students. These notes are collection of those solutions of exercises.

Mahmut Kuzucuoğlu
METU, Ankara
March 14, 2015

M. Kuzucuoğlu

1
 Name : Answer Key

METU
.......
...... .... ID Number : 2360262
.................... .................... DEPARTMENT OF MATHEMATICS
.............. ............ Signature : 0000000
 ........
.. Math 262 Quiz I Duration : 60 minutes
1956
(05.03.2008)

Show all your work. Unsupported answers will not be graded.

 
2 4 −2
1.) Let A =  0 −4 3 .
 

−3 −7 4

(a) Find the characteristic polynomial of A.

Solution. The characteristic polynomial of A is f (x) = det(xI − A). So,
 
 x − 2 −4 2 

f (x) =  0 x + 4 −3 
 
 
 3 7 x−4 
   
 x + 4 −3   −4 2 
= (x − 2)   + 3
   

 7 x−4   x + 4 −3 
= (x − 2)(x2 − 16 + 21) + 3(12 − 2x − 8)
= (x − 2)(x2 + 5) + 3(4 − 2x)
= (x − 2)(x2 + 5 − 6)
= (x − 2)(x − 1)(x + 1)

(b) Find the minimal polynomial of A.

Solution. We know that the minimal polynomial divides the characteristic polynomial and they same
the same roots. Thus, the minimal polynomial for A is mA (x) = f (x) = (x − 2)(x − 1)(x + 1).

(c) Find the characteristic vectors and a basis B such that [A]B is diagonal.
Solution. The characteristic values of A are c1 = 2, c2 = 1, c3 = −1.
   
0 4 −2 −3 −7 2
−3x − 7y + 2z = 0 z = 2y
A − 2I =  0 −6 3  −→  0 2 −1  , ⇒
   
2y − z = 0 x = −y
−3 −7 2 0 0 0

Thus, α1 = (−1, 1, 2) is a characteristic vector associated with the characteristic value c1 = 2.

   
1 4 −2 1 4 −2 y = 3k
x + 4y − 2z = 0
A − I =  0 −5 3  −→  0 −5 3 , ⇒ z = 5k
   
−5y + 3z = 0
−3 −7 3 0 0 0 x = −2k

Thus, α2 = (−2, 3, 5) is a characteristic vector associated with the characteristic value c2 = 1.

   
3 4 −2 3 4 −2 x = −2t
3x + 4y − 2z = 0
A + I =  0 −3 3  −→  0 1 −1  , ⇒ y = 3t
   
y−z =0
−3 −7 5 0 0 0 z = 3t
 Thus, α3 = (−2, 3, 3) is a characteristic vector associated with the characteristic value c3 = −1.
 
2 0 0
Now, B = {α1 , α2 , α3 } is a basis and [A]B =  0 1 0  is a diagonal matrix.
 

0 0 −1

(d) Find A -conductor of the vector α = (1, 1, 1) into the invariant subspace spanned by (−1, 1, 2).
Solution. Set W =< (−1, 1, 2) > and denote the A -conductor of α into W by g(x). Since
mA (A) = 0 we have mA (A)α ∈ W. Thus, g(x) divides mA (x). Hence, the possibilities for g(x)
are x − 2, x − 1, x + 1, (x − 2)(x − 1), (x − 2)(x + 1), (x − 1)(x + 1). We will try these polynomials.
(Actually, the answer could be given directly.) Now,
     
0 4 −2 1 2
(A − 2I)α =  0 −6 3  ·  1  =  −3  ∈ / W ⇒ g(x) 6= x − 2,
     

−3 −7 2 1 −8
     
1 4 −2 1 3
(A − I)α =  0 −5 3  ·  1  =  −2  ∈ / W ⇒ g(x) 6= x − 1,
     

−3 −7 3 1 −7
     
3 4 −2 1 5
(A + I)α =  0 −3 3 · 1 = 0 ∈ / W ⇒ g(x) 6= x + 1,
     

−3 −7 5 1 −5
     
0 4 −2 3 6
(A − 2I)(A − I)α =  0 −6 3  ·  −2  =  −9  ∈ / W ⇒ g(x) 6= (x − 2)(x − 1),
     

−3 −7 2 −7 −9
     
0 4 −2 5 10
(A − 2I)(A + I)α =  0 −6 3  ·  0  =  −15  ∈ / W ⇒ g(x) 6= (x − 2)(x + 1),
     

−3 −7 2 −5 −25
     
1 4 −2 5 15
(A − I)(A + I)α =  0 −5 3  ·  0  =  −15  = −15α1 ∈ W ⇒ g(x) = x2 − 1.
     

−3 −7 3 −5 −30

2.) Find a 3 × 3 matrix whose minimal polynomial is x2 .

 
0 0 1
Solution. For the matrix A =  0 0 0  we have A 6= 0 and A2 = 0. Thus, A is a 3 × 3 matrix
 

0 0 0
2
whose minimal polynomial is x .

3.) Prove that similar matrices have the same minimal polynomial.
Solution. Let A and B be similar matrices, i.e., B = P −1 AP for some invertible matrix P. For
any k > 0 we have B k = (P −1 AP )k = P −1 Ak P which implies that f (B) = P −1 f (A)P for any
polynomial f (x). Let fA and fB be the minimal polynomials of A and B, respectively. Then fA (B) =
P −1 fA (A)P = P −1 OP = O implies that fB divides fA . On the other hand, O = fB (B) = P −1 fB (A)P
gives us fB (A) = O. Hence, fA divides fB . Therefore, we have fA = fB .
2 M. KUZUCUOĞLU

1. Math 262 Exercises and Solutions

(1) Let A be a 3 × 3 matrix with real entries. Prove that if A is

not similar over R to a triangular matrix then A is similar over
C to a diagonal matrix.
Proof. Since A is a 3 × 3 matrix with real entries, the
characteristic polynomial, f (x), of A is a polynomial of degree
3 with real coefficients. We know that every polynomial of
degree 3 with real coefficients has a real root, say c1 .
On the other hand, since A is not similar over R to a tri-
angular matrix, the minimal polynomial of A is not product
of polynomials of degree one. So one of the irreducible factor,
h, of the minimal polynomial of A is degree 2. Then h has
two complex roots, one of which is the conjugate of the other.
Thus, the characteristic polynomial has one real root and two
complex roots, c1 , λ and λ̄.
The minimal polynomial over complex numbers is (x −
c1 )(x − λ)(x − λ̄) which implies that A is diagonalizable over
complex numbers.

(2) Let T be a linear operator on a finite dimensional vector space

over an algebraically closed field F. Let f be a polynomial over
F. Prove that c is a characteristic value of f (T ) if and only if
f (t) = c where t is a characteristic value of T.
Proof. Let t be a characteristic value of T and β be a non-
zero characteristic vector associated with the characteristic
value t. Then, T β = tβ, T 2 β = T (T β) = T (tβ) = tT β = t2 β,
and inductively we can see that T k β = tk β for any k ≥ 1.
Thus, for any polynomial f (x) we have f (T )β = f (t)β which
means, since β ̸= 0, that f (t) is a characteristic value of the
linear operator f (T ).
Assume that c is a characteristic value of f (T ). Since F is
algebraically closed, the minimal polynomial of T is product
of linear polynomials, that is, T is similar to a triangular op-
erator. If [P −1 T P ]B is triangular matrix, then [P −1 f (T )P ]B is
 EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 3

also triangular and on the diagonal of [P −1 f (T )P ]B we have

f (ci ), where ci is a characteristic value of T.

(3) Let c be a characteristic value of T and let W be the space of

characteristic vectors associated with the characteristic value
c. What is the restriction operator T |W .
Solution. Every vector v ∈ W is a characteristic vector.
Hence, T v = cv for all v ∈ W. Therefore, T |W = cI.

(4) Every matrix A satisfying A2 = A is similar to a diagonal

matrix.
Solution. A satisfies the polynomial x2 −x. Thus, the min-
imal polynomial, mA (x), of A divides x2 −x, that is mA (x) = x
or mA (x) = x − 1 or mA (x) = x(x − 1).
If mA (x) = x, then A = 0.
If mA (x) = x − 1, then A = I.
If mA (x) = x(x − 1), then the minimal polynomial of A
is product of distinct polynomials of degree one. Thus, by
a Theorem, the matrix A is similar to diagonal matrix with
diagonal entries consisting of the characteristic values, 0 and
1.

(5) Let T be a linear operator on V. If every subspace of V is

invariant under T then it is a scalar multiple of the identity
operator.
Solution. If dim V = 1 then for any 0 ̸= v ∈ V, we have
T v = cv, since V is invariant under T. Hence, T = cI.
Assume that dim V > 1 and let B = {v1 , v2 , · · · , vn } be
a basis for V. Since W1 = ⟨v1 ⟩ is invariant under T, we have
T v1 = c1 v1 . Similarly, since W2 = ⟨v2 ⟩ is invariant under T, we
have T v2 = c2 v2 . Now, W3 = ⟨v1 +v2 ⟩ is also invariant under T.
Hence, T (v1 +v2 ) = λ(v1 +v2 ) or c1 v1 +c2 v2 = λ(v1 +v2 ), which
gives us (c1 − λ)v1 + (c2 − λ)v2 = 0. However, v1 and v2 are
linearly independent and hence we should have c1 = c2 = λ.
Similarly, one can continue with the subspace ⟨v1 + v2 + v3 ⟩
4 M. KUZUCUOĞLU

and observe that T (v3 ) = λv3 . So for any vi ∈ B, we have

T vi = λvi . Thus, T = λI.

(6) Let V be the vector space of n × n matrices over F. Let A be

a fixed n × n matrix. Let T be a linear operator on V defined
by T (B) = AB. Show that the minimal polynomial of T is the
minimal polynomial of A.
Solution. Let mA (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be
the minimal polynomial of A, so that mA (A) = 0. It is easy
to see that T k (B) = Ak B for any k ≥ 1. Then, for any B ∈ V
we have

mA (T )B = (T n + an−1 T n−1 + · · · + a1 T + a0 I)B

= T n (B) + an−1 T n−1 (B) + · · · + a1 T (B) + a0 B
= An B + an−1 An−1 B + · · · + a1 AB + a0 B
= (An + an−1 An−1 + · · · + a1 A + a0 I)B
= mA (A)B = 0.

Thus, we obtain mA (T ) = 0, which means that mT (x) divides

mA (x).
Now, let mT (x) = xm + cm−1 xm−1 + · · · + c1 x + c0 be the
minimal polynomial of T, so that mT (T ) = 0. Then, for any
B ∈ V we have

mT (A)B = (Am + cm−1 Am−1 + · · · + c1 A + c0 I)B

= Am B + cm−1 Am−1 B + · · · + c1 AB + c0 B
= T m (B) + cm−1 T m−1 (B) + · · · + c1 T (B) + c0 B
= (T m + cm−1 T m−1 + · · · + c1 T + c0 I)B
= mT (T )B = 0,

which leads to mT (A) = 0, meaning that mA (x) divides mT (x).

Since, monic polynomials dividing each other are the same we
have mT (x) = mA (x).
 EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 5

(7) If E is a projection and f is a polynomial, then show that

f (E) = aI + bE. What are a and b in terms of the coefficients
of f ?
Solution. Let f (x) = c0 + c1 x + · · · + cn xn . Then, f (E) =
c0 I + c1 E + · · · + cn E n . Since E is a projection, (E 2 = E), we
have E k = E for any k ≥ 1. Then,

f (E) = c0 I + c1 E + · · · + cn E n
= c0 I + c1 E + · · · + cn E
= c0 I + (c1 + · · · + cn )E.

Thus, a is the constant term of f and b is the sum of all other

coefficients.

(8) Let V be a finite dimensional vector space and let W1 be any

subspace of V. Prove that there is a subspace W2 of V such
that V = W1 ⊕ W2 .
Proof. Let BW1 = {β1 , · · · , βk } be a basis for W1 . We may
extend BW1 to a basis BV of V, say BV = {β1 , · · · , βk , βk+1 , · · · , βn }.
Let W2 be the subspace spanned by βk+1 , · · · , βn . Then, as
they are linearly independent in V, we have BW2 = {βk+1 , · · · , βn }.
Clearly W1 + W2 = V as W1 + W2 contains a basis of V
and so spans V. Let β ∈ W1 ∩ W2 . Then, β ∈ W1 implies
that β = c1 β1 + · · · + ck βk , and β ∈ W2 implies that β =
ck+1 βk+1 + · · · + cn βn . The last two equalities give us c1 β1 +
· · ·+ck βk −ck+1 βk+1 −· · ·+cn βn = 0, but since βi ’s are linearly
independent, we obtain ci = 0 for all i = 1, · · · , n which means
that β = 0. That is W1 ∩ W2 = {0}, and hence V = W1 ⊕ W2 .

(9) Let V be a real vector space and E be an idempotent lin-

ear operator on V, that is a projection. Prove that I + E is
invertible. Find (I + E)−1 .
Proof. Since E is an idempotent linear operator it is
diagonalizable by Question 4. So there exists a basis of V
consisting of characteristics vectors of E corresponding to the
characteristic values 0 and 1. That is, there exists a basis
6 M. KUZUCUOĞLU

B = {β1 , · · · , βn } such that Eβi = βi for i = 1, · · · , k, and

Eβi = 0 for i = k + 1, · · · , n. Then (I + E)βi = 2βi for
i = 1, · · · , k and (I + E)βi = βi for i = k + 1, · · · , n, that is,
[ ]
2I1 0
[I + E]B = ,
0 I2

where I1 stands for k × k identity matrix, I2 is (n − k) × (n −

k) identity matrix and each 0 represents the zero matrix of
appropriate dimension. It is now easy to see that [I + E]B is
invertible, since det(I + E) = 2k ̸= 0.
To find the inverse of (I + E), we note that
[ ] [ ] [ ]
−1
1
I1 0 I 1 0 − 1
I1 0 1
([I + E]B ) = 2 = + 2 = I − [E]B .
0 I2 0 I2 0 0 2

Therefore, (I + E)−1 = I − 12 E. (You may verify that really

this is the inverse, by showing that (I + E)(I − 12 E) = (I −
1
2
E)(I + E) = I.)

(10) Let T be a linear operator on V which commutes with every

projection operator on V. What can you say about T ?
Solution. Let B be a basis for V and βi ∈ B, i ∈ I where
I is some index set. We can write V as a direct sum V =
Wi ⊕ U where Wi = ⟨βi ⟩. Then there exists a projection Ei of
V onto the subspace Wi for each i ∈ I. Note that Ei v ∈ Wi
for all v ∈ V, and Ei βi = βi . Now, by assumption, the linear
operator T commutes with Ei for all i ∈ I, that is, T Ei = Ei T.
Then, for βi ∈ Wi , we have T Ei βi = Ei T βi ∈ Wi implies that
T βi = T (Ei βi ) = ci βi for some constant ci ∈ F. Thus, βi is a
characteristic vector of T. Hence, V has a basis consisting of
characteristic vectors of T. It follows that T is a diagonalizable
linear operator on V.

(11) Let V be the vector space of continuous real valued functions

on the interval [−1, 1] of the real line. Let We be the space
 EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 7

of even functions, f (−x) = f (x), and Wo be the space of odd

functions, f (−x) = −f (x).
a) Show that V = We ⊕ Wo . ∫ x
b) If T is the indefinite integral operator (T f )(x) = f (t)dt,
0
are We and Wo invariant under T ?
Solution. a) Let f ∈ V. Then, we may write

f (x) + f (−x) + f (x) − f (−x) f (x) + f (−x) f (x) − f (−x)

f (x) = = + .
2 2 2

f (x) + f (−x)
Observe that fe (x) = is a continuous even func-
2
f (x) − f (−x)
tion and fo (x) = is a continuous odd function.
2
Hence, f = fe + fo , that is V = We + Wo . To show that
V = We ⊕ Wo , we need to show that We ∩ Wo = {0}. To see
this, let g ∈ We ∩Wo . Then, g ∈ We implies that g(−x) = g(x),
and g ∈ Wo implies that g(−x) = −g(x). Thus, we have
g(x) = −g(x) or g(x) = 0 for all x ∈ [−1, 1], which means
that g = 0.
b) For f (x) = x ∈ Wo , we have (T f )(x) = x2 /2 ∈ / Wo ,
and for g(x) = x ∈ We , we have (T g)(x) = x /3 ∈
2 3
/ We . Thus,
neither We nor Wo are invariant under T.

(12) Let V be a finite dimensional vector space over the field F, and
let T be a linear operator on V, such that rank(T ) = 1. Prove
that either T is diagonalizable or T is nilpotent, but not both.
Proof. Since rank(T ) = dim(Im(T )) = 1, we have dim(Ker(T )) =
n − 1. Let 0 ̸= β ∈ Im(T ). So, Im(T ) = ⟨β⟩. Since β ∈
Im(T ), there exists a vector α0 ∈ V such that T α0 = β.
Let {α1 , α2 , · · · , αn−1 } be a basis for Ker(T ). Then, B =
{α0 , α1 , α2 , · · · , αn−1 } is a basis for V.
We have T αi = 0 for all i = 1, 2, · · · , n − 1.
8 M. KUZUCUOĞLU

If T α0 ∈ Ker(T ), then T α0 = c1 α1 + · · · + cn−1 αn−1 and

 
0 0 ··· 0
0
 ··· 0 
 c1 0 0 
 
[T ]B =  c2 0 ··· 0 
0
 . . .. 
 .. .. ..
. . 
 . . . 
..
cn−1 0 0 ··· .

and it is easily seen that T 2 = 0 meaning that T is nilpotent.

Note that at least one of ci ’s is nonzero, since otherwise, α0
would be in Ker(T ) which contradicts with the choice of B.
If T α0 ∈
/ Ker(T ), then T β ∈ Im(T ) and T β = c0 β. In this
case we construct a new basis B ′ = {β, α1 , α2 , · · · , αn−1 } and
 
c0 0 ··· 0
 0 0 ··· 0 
 
[T ]B′ =  .. .. . . .. 
 . . . . 
0 0 ··· 0

which means that T is diagonalizable.

(13) Let T be a linear operator on the finite dimensional vector

space V. Suppose T has a cyclic vector. Prove that if U is any
linear operator which commutes with T , then U is a polyno-
mial in T.
Proof. Let B = {α, T α, · · · , T n−1 α} be a basis for V
containing the cyclic vector α and let m(x) = xn + an−1 xn−1 +
· · · + a1 x + a0 be the minimal polynomial of T. Since U α is in
V, it can be written as a linear combination of basis vectors.
Then, U α = b0 α+b1 T α+· · ·+bn−1 T n−1 α where b0 , b1 , · · · , bn−1
are elements of the field F. That is, (b0 I +b1 T +· · ·+bn−1 T n−1 −
U )α = 0. Now, since U and T commute, we have

U T (α) = T U (α) = T (b0 α + b1 T α + · · · + bn−1 T n−1 α)

= b0 T α + b1 T 2 α + · · · + bn−1 T n α
= (b0 I + b1 T + · · · + bn−1 T n−1 )T α
 EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 9

which means that

(b0 I + b1 T + · · · + bn−1 T n−1 − U )T α = 0.

Similarly, we can show that (b0 I + b1 T + · · · + bn−1 T n−1 −

U )T i α = 0 for all i = 2, 3, · · · , n − 1. Since the transformation
b0 I + b1 T + · · · + bn−1 T n−1 − U maps each basis vector to the
zero vector, it is identically equal to zero on the whole space.
Thus, we obtain

U = b0 I + b1 T + · · · + bn−1 T n−1 .

(14) Give an example of two 4 × 4 nilpotent matrices which have

the same minimal polynomial
 but which
 are not similar. 
0 0 0 0 0 0 0 0
 1 0 0 0   1 0 0 0 
   
Solution. Let A =   and B =  .
 0 0 0 0   0 0 0 0 
0 0 0 0 0 0 1 0
It is easy to see that mA (x) = mB (x) = x2 but they are
not similar since, A has 3 distinct characteristic vectors corre-
sponding to the characteristic value zero, but B has only two
characteristic vectors corresponding to the characteristic value
zero.

(15) Show that if N is a nilpotent linear operator on an n−dimensional

vector space V, then the characteristic polynomial for N is xn .
Solution. Recall that N is nilpotent, if N k = 0 for some
k ∈ N+ . Since, N is a nilpotent linear operator on V, the
minimal polynomial for N is of the form xm for some m ≤
n. Then, all characteristic values of N are zero. Since the
minimal polynomial is a product of linear polynomials, N is a
triangulable operator. It follows that there exists a basis B of
10 M. KUZUCUOĞLU

V such that
 
0 0 ··· 0
 ⋆ 0 ··· 0 
 
[N ]B =  . .
 ⋆ ⋆ . . . .. 
⋆ ⋆ ··· 0
since, similar matrices have the characteristic polynomial, it
follows that the characteristic polynomial of N is xn where
n = dimV.

(16) Let T be a linear operator on R3 which is represented in the

standard ordered basis by the matrix
 
2 0 0
 
 0 2 0 .
0 0 −1
Prove that T has no cyclic vector. What is the T cyclic sub-
space generated by the vector β = (1, −1, 3)?
Solution. Assume that T has a cyclic vector α = (a1 , a2 , a3 ).
Then B = {α, T α, T 2 α} will be a basis for R3 . That is, the vec-
tors α = (a1 , a2 , a3 ), T α = (2a1 , 2a2 , −a3 ), T 2 α = (4a1 , 4a2 , a3 )
must be linearly independent, or the matrix
 
a1 a2 a3
 
 2a1 2a2 −a3 
4a1 4a2 a3
must be invertible. Applying elementary row operations, we
obtain
     
a1 a2 a3 −2R1 + R2 a1 a2 a3 −2R2 + R3 a1 a2 a3
     
 2a1 2a2 −a3  −→  0 0 −3a3  −→  0 0 a3 
4a1 4a2 a3 −4R1 + R3 0 0 −3a3 − 31 R2 0 0 0
which is not invertible. Hence, T has no cyclic vector.
To find the cyclic subspace generated by β, it is enough to
check if β and T β are independent since we have already shown
that the set {α, T α, T 2 α} can not be linearly independent for
any α ∈ R3 . Clearly, β = (1, −1, 3) and T β = (2, −2, −3) are
 EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 11

linearly independent since, otherwise, one of them would be a

multiple of the other one which is not the case here. Thus, the
cyclic subspace generated by β is
Z(β; T ) = ⟨(1, −1, 3), (2, −2, −3)⟩ = {λ(1, −1, 3)+µ(2, −2, −3) : λ, µ ∈ R}.

(17) Find the minimal polynomial and rational form of the matrix
 
c 0 −1
 
T = 0 c 1 .
−1 1 c
Solution. The characteristic polynomial of T is
 
 x−c 
 0 1 
 
fT (x) = det(xI − T ) =  0 x − c −1 
 
 1 −1 x − c 
   
 x − c −1   0 
   1 
= (x − c)  + 
 −1 x − c   x − c −1 
= (x − c)((x − c)2 − 1) − (x − c)
= (x − c)((x − c)2 − 2)
√ √
= (x − c)(x − c − 2)(x − c + 2).
Since the characteristic polynomial and the minimal polyno-
mial have the same roots and the minimal polynomial di-
vides the characteristic polynomial we have mT (x) = fT (x) =
(x − c)((x − c)2 − 2) = (x − c)3 − 2(x − c) = x3 + (−3c)x2 +
(3c2 − 2)x + (−c3 + 2c). Thus the rational form of T is
 
0 0 c3 − 2c
 
R =  1 0 −3c2 + 2  .
0 1 3c