Bachelor in Statistics and Business Mathematical Methods II

Universidad Carlos III de Madrid Marıa Barbero Linan

Name Student number

Solution: Quiz 2 (Units 3 and 4)

November 9, 2011

Instructions

• You have 90 minutes to answer the quiz.

• Marks per question are given in bold. Total marks: 10 points

• The mark of this quiz is part of the 40% corresponding with the continuous evaluation

of this course.

• Write your name in all the sheets.

• It is not allowed to use lecture notes, scientific calculators or cellphones during the

exam.

1. Let v1 =

1

0

−1

2

, v2 =

1

2

0

3

, v3 =

2

2

−1

5

, v4 =

0

2

1

1

be vectors in R4.

(a) Are the vectors v1, v2, v3 and v4 linearly independent? Why? [1]

(b) Find a basis for the subspace spanned by v1, v2, v3 and v4. [0.5]

Solution: (a) We take the matrix whose columns are the vectors v1, v2, v3 and v4. We

compute the equivalent matrix in row echelon form to identify the dependences among

all the columns.1 1 2 0

0 2 2 2

−1 0 −1 1

2 3 5 1

∼

R1 +R3 → R3

−2R1 +R4 → R4

1 1 2 0

0 2 2 2

0 1 1 1

0 1 1 1

∼

−2R3 +R2 → R3

2R4 +R2 → R4

1 1 2 0

0 2 2 2

0 0 0 0

0 0 0 0

The matrix has two pivto columns, less than the number of vectors given. Hence the

vectors are linearly dependent.

(b)A basis for the space spanned by v1, v2, v3 and v4 is given by the columns in the

original matrix corresponding with the pivot columns in the matrix in row echelon

form, that is, v1 and v2. Thus, span{v1, v2, v3, v4} = span{v1, v2}.

2. Let A =

−2 4 6

−1 2 3

2 −4 −6

and w =

−1

1

−1

.

(a) Find a basis for the subspaces ColA and NulA. [1]

(b) Is w in NulA? Is w in ColA? [0.5]

Solution: (a) Let us compute the row echelon form of the matrix A to find a basis

for both the null space and the column space:−2 4 6

−1 2 3

2 −4 −6

∼R1 − 2R2 → R2

R1 +R3 → R3

−2 4 6

0 0 0

0 0 0

∼R1/(−2)→ R1

1 −2 −3

0 0 0

0 0 0

The column space of A is spanned by the columns in the matrix A that correspond with

the pivots columns in the matrix in row echelon form, that is, ColA = span

−2

−1

2

.

The null space of A is given by the solutions to the system Ax = 0. From the augmented

matrix of the system1 −2 −3 0

0 0 0 0

0 0 0 0

we have

x = 2y + 3z

y = y

z = z

Thus, NulA = span

2

1

0

,

3

0

1

.

(b) To check if w =

−1

1

−1

lies in NulA we just have to compute

Aw =

−2 4 6

−1 2 3

2 −4 −6

−1

1

−1

=

2 + 4− 6

1 + 2− 3

−2− 4 + 6

=

0

0

0

.

Thus, w ∈ NulA.

To see if w lies in ColA we must solve the system Ax = w. We use the same elementary

row operations as in the first part, but fot the augmented matrix of the system (A w).−2 4 6 −1

−1 2 3 1

2 −4 −6 −1

∼R1 − 2R2 → R2

R1 +R3 → R3

−2 4 6 −1

0 0 0 −3

0 0 0 −2

∼R1

−2 → R1

1 −2 −3 1/2

0 0 0 −3

0 0 0 −2

As the last column of the augmented matrix of the systems is a pivot column, the

system Ax = w is inconsistent. Thus w is not in ColA.

3. Let B1 =

{(0

1

),

(1

0

)}and B2 =

{(2

1

),

(0

1

)}be bases for R2.

(a) Find the change-of-basis matrix from B1 to the standard basis Bc for R2 and from

B2 to the standard basis Bc for R2. [0.5]

(b) Find the change-of-basis matrix from B1 to B2. [1]

(c) Find the coordinate vectors [x]B2 of [x]B1 =

(2

4

)in the basis B2. [0.5]

Solution: (a) Note that B1 =

[b1]Bc︷ ︸︸ ︷(

0

1

),

[b2]Bc︷ ︸︸ ︷(1

0

). The change-of-basis matrix from B1 to

the standard basis Bc for R2 is given by

PBcB1 = ([b1]Bc [b2]Bc) =

(0 1

1 0

).

Note that B2 =

[c1]Bc︷ ︸︸ ︷(

2

1

),

[c2]Bc︷ ︸︸ ︷(0

1

). The change-of-basis matrix from B2 to the standard

basis Bc for R2 is given by

PBcB2 = ([c1]Bc [c2]Bc) =

(2 0

1 1

).

(b) The change-of-basis matrix from B1 to B2 is given by

PB2B1 = PB2BcPBcB1 = (PBcB2)−1 PBcB1 .

Thus, we have to compute the inverse matrix of PBcB2 =

(2 0

1 1

)that appeared in the

first question.

detPBcB2 =

∣∣∣∣∣2 0

1 1

∣∣∣∣∣ = 2.

Then,

(PBcB2)−1 =

1

2

(1 0

−1 2

),

because A−1 =

(a b

c d

)−1= 1

detA

(d −b−c a

). Thus,

PB2B1 = (PBcB2)−1 PBcB1 =

1

2

(1 0

−1 2

)(0 1

1 0

)=

(0 1/2

1 −1/2

).

(c)

[x]B2 = PB2B1 [x]B1 =

(0 1/2

1 −1/2

)(2

4

)=

(2

0

).

4. Let A =

2 0 0 0

0 1 0 1

0 0 2 0

0 1 0 1

.

(a) Find the characteristic polynomial of A and the eigenvalues of A. Give the alge-

braic multiplicity of each eigenvalue. [0.75]

(b) Having in mind that the characteristic polynomial of A is λ(λ−2)3, can the matrix

A be diagonalized? Why? Give the geometric multiplicity of each eigenvalue. If

A can be diagonalized, give the matrices P and D. [1.75]

Solution: (a) To compute the characteristic polynomial of the matrix A we have to

compute the following determinant, we will do it as cofactor expansion across the first

row and then across the second row:

det(A− λId) =

∣∣∣∣∣∣∣∣∣2− λ 0 0 0

0 1− λ 0 1

0 0 2− λ 0

0 1 0 1− λ

∣∣∣∣∣∣∣∣∣ = (2− λ)

∣∣∣∣∣∣∣1− λ 0 1

0 2− λ 0

1 0 1− λ

∣∣∣∣∣∣∣= (2− λ)2

∣∣∣∣∣1− λ 1

1 1− λ

∣∣∣∣∣ = (2− λ)2((1− λ)2 − 1)

= (2− λ)2(1− 2λ+ λ2 − 1) = (2− λ)2λ(−2 + λ) = λ(λ− 2)3.

The eigenvalues are 0 and 2. The algebraic multiplicities are n0 = 1 and n2 = 3.

(b) We first have to compute the eigenvectors of each eigenvalue to decide if the matrix

is diagonalizable or not. For λ = 0,

A− 0I = A =

2 0 0 0

0 1 0 1

0 0 2 0

0 1 0 1

∼R2 − 2R4 → R4

2 0 0 0

0 1 0 1

0 0 2 0

0 0 0 0

The vectors in the null space of the matrix A− 0I = A verify that

a = 0

b = −dc = 0

d = d

.

Thus V (0) = span

0

−1

0

1

. The geometric multiplicity of λ = 0 is 1 = m0.

For λ = 2, we have

A− 2I =

0 0 0 0

0 −1 0 1

0 0 0 0

0 1 0 −1

∼R2 +R4 → R4

0 0 0 0

0 −1 0 1

0 0 0 0

0 0 0 0

.

The vectors int he null space of A− 2I verify that

a = a

b = d

c = c

d = d

.

Thus V (2) = Gen

1

0

0

0

,

0

1

0

1

,

0

0

1

0

. The geometric multiplicity of λ = 2 is

3 = m2.

As the algebraic and geometric multiplicities are the same for both eigenvalues, and

they add 4, the matrix A is diagonalizable.

D =

0 0 0 0

0 2 0 0

0 0 2 0

0 0 0 2

P =

0 −1 0 1

1 0 0 0

0 1 0 1

0 0 1 0

.

5. Answer the following questions and always reason your answers.

(a) Is the set W = {(x, y, z) : 2x+ y − z + 1 = 0} a subspace of R3? Why? [0.5]

Solution: No, it is not a subspace because (0, 0, 0) is not in W . The point (0, 0, 0)

does not fulfill the equation defining W : 2 · 0 + 0− 0 + 1 = 1 6= 0.

(b) Find the matrix associated with the linear transformation T : R3 → R2 defined

by T (x, y, z) = (x,−y + z). [0.5]

Solution: As T goes from R3 to R2, its associated matrix has size 2× 3:(1 0 0

0 −1 1

)

(c) If the rank of a 3× 7 matrix A is 2, what is the dimension of the null space of A?

Why? [0.5]

Solution: By Rank Theorem, we know that

rankA+ dim NulA = number of columns of A.

In this case we have

2 + dim NulA = 7, dim NulA = 7− 2 = 5.

(d) Let A be a 5 × 5 matrix with two different eigenvalues. If one eigenspace has

dimension 3 and the other one has dimension 2, is A diagonalizable? Why? [0.5]

Solution: The matrix A is diagonalizable because there exists a basis of egein-

vectors for R5, 3 + 2 = 5.

(e) Is

(1

1

)an eigenvector of

(1 −1

6 −4

)? If so, find the eigenvalue. [0.5]

Solution: Let us check if the vector v =

(1

1

)satisfies Av = λv for any λ and

A =

(1 −1

6 −4

):

Av =

(1 −1

6 −4

)(1

1

)=

(0

2

)?= λ

(1

1

).

From

(0

2

)= λ

(1

1

), we have λ = 0 and λ = 2. Thus the system is inconsistent

because there does not exist a value λ such that it satisfies both equations in the

system. Hence, the vector v is not an eigenvector of the matrix A.