Solving Matrix Eigenvalues and Finding Taylor Approximations, Exams of Mathematics

Solutions to exercises involving finding eigenvalues and eigenvectors of matrices, as well as third-order taylor approximations for functions such as e^x and sin(x).

Typology: Exams

Pre 2010

Uploaded on 09/17/2009

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2J - Seond Midterm
05/26/06
Notes, books and alulators are not allowed. Explain all your answer.
Undetailed answer will not reeived full redit
1 Exerise 1 (25 p oints)
Let us onsider the following matrix:
A
=
0
4
5 1
1 0
1
0 1
1
1
A
:
Find the eigenvalues and orresponding eigenvetors.
Solution:
The harateristi polynomial of
A
is:
p
(
) = det(
A
I
)
=
3
+ 3
2
2
=
(
2
3
+ 2)
=
(
1)(
2)
:
The eigenvalues are
1
= 0,
2
= 1 and
3
= 2. The orresponding eigenvetors are given by
(
A
i
)
v
i
= 0 (with
i
= 1
;
2
;
3). This gives
v
1
=
0
1
1
1
1
A
,
v
2
=
0
3
2
1
1
A
,
v
3
=
0
7
3
1
1
A
.
1
pf3
pf4

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2J - Se ond Midterm

Notes, b o oks and al ulators are not allowed. Explain all your answer.

Undetailed answer will not re eived full redit

1 Exer ise 1 (25 p oints)

Let us onsider the following matrix:

A =

A

Find the eigenvalues and orresp onding eigenve tors.

Solution: The hara teristi p olynomial of A is:

p() = det(A I )

The eigenvalues are  1

= 1 and  3

= 2. The orresp onding eigenve tors are given by

(A 

i

)v i

= 0 (with i = 1 ; 2 ; 3). This gives v 1

A

, v 2

A

, v 3

A

Let A b e the following matrix:

A =

  1. ( 10 p oints) Find the eigenvalues and eigenve tors of A. Can A b e diagonalized?
  2. ( 10 p oints) Find A
  1. ( 10 p oints) Without any omputation, an you explain why the matrix A do es not have

an inverse?

Solution: The hara teristi p olynomial of the matrix A is p() = ( + 1). There ex-

ist 2 eigenvalues  1 = 0 and  2 = 1. The orresp onding eigenve tors are v 1

and

v 1

. The matrix A is a n  n matrix and we have n eigenve tors, therefore the matrix

an b e diagonalized (another way to say this is to noti e that, the 2 eigenvalues have multipli ity

1 and b oth have 1 eigenve tor).

Let us de ne the matrix X = (v 1

; v 2

. Therefore X

. We

an then write A as A = X D X

with D the diagonalized matrix D =

This gives that A

= X D

X

. D b eing diagonal, we have D

. We an

see that D

= D. Then A

= X D

X

= X D X

, that is A

= A.

We an easily on lude that A an not have an inverse, b e ause  = 0 is an eigenvalue.

  1. ( 10 p oints) What are the 3rd order Taylor approximation of e

x

and sin(x) ab out = 0?

  1. ( 15 p oints) What is the 3rd order Taylor approximation of f (x) =

e

x

1 sin(x)

ab out = 0?

Solution: The 3rd order approximation of e

x

is e

x

' 1 + x +

x

x

, and sin(x) ' x

x

(Also 3! = 6).

The Taylor approximation of

1 x

is

1 x

' 1 + x + x

  • x

. Repla ing x by sin(x), we get

1 sin(x)

' 1 + sin(x) + sin

(x) + sin

(x)

' 1 + (x

x

) + (x

x

  • (x

x

' 1 + x + x

x

Remark: in the al ulation ab ove, we dropp ed the x

, x

,... terms sin e we are only interested

in the 3rd order approximation.

Then

e

x

1 sin(x)

= e

x

1 sin(x)

' e

x

(1 + sin(x) + sin

(x) + sin

(x))

' (1 + x +

x

x

)(1 + x + x

x

' 1 + 2 x +

x

  • 2 x