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Linear Algebra I - Test 2 with Solutions | 1016 331, Exams of Linear Algebra

Material Type: Exam; Class: 1016 - Linear Algebra I; Subject: Mathematics & Statistics; University: Rochester Institute of Technology; Term: Unknown 2009;

Typology: Exams

2009/2010

Uploaded on 03/28/2010

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1016-331 RIT, 20091 1
Linear Algebra I 1016-331
Test 2 Solution
1. (6p) We know that the following matrix Ahas the reduced row echelon form R:
A=
1 2 34
24 7 9
1 2 67
;
1201
0 0 1 1
0 0 0 0
=R.
(a) Find a basis for row(A).
The nonzero rows from R:
{£1 2 0 1¤,£0 0 1 1 ¤}
(b) Find a basis for col(A).
The pivot columns from A:
{
1
2
1
,
3
7
6
}
(c) Find a basis for null(A).
We solve the system given by Ax = 0: if the variables are x, y, z and u, then from
Rwe see that yand uare free and z=uand x=u2y. This means that the
solutions are
x
y
z
u
=
u2y
y
u
u
.
This shows that the answer is
{
2
1
0
0
,
1
0
1
1
}
pf3
pf4
pf5

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Linear Algebra I 1016-

Test 2 Solution

  1. (6p) We know that the following matrix A has the reduced row echelon form R:

A =

 = R.

(a) Find a basis for row(A).

The nonzero rows from R:

[

]

[

]

(b) Find a basis for col(A).

The pivot columns from A:

(c) Find a basis for null(A).

We solve the system given by Ax = 0: if the variables are x, y, z and u, then from

R we see that y and u are free and z = −u and x = u − 2 y. This means that the

solutions are (^) 

x

y

z

u

u − 2 y

y

−u

u

This shows that the answer is

  1. (6p) Find the dimension of

span{

This span is just the row space of the previous A matrix. The dimension is 2.

  1. (6p) Find the rank of A

T A, if

A =

[

]

In class, we proved that the rank of A

T A is the same as the rank of A. So the answer

is 2 , as we can see from here:

A =

[

]

R 2 − 2 R 1 ;

[

]

  1. (6p) Find the matrix of the linear transformation which projects a vector v in R

3

onto the vector u =

As we know, this is

P =

uu

T

uT^ u

[

]

[

]

  1. (6p) Find the inverse of the matrix

a b c d

We compute:

A =

t 1 1 1

1 t 1 1

1 1 t 1

1 1 1 t

R 1 ↔R 4 ;

1 1 1 t

1 t 1 1

1 1 t 1

t 1 1 1

R 2 −R 1 ,R 3 −R 1 ,R 4 −tR 1 ;

1 1 1 t

0 t − 1 0 1 − t

0 0 t − 1 1 − t

0 1 − t 1 − t 1 − t

2

R4+R 2 ;

1 1 1 t

0 t − 1 0 1 − t

0 0 t − 1 1 − t

0 0 1 − t 2 − t − t

2

R4+R 3 ;

1 1 1 t

0 t − 1 0 1 − t

0 0 t − 1 1 − t

0 0 0 3 − 2 t − t

2

Now 3 − 2 t − t

2 = −(t − 1)(t + 3), so when t = −3 , then the rank is 3. (For the

other possible value, t = 1, the rank is 1.)

  1. (6p) Let A be an n × m matrix. Show that for any choice of b ∈ R

n , the linear

system of equations

A

T Ax = A

T b

is consistent (x ∈ R

m ).

We know that a system Ax = b is consistent if rank(A) = rank(A|b). We also proved

in class that rank(AB) ≤ rank(A) and that rank(A

T A) = rank(A) = rank(A

T ). Now

rank(A

T ) = rank(A

T A) ≤ rank(A

T A|A

T b) = rank(A

T (A|b)) ≤ rank(A

T ).

Looking at the two ends shows that all these ranks are equal, and we are done.

  1. (6p) Suppose that A is an n × n matrix and A

3 = O (the zero matrix). Show

that I − A is invertible.

We compute:

(I − A)(I + A + A

2 ) = I + A + A

2 − A − A

2 − A

3 = I.

We found the inverse.

  1. (6p) Suppose that A and B are n × n matrices such that AB = O (the zero

matrix). Show that

rank(A) + rank(B) ≤ n.

As we know, elements in the column space of B are given by the form Bx. Now

A(Bx) = ABx = Ox = 0, which shows that col(B) ⊆ null(A), and then the dimen-

sion of the column space of B (which is just rank(B)) is less than or equal to the

dimension of the null space of A, which is the nullity of A. So we obtain that

rank(A) + rank(B) ≤ rank(A) + nullity(A) = n

and we are done.