Exam 2 — Version 2 — Solutions
Math 340L, Spring 2007, Purcell
1. (8 points total) Let
A
be the matrix:
4
1
2

1
6
0
2
3

3

1
0
0

1
0
4
0
0
0
0
0
0
0
0
0
6
(a) (2 points) What is the rank of
A
? Explain.
Solution:
The rank is 4. This is because
A
has 4 pivot columns. (If this isn’t
obvious, swap the last two rows to put
A
into echelon form. Now count: 4 pivots.)
The number of pivot columns is the dimension of the column space, which is the
rank.
(b) (2 points) Is
A
diagonalizable? Why or why not?
Solution:
Yes
A
is diagonalizable. It has 5 distinct eigenvalues, which means it
has 5 linearly independent eigenvectors.
(c) (4 points) Find a basis for Col(
A
).
Solution:
A basis consists of columns corresponding to pivots.
These are the
first, second, third, and fifth columns.
4
0
0
0
0
,
1
2
0
0
0
,
2
3

1
0
0
,
6

1
4
0
6
1
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2. (9 points total, 3 points each)
(a) If the null space of a 4
×
6 matrix
A
is 3–dimensional, what is the dimension of
the row space of
A
? Why?
Solution:
The row space is 3dimensional.
Dimension of the row space is the
dimension of the column space is the rank of the matrix. Rank plus dimension of
the null space equals the number of columns = 6. So dimension of the row space
is 6

3 = 3.
(b) If
A
is a 5
×
7 matrix, what is the largest possible dimension of the column space
of
A
? Why?
Solution:
The column space has dimension equal to the number of pivots. Since
A
can’t have more pivots than it has rows, the largest dimension is 5.
(c) If
A
is a 2
×
4 matrix, then the null space of
A
is a subspace of
R
n
. What is
n
?
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 Spring '08
 PAVLOVIC
 Math, Linear Algebra, Matrices, NulA

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