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
This
preview
has intentionally blurred sections.
Sign up to view the full version.
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 3-dimensional.
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
?
This is the end of the preview.
Sign up
to
access the rest of the document.
- Spring '08
- PAVLOVIC
- Math, Linear Algebra, Matrices, NulA
-
Click to edit the document details
