MATH 415 G83 - WORKSHEET 08
The purpose of this worksheet is to help you get accustomed to doing linear algebra by hand, not
just with
. Please note that it is NOT the purpose of the worksheet to explain what
you are doing in your homework assignments. You are still expected to read and know the Basics
and Tutorials in the courseware.
Mathematica
Denition 1 (Subspace - True Denition). A subspace
S of Rn is a set of vectors in Rn such that
for any two vectors X and Y in S , their sum X + Y is in S , and for any real number k, the vector
kX is in S .
Denition 2 (Span). Consider the set of vectors X1 , X2 , ..., Xm in Rn . The span of these vectors,
denoted span{X1 , ..., Xm }, is the set of all linear combinations of X1 , ..., Xm :
a1 X1 + a2 X2 + ... + am Xm
where ai ∈ R.
Exercise 1 (Spans are Subspaces). Show that span{X1 , ..., Xm } is a subspace.
Denition 3 (Linear Independence). A set of vectors {X1 , ..., Xm } is linearly
span of these vectors yields a subspace of dimension m.
independent if the
Exercise 2. What is the span of the following set of vectors?
   
2 
 1
 3 , 3 


5
3
Is this set linearly independent?
Exercise 3. What is the span of the following set of vectors?
   
2 
 4
 6 , 3 


6
3
Is this set linearly independent?
Exercise 4 (Images of Matrices are Subspaces). Show that if A is an m × n matrix, then its image
is a subspace of Rm .
Denition 4 (Null Space). Let A be an m × n matrix. Then the null space of A is the set of all
vectors X in the domain Rn such that AX = 0.
Exercise 5 (Null Spaces are Subspaces). Show that if N is the null space of an m × n matrix N ,
then it is a subspace of Rn .
Date
: April 1, 2013.
1
Mathematica
If you do not have access to
, and you need to solve a determinant or nd the
eigenvalues and eigenvectors of a matrix, then there are some ways of doing this by hand.
Denition 5 (Determinant). Let A be an n × n matrix:

a11 a12
 a21 a22

A =  ..
..
 .
.
an1 an2
Then the

a1n
a2n 

.
. . . .. 
. 
... ann
...
...
determinant of A, denoted det(A), is given by
det(A) =
=
n
X
j=1
n
X
(−1)i+j aij Mij
(along row i)
(−1)i+j aij Mij
(along column j ).
i=1
(i, j)th
minor
Here, Mij is the
of A, which is the determinant of the (n−1)×(n−1) matrix obtained
by removing the ith row and j th column from A.
Remark 6. The denition of determinant above is inductive; that is, you reduce the determinant
of an n × n matrix to a linear combination of determinants of (n − 1) × (n − 1) matrices. You
continue breaking each determinant down in this way until you reach small determinants that you
can handle (like 2 × 2 or 3 × 3 determinants).
Exercise 6 (Calculate Determinants). Find determinants of the following matrices.
(a)
(b)
(c)
a b
c d


1 2 3
 0 1 2 
−1 0 1


2 3
5
7
 0 −1 0
2 


 1 1
2
3 
0 4 −3 −2
2