Sections 2.3, 2.4
Definitions
• An n × n matrix is invertible if A has left and right inverses. That is, A is
invertible if there exist matrices B and C such that AB = I and CA = I.
• A matrix is called an elementary matrix if it comes from performing a
single elementary row operation on the identity matrix I.
Lemmas and theorems with direct proofs
• Let A be an m × n matrix. Then rank A = m if and only if A has a right
inverse.
• Let A be an m × n matrix. Then rank A = n if and only if A has a left
inverse.
• Elementary matrices are invertible.
• If A is invertible with right and left inverses B and C, then B = C.
• If A, B are invertible, then AB is invertible.
Theorems that follow
• Let A be an n × n matrix. Then A is nonsingular if and only if A is
invertible.
• If A is a square matrix with a right inverse then A is invertible.
• If A is a square matrix with a left inverse then A is invertible.
Computations
• Deduce whether a matrix is invertible by row reduction. If it is invertible,
find its inverse by row reduction.
• Find inverse of a 2 × 2 matrix via the formula
1
d −b
.
A−1 =
a
ad − bc −c
1
Section 2.5
Definitions
• Let A be an m × n matrix. The transpose of A is the n × m matrix whose
rows are the columns of A.
• Let A be an n × n matrix. Then A is orthogonal if AT A = I.
Theorems
• Let A be an m × n matrix. Then (AB)T = B T AT .
• Let A be an invertible n × n matrix. Then AT is invertible and its inverse
is (A−1 )T .
• Let A be an n × n matrix. Then for any x, y ∈ Rn
Ax · y = x · AT y.
• Let A be an orthogonal matrix. Then the columns of A are orthogonal.
Moreover, if x, y ∈ Rn are two vectors with angle θ between them, then
the angle between Ax and Ay is also θ.
Section 3.1
Definitions
• A set V ⊂ Rn is a subspace if the following conditions hold:
1. 0 ∈ V ,
2. if u, v ∈ V then u + v ∈ V ,
3. if u ∈ V and c ∈ R then cu ∈ V .
• Let V ⊂ Rn . Then the orthogonal complement of V is the set V ⊥ = {w ∈
Rn : v · w = 0 for all v ∈ V }
• Let V, W ⊂ Rn be subspaces. Then the sum of V and W is the set
V + W = {v + w : v ∈ V, w ∈ W }.
2
Theorems
• Let v1 , . . . , vk ∈ Rn . Then Span(v1 , . . . , vk ) is a subspace.
• Let A be an m × n matrix. The solution set to Ax = 0 is a subspace of
Rn .
• Let V ⊂ Rn . Then V ⊥ is a subspace.
• Let V, W ⊂ Rn be subspaces. Then V + W is a subspace.
Section 3.3
Definitions
• A set of vectors {v1 , . . . , vk } ⊂ Rn is linearly independent if the equation
x1 v1 + · · · + xk vk = 0
has only the trivial solution. Otherwise, {v1 , . . . , vk } is said to be linearly
dependent.
• Let V ⊂ Rn be a subspace. Then {v1 , . . . , vk } ⊂ V is a basis for V if
1. Span{v1 , . . . , vk } = V ,
2. {v1 , . . . , vk } is linearly independent.
• Let V ⊂ Rn be a subspace. Then the dimension of V , denoted dim V , is
the cardinality of any basis of V .
Lemmas
• Let {v1 , . . . , vk } ⊂ Rn be a linearly independent set and let v ∈ Rn . Then
{v1 , . . . , vk , v} is linearly independent if and only if v 6∈ Span{v1 , . . . , vk }.
• Let V ⊂ Rn be a subspace. If {v1 , . . . , vk } ⊂ V is linearly independent
and {w1 , . . . , w` } ⊂ V spans V , then k ≤ `.
Theorems that follow
• Let V ⊂ Rn be a nontrivial subspace (that is, V 6= {0}). Then V has a
basis.
• Let V ⊂ Rn be a nontrivial subspace. Then every basis of V has the same
cardinality.
3
Theorems with direct proofs
• Let A be an n × n matrix with columns v1 , . . . , vn ∈ Rn . Then A is
nonsingular if and only if {v1 , . . . , vn } is a basis for Rn .
Sections 3.2, 3.4
Definitions
• Let A be an m × n matrix with columns denoted a1 , . . . , an ∈ Rm .
Then the column space of A is the subspace of Rm given by C(A) =
Span(a1 , . . . , an ).
• Let A be an m × n matrix with row denoted A1 , . . . , Am ∈ Rn . Then the
row space of A is the subspace of Rn given by R(A) = Span(A1 , . . . , Am ).
• Let A be an m × n matrix. Then the null space of A is the subspace of
Rn given by N (A) = {x ∈ Rn : Ax = 0}.
• Let A be an m × n matrix. Then the left null space of A is the null space
of AT (and so is a subspace of Rm ).
Lemmas and theorems with direct proofs
• Let v1 , . . . , vk ∈ Rn . If v1 ∈ Span(v2 , . . . , vk ) then Span(v1 , v2 , . . . , vk ) =
Span(v2 , . . . , vk ).
• Let A be an m × n matrix of rank k with columns a1 , . . . , an ∈ Rm .
Then C(A) is a k-dimensional subspace of Rm with a basis given by the k
element subset of {a1 , . . . , an } corresponding to the pivot columns of the
reduced echelon form of A. [WARNING: the pivot columns of the reduced
echelon form are not necessarily a basis of C(A), rather it’s the columns
of A that correspond to pivot columns that are the basis.]
• Let A be an m × n matrix of rank k with rows A1 , . . . , Am ∈ Rn . Then
R(A) is a k-dimensional subspace of Rn with a basis given by the nonzero
rows of the reduced echelon form of A.
• Let A be an m × n matrix of rank k. Then N (A) is an (n − k)-dimensional
subspace of Rn with a basis found by setting each free variable equal to 1
and the other free variables equal to 0 in the general solution of Ax = 0.
• Let A be an m × n matrix. Then N (A) = R(A)⊥ .
• Let V, W ⊂ Rn be subspaces. If V ⊆ W and dim V = dim W then V = W .
4
• Let V be a subspace. Then V ⊂ (V ⊥ )⊥ .
Lemmas and theorems that follow
• Let A be an m × n matrix. Then
n = dim C(A) + dim N (A) = rank(A) + null(A).
• Let V ⊂ Rn be a subspace and suppose dim V = k. Then dim V ⊥ = n−k.
• Let V ⊂ Rn be a subspace. Then Rn = V + V ⊥ .
• Let V be a subspace. Then V = (V ⊥ )⊥ .
• Let A be an m × n matrix. Then N (A)⊥ = R(A).
• Let A be an m × n matrix. Then N (AT ) = C(A)⊥ and N (AT )⊥ = C(A).
Computations
• Given a matrix A find bases for C(A), R(A), and N (A).
• Given vectors v1 , . . . , vk ∈ Rn , find a basis for V = Span(v1 , . . . , vk )
and a basis for V ⊥ by finding bases for C(A) or R(AT ) and N (A) for an
appropriate matrix A.
5
Comments
Your exam will consist of roughly six questions, some with multiple parts. There
will be computational questions (find the inverse of a matrix via row reduction
or using the 2 × 2 matrix formula, find bases for fundamental subspaces, find
coordinate vector with respect to a basis), state the definition questions, proof
questions, and true false questions. The questions will be similar to homework
questions or the questions in the sample exam below. In the proof questions,
you may be given a lemma (like in Problem 9 below) which you may cite without
proof to prove the given statement. If no lemma is given, you’ll be expected
to prove the given statement directly and not simply cite a result from the
book/class. The sample exam below is longer than the actual exam will be and
is not intended to be a totally comprehensive study tool, but similar questions
could certainly appear on the exam.
Sample Exam

1 0 1
1. Find the inverse of the matrix A =  0 2 1  or state why A is not
−1 3 1
invertible.


1 1 1
2. Consider the matrix A =  2 2 2 . Find dim C(A). Explain, using
3 3 3
dim C(A), why A cannot be invertible.
2
3
3. Show that B =
,
is a basis for R2 and find the B-coordinate
3
5
3
vector [v]B of v =
.
4

4. Let

1
 1

 2
−1

1 0
1 −1
1 2 −1
1 
.
2 2
0
0 
−1 2 −3
3
Find bases for C(A), R(A), N (A) and state the dimensions of each of these
subspaces.
5. Suppose A is an m × n matrix of rank m. Prove that A has a right inverse
by constructing an n × m matrix B so that AB = I.
6. Recall that the reduced echelon form of an n × n matrix of rank n is I.
Use this fact to prove that if A is n × n matrix with rank n then A has a
left inverse.
6
7. Let V be a subspace of Rn . State the definition of V ⊥ . Prove that V ⊥ is
a subspace. Show that V ∩ V ⊥ = {0}.
8. Prove that if {v1 , v2 , v3 , v4 } is linearly independent then {v1 , v2 , v3 } is
linearly independent.
9. Suppose V is a k-dimensional subspace of Rn . Recall that if a set of vectors
{v1 , . . . , v` } ⊂ V is linearly independent and v ∈ V , then {v1 , . . . , v` , v}
is linearly dependent if and only if v ∈ Span(v1 , . . . , v` ). Use this fact to
prove
(a) Any set of k vectors that span V must be linearly independent.
(b) Any linearly independent set of k vectors must span V .
10. True/False:
(a) If A is a square matrix and dim N (A) = 0 then A is invertible.


 
   
 
1 
1
0
1
1




 1   2   −1   0 
1 










is a basis for
,
,
,
,
(b) The set 
2   2   2   0   2 





−1
3
2
1
−1
R4 .

 
 
0
 1
(c) There exists a matrix A whose column space has basis  0  ,  1  , 

0
0
 
0


and whose null space has basis  0  .


1

 
 
0
1



 

0 
  1  
(d) There exists a matrix A whose column space has basis 
 0 , 0 ,



0
0
 
 0 
and whose null space has basis  0  .


1

 
 
1
0




 
0 
  1  
(e) There exists a matrix A whose column space has basis 
 0 , 0 ,



0
0



0 





0 


and whose null space has basis 
.
0 





1
(f) There exists a matrix whose row space contains

 the vector (1, 1, 1)
1
and whose null space contains the vector  0  .
1
7

0 
0 

1

0 


0 

1 


0

0 


0 

1 


0