Math 317: Linear Algebra
Exam 2
Fall 2015
Name:
*No notes or electronic devices. You must show all your work to receive full credit. When
justifying your answers, use only those techniques that we learned in class up to Section
3.4. Good luck!
1. (a) (10 points)
Let A and B be 3 × 3 matrices such that the first column of A is
 
3
given by 2, and the first row of B is given by 1 0 −1 . Prove that A and
1
B cannot be inverses of each other. That is B 6= A−1 .
Proof : To show that A and B are not inverses of each other, we show
that AB 6= I or BA 6= I. By definition of matrix multiplication
[BA]11 = the dot product of the first row of B with the first column
of A. This gives us (1, 0, −1) · (3, 2, 1) = 1(3) + 0(2) + 1(−1) = 2 6=
[I]11 = 1. Thus BA 6= I and so A and B cannot be inverses of each
other.
(b) (10 points) Suppose that A is an n × n matrix satisfying A2 − 2A + I = 0,
where I denotes the identity matrix. Prove that A is invertible.
Proof : We look for a matrix B such that AB = I or BA = I. We see
that A2 − 2A + I = 0 =⇒ A2 − 2A = −I =⇒ 2A − A2 = I =⇒
A(2I − A) = I. Thus A−1 = 2I − A and hence A is invertible.
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Math 317: Linear Algebra
Exam 2
Fall 2015
2. Consider the following matrix:


2
1 3
A =  4 −1 3 .
−2 5 5
(a) (10 points) Find the LU factorization of A.
We row reduce A to upper triangular form as follows:






2
1 3
2 1
3
2 1
3
+R2 →R2
2 +R3
0 −3 −3 ,
 4 −1 3 −2R1−→
0 −3 −3 2R−→
1R1 +R3 →R3
−2 5 5
0 6
8
0 0
2


2 1
3

so U = 0 −3 −3. The entries below the diagonal entry of L
0 0
2
is
made
up
of
the
multipliers used in reducing A to U . Thus L =


1
0 0
2
1 0.
−1 −2 1
     
1
3 
 2





4 , −1 , 3 is a basis for R3 .
(b) (5 points) Prove that V =


−2
5
5
Proof: We observe that V is made up of the columns of A in part (a).
Noting that A is nonsingular (since there are no zero rows in the row
echelon form of A and A is square) we recall that the columns of A
form a basis for R3 as proven in class. Thus V is a basis for R3 .
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Math 317: Linear Algebra
Exam 2
Fall 2015
3. (20 points) Consider the following matrix A and the echelon form of the augmented
matrix [A|b].

1
0
A=
1
2
1
1
1
1

2 0
0
1 −1 −1
,
2 1
2
3 −1 −3

1
 0
[U |b̂] = 
 0
0
1
1
0
0

2 0
0
b1

b2
1 −1 −1


0 1
2
−b1 + b3
0 0
0 −4b1 + b2 + 2b3 + b4
(a) Find a basis and the dimension of C(A).
A basis for C(A) is given by the pivot columns of A. Consulting the
row echelon form of A, we find the columns 1, 2, and 4 all correspond
to pivot columns. Thus, the 1st, 2nd, and 4th
columns
   ofA will
 form
1
1
0 



      
0
1
−1
 ,   ,   and
a basis for C(A). That is, C(A) = span 
0 0  1 





0
0
0
dim(C(A)) = 3.
(b) Find a basis and the dimension of R(A).
A basis for R(A)
rows in any echelon form of A. Thus,
 are
 the
 nonzero
  
1
0
0






  1  0

1

     
    
R(A) = span 
2 ,  1  , 0 and dim(R(A)) = 3.


0 −1 1






0
−1
2
(c) Find a basis and the dimension of N (A).
To find a basis for N (A), we solve Ax = 0. From the row echelon
form of A, we see that x3 and x5 are free variables. We also obtain the
following reduced system of equations to solve:
x1 + x2 + 2x3 = 0 =⇒ x1 = −x3 + x5
x2 + x3 − x4 − x5 = 0 =⇒ x2 = −x3 − x5
x4 + 2x5 = 0 =⇒ x4 = −2x5 .
  

x1
−x3 + x5
x2  −x3 − x5 
  

 =  x3
=
x
In standard form we may write the solution as 
3
  

x4   −2x5 
x5
x5
3
Math 317: Linear Algebra
Exam 2
Fall 2015
 
 
−1
1
−1
−1
 
 
 + x5  0 . Thus a basis for N (A) is given by N (A) =
1
x3 
 
 
0
−2
0
1

   
−1
1 





 −1

−1

   
  
span 
 1  ,  0  and dim(N (A)) = 2.


 0  −2





0
1
(d) Find a basis and the dimension of N (AT ).
A basis for N (AT ) is made up of the coefficients in the constraint
equations for b so that Ax = b. Since we only have one zero row in
our echelon form, we have one constraint
and thus a basis for
 equation

−4 


 

1
T
N (AT ) is given by N (AT ) = span 
 2 , and dim(N (A )) = 1.





1
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Math 317: Linear Algebra
Exam 2
Fall 2015
4. Suppose that V and W are subspaces of Rn .
(a) (5 points) State the definition of W ⊥ .
W ⊥ = {x ∈ Rn | x · w = 0 for all w ∈ W }.
(b) (5 points) Prove that W ⊥ is a subspace of Rn .
Proof : We note that 0 ∈ W ⊥ since 0 · w = 0 for every w ∈ W .
Suppose that x, y ∈ W ⊥ . Then x · w = 0 and y · w = 0 for every
w ∈ W . Thus, x · w + y · w = 0 + 0 = (x + y) · w for every w ∈ W .
Thus x + y ∈ W ⊥ . Let c ∈ R and x ∈ W ⊥ . Then x · w = 0 for every
w ∈ W =⇒ c(x · w) = c(0) = 0 = (cx) · w. Thus, cx ∈ W ⊥ . Since all
three properties of a subspace are satisfied, W ⊥ is a subspace of Rn .
(c) (5 points) Suppose that V ⊂ W . Show that W ⊥ ⊂ V ⊥ .
Proof : Suppose that x ∈ W ⊥ . Then x · w = 0 for every w ∈ W . If
v ∈ V ⊂ W , then x · v = 0 since v ∈ W . Since this is true for every
v ∈ V , x ∈ V ⊥ . Thus, W ⊥ ⊂ V ⊥ .
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Math 317: Linear Algebra
Exam 2
Fall 2015
5. (a) (10 points) Let A be an m × n matrix, v1 , v2 , . . . , vk ∈ Rn and suppose that
rank(A) = n. Prove that if {v1 , v2 , . . . , vk } is linearly independent, then
{Av1 , Av2 , . . . , Avk } is linearly independent.
Proof : To show that {Av1 , Av2 , . . . , Avk } is linearly independent, we
start with c1 (Av1 ) + c2 (Av2 ) + . . . + ck (Avk ) = 0, and show that c1 =
c2 = . . . = ck = 0. By linearity of A we have that c1 (Av1 ) + c2 (Av2 ) +
. . . + ck (Avk ) = A(c1 v1 + . . . + ck vk ) = 0. Since rank(A) = n for an
m × n matrix, we know that Ax = 0 has only the trivial solution, i.e.
x = 0. Thus, A(c1 v1 +. . .+ck vk ) = 0 =⇒ c1 v1 +. . .+ck vk = 0. Since
{v1 , v2 , . . . , vk } is linearly independent, then c1 = c2 = . . . = ck = 0.
Thus, {Av1 , Av2 , . . . , Avk } is linearly independent.
(b) (10 points) Let A be an n × n matrix, v1 , v2 , . . . , vk ∈ Rn and suppose that
N (A) = {0}. Furthermore, suppose that {v1 , v2 , . . . , vk } is a linearly independent spanning set for Rn . Prove that {Av1 , Av2 , . . . , Avk } is a basis for Rn .
Hint: It may be helpful to recall that An×n is nonsingular if and only if for
every b ∈ Rn there is an x ∈ Rn such that Ax = b.
Proof : To show that {Av1 , Av2 , . . . , Avk } is a basis for Rn , we
must show that it is a linearly independent spanning set for Rn .
We begin by noting that {Av1 , Av2 , . . . , Avk } is a linearly independent set from part (a). Thus, all we need to show is that Rn =
span {Av1 , Av2 , . . . , Avk }. By definition, span {Av1 , Av2 , . . . , Avk } ⊂
Rn , since Avi ∈ Rn for each i.
To show that Rn ⊂
n
span {Av1 , Av2 , . . . , Avk }, take b ∈ R . Since A is nonsingular (A
is a square matrix and N (A) = {0} means that A is full rank), there is
a unique x ∈ Rn such that Ax = b. Using the fact that {v1 , v2 , . . . , vk }
is a linearly independent spanning set for Rn , we know that there are
scalars c1 , c2 , . . . , ck ∈ R such that x = c1 v1 + c2 v2 + . . . + ck vk , and
thus, b = Ax = A(c1 v1 + c2 v2 + . . . + ck vk ) = c1 Av1 + . . . + ck Avk ∈
span {Av1 , Av2 , . . . , Avk }. Thus {Av1 , Av2 , . . . , Avk } is a linearly independent spanning set for Rn and is hence a basis for Rn .
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Math 317: Linear Algebra
Exam 2
Fall 2015
6. (5 points each) Determine if the following statements are true or false. If the statement is true, give a brief justification. If the statement is false, provide a counterexample or explain why the statement is not true.
(a) If A is a skew-symmetric matrix, then the diagonal entries of A are 0.
True. Proof : Suppose that A is skew-symmetric. Then A = −AT =⇒
aij = −aji for i, j = 1, 2, . . . n. On the diagonal, this means that
aii = −aii =⇒ 2aii = 0 =⇒ aii = 0 for i = 1, 2, . . . , n.
(b) If A is an m × n matrix (m 6= n) and rank(A) = n, then R(A) = Rn , where
R(A) denotes the row space of A.
True. Proof : R(A) ⊂ Rn since A is m × n. Furthermore, rank(A) =
n =⇒ dim(R(A)) = n = dim(Rn ). Thus, R(A) = Rn . (Recall that if
V ⊂ W and dim V = dim W then V = W .)
(c) There exist a matrix
  A such that the row space contains (1, 1, 1) and the null
−1

space contains −1.
−1
False. Recall that R(A) and N (A) are orthogonal subspaces and so
if x ∈ R(A) and y ∈ N (A) then x · y = 0. However, we have that
(1, 1, 1) · (−1, −1, −1) = 1(−1) + 1(−1) + 1(−1) = −3 6= 0.
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Math 317: Linear Algebra
Exam 2
Fall 2015
7. (5 points) (Bonus): What is the definition of linear algebra?
8