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 B is
 
3
given by 2, and the first row of A 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
[AB]11 = the dot product of the first row of A with the first column
of B. This gives us (1, 0, −1) · (3, 2, 1) = 1(3) + 0(2) + 1(−1) = 2 6=
[I]11 = 1. Thus AB 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.
1
Math 317: Linear Algebra
Exam 2
Fall 2015
2. Consider the following matrix:


2
1 3
A =  4 −1 3 .
−2 5 5
(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 is made
0 0
2


1
0 0
1 0.
up of the multipliers used in reducing A to U . Thus L =  2
−1 −2 1
2
Math 317: Linear Algebra
Exam 2
Fall 2015
3. (25 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






a basis for C(A). That is, C(A) = span   ,   ,   and
1
1
1 





2
1
−1
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 .
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Math 317: Linear Algebra
Exam 2
Fall 2015
  

x1
−x3 + x5
x2  −x3 − x5 
  

 =  x3
=
x
In standard form we may write the solution as 
3
  

x4   −2x5 
x5
x5
 
 
−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
(e) Find a matrix X so that N (X) = C(A).
Recalling that we can write C(A) in terms of the vectors b which
satisfies the constraint equations associated with Ax = b, we have
that X = [−4 1 2 1].
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Math 317: Linear Algebra
Exam 2
Fall 2015
4. (a) (5 points) State the definition of V ⊥ for a subset V ⊂ Rn .
V ⊥ = {x ∈ Rn | x · v = 0 for all v ∈ V }.
(b) (5 points) Suppose that A is a symmetric n × n matrix. Let V ⊂ Rn be a
subspace with the property that if x ∈ V then Ax ∈ V . Suppose that y ∈ V ⊥ .
Prove that Ay ∈ V ⊥ .
Proof : We show that Ay · x = 0 for every x ∈ V . Using the fact that
A is symmetric (i.e. A = AT ), x ∈ V =⇒ Ax ∈ V , and that y ∈ V ⊥ ,
we obtain:
Ay · x = y · AT x = y · Ax = 0.
Thus Ay ∈ 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 x1 , x2 , . . . , xk be vectors in Rn . Suppose there is a matrix Am×n
for which
{Ax1 , Ax2 , . . . , Axk } is linearly independent. Prove that {x1 , x2 , . . . , xk } is
linearly independent.
Proof : To show that {x1 , x2 , . . . , xk } is linearly independent, we start
with c1 x1 +c2 x2 +. . .+ck xk = 0, and show that c1 = c2 = . . . = ck = 0.
By multiplying both sides by A we have that A(c1 x1 + c2 x2 + . . . +
ck xk ) = A(0) = 0 =⇒ c1 Ax1 + c2 Ax2 + . . . + ck Axk = 0. Since
{Ax1 , Ax2 , . . . , Axk } is linearly independent, then c1 = c2 = . . . =
ck = 0. Thus, {x1 , x2 , . . . , xk } is linearly independent.
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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) Let V = {x = (x1 , x2 , x3 )|x3 ≥ 0} ⊂ R3 . Then V is a subspace in R3 .
False. As a counterexample, let x = (1, 1, 1) ∈ V , since x3 = 1 ≥ 0.
Let c = −1. Then cx = (−1, −1, −1) 6∈ V since x3 = −1 6≥ 0.
 
−1
(c) There is a 3×3 matrix, A, such that [1, −1, 1] ∈ R(A) and  0  ∈ N (A).
1


 
1 −1 1
−1
True. Let A = 0 0 0. Then [1, −1, 1] ∈ R(A) and  0  ∈
1
 0 0 0
−1
N (A) since A  0  = 0.
1
(d) Suppose
 that
  A is a3× 3 matrix such that a basis for N (A) is given by
0
0 
 1
B = 0 , 1 , 0 . Then A is the zero matrix.


0
1
0
True. By the rank + nullity theorem, rank(A) = 3 − dimN (A) =
3 − 3 = 0 which implies that rank(A) = 0 =⇒ A = 0.
7