Math 3260
Practice Exam 3 Answer key
1. Since the sum of two symmetric matrices is a symmetric matrix, and a scalar multiple
of a symmetric matrix is a symmetric matrix, it follows that the set of symmetric
matrices of order three is a subspace of the space of all matrices of order 3. Since a
symmetric matrix of order 3 has aij  a ji , the matrix will depend only on the entries
on the diagonal and above the diagonal. Let us write such a matrix:
a b c 
1 0 0 
0 1 0 
0 0 1 






A   b d e   a 0 0 0   b 1 0 0   c 0 0 0 
 c e f 
0 0 0 
0 0 0 
1 0 0 
0 0 0 
0 0 0
0 0 0




 d 0 1 0   e 0 0 1   f 0 0 0   aE1  bE2  cE3  dE4  eE5  fE6
0 0 0 
0 1 0 
0 0 1 
It can be seen from the above decomposition that the six matrices, E1 , E2 , E3 , E4 , E5 , E6
span the space of symmetric matrices of order three, and they form a linearly independent
set since if a linear combination like the above one is zero, then all coefficients must be
zero. Therefore the matrices E1 , E2 , E3 , E4 , E5 , E6 form a basis in the space of symmetric
matrices of order three, and the dimension of the space is 6.
2. The set is not a linear vector space with the given operations since there does not
exists a zero vector for the addition. Indeed, we need to find a vector (a, b) such that
for any vector ( x, y ) : (a, b)  ( x, y)  ( x, y) . Take for example the vector (1,1) . If
(a, b) were the zero vector, then (a, b)  (1,1)  (1,1) . But the second component of the
sum is always zero, therefore no zero vector exists.
3. S  {( x, y) : x 2  y 2 } is not a subspace of R2 under the usual operations of addition
and multiplication by scalars. We will show that S is not closed under addition.
Indeed, let (2, 2), (3,3)  S . We have (2, 2)  (3,3)  (5,1) and 52  12 .
4. We will use the linearity of the operator to determine the value of T on the standard
basis vectors, T (1, 0), and T (0,1). Refer to the standard basis vectors as e1 , e2 . We
have:
T (1, 2)  T (e1  2e2 )  T (e1 )  2T (e2 )  (2,3)
T (1, 1)  T (e1  e2 )  T (e1 )  T (e2 )  (5, 2)
We can solve the above system to obtain: 3T (e2 )  (7,1), 3T (e1 )  (8,7) . Therefore
1 8 7 
1 8 7  7  7 
T 

, and T (7,5)  
.

3 7 1 
3 7 1  5  18
5. a) Let u1  ( x1 , y1 ), u2  ( x2 , y2 )  R 2 , c1 , c2  R . We have
T (c1u1  c2u2 )  (c1 x1  c2 x2  c1 y1  c2 y2 , c1 x1  c2 x2  c1 y1  c2 y2 ,3(c1 x1  c2 x2 )  2(c1 y1  c2 y2 ))
 c1 ( x1  y1 , x1  y1 ,3x1  2 y1 )  c2 ( x2  y2 , x2  y2 ,3x2  2 y2 )  c1T (u1 )  c2T (u2 )
so the transformation is linear.
b) In order to decide whether T is one-to-one we need to see whether T ( x, y )  (0, 0, 0)
has only the trivial solution ( x, y )  (0, 0) . We write the matrix of the transformation and
reduce it to its RREF:
1 1  1 0 
T  1 1 ~ 0 1  . Since the matrix has two leading ones, and we solve for two
3 2  0 0 
unknowns, the system will have only the trivial solution, and therefore the transformation
is one-to-one.
c) Let now ( w1 , w2 , w3 )  R 3 . We need to see whether there exists
( x, y )  R 2 : T ( x, y )  ( w1 , w2 , w3 ) . We write the augmented matrix of the system and
reduce it:


1 1

w1
w1  
1 1 w1  1 1

1
1 1 w  ~ 0 2 w  w  ~ 0 1

 ( w2  w1 )
2
2
1 




2
3 2 w3  0 1 w3  3w1  

1
0 0 w3  3w1  ( w2  w1 ) 

2

1
The system will be consistent if and only if w3  3w1  ( w2  w1 )  0, 2w3  5w1  w2  0 .
2
A vector that is not in the range(T) is (1, 0,1) .
6. Of course, since one vector is a multiple of the other the two vectors cannot span R2 .
We will actually find a vector in R2 that is not in span{(1,1), (2, 2)}. Consider the
vector (1,0). If it were in the subspace spanned by the two given vectors, then there
must exist c1 , c2 such that c1 (1,1)  c2 (2, 2)  (1,0). But if we write the system
corresponding to this vector equation we obtain c1  2c2  1  0 . Therefore there do
not exist such scalars, and thus (1,0) is not in span{(1,1), (2, 2)}.
7. a) Since S has three vectors, the dimension of P2 , if it were a spanning set for P2 it
will have to be linearly independent as well. But it can be seen that the three vectors
are not linearly independent, since he third polynomial can be written as a linear
combination of the first two: x 2  2  x 2  2 1 .
b) We prove that this set spans P2 . Since we have three vectors again, we will check that
they are linearly independent. Consider:
c1 ( x  1)  c2 ( x  2)  c3 ( x 2  1)  0 . Then c1 , c2 , c3 are solutions to the system whose
coefficient matrix is:
1 2 1 1 0 0 
1 1 0  ~ 0 1 0  ,

 

0 0 1  0 0 1 
therefore the system has only the trivial solution, so the vectors are linearly independent.
Since dim( P2 )  3 , they form a basis in P2 , therefore they span the space.
1 0 1
8. We find the reduced row echelon form of the matrix 0 1 2  . Therefore a basis


0 0 0 
for the row space is {(1, 0, 1), (0,1, 2)}, and a basis for the column space is
{(3,-5,-6), (0,1,0)} .
9. Let us assume that there are two null vectors, 0 and 0'. If they are both zero vectors
they must satisfy the axiom 0  x  x and 0 ' x  x, for all x , in particular when
x  0' or x  0, respectively . We then have 0  0 '  0, and also 0 ' 0  0 ' , and since
the two expressions must be equal, it follows 0  0' .
10. a) The set is obviously a subspace, since it is the equation of a plane going through
the origin.
b) So we have the set of vectors {( x, y, z ) : x  0, or y  0}. This set is not a subspace
since for example both (1, 0, 2) and (0,1, 0) are in the set, but their sum (1,1, 2) is not.
c) The set can be described as {( x, y, z ) : x  0 or y=z}. This set is not a subspace since it
contains the vectors (0,1, 2) and (1,3,3) , but not their sum (1,4,5).
d) The set is not a subspace because it does not contain the zero vector (0,0,0).
e) Not a subspace since it does not contain the zero vector.
11. Since the sum of two matrices with diagonal elements zero has the diagonal zero, and
multiplication by a scalar will still leave the diagonal zero, the set is closed under
addition and scalar multiplication, and therefore it is a subspace of M nn .
12. In order to determine whether the four vector span R 3 , we will define a matrix with
these vectors as columns and find its rank. We get that the row echelon form of the
matrix formed with the four vectors has only two leading ones, and therefore the rank
is 2, and there are only two linearly independent vectors among the given four.
Therefore they cannot span R 3 .
13. Arrange the vectors as rows in a matrix,
r1  (1,3, 1, 4), r2  (1,3,0,6), r3  (1, 3,0,8) . The row echelon form of the matrix has
1 3 1 4 
the form 0 0 1 2  , therefore the three vectors are linearly independent, so they


0 0 0 1 
form a basis in the subspace they span. We may as well choose the rows of the RREF
matrix to form the basis for the space spanned by the vectors.
14. We have to find all 2  2 matrix having 3 x  5 y  0 as its nullspace. Then the matrix
must be the coefficient matrix of a system with two equations in two unknowns, in
which the first equation is given by the plane above. We can choose for the second
equation in the system any equation proportional to the first one. Therefore one such
3 5 
matrix can be: A  
 . Since we are asked to find all such matrices, they mus
6 10
3
be of the form: A  
3k
5 
3k
or A  

5k 
3
5k 
, where k is any real number.
5 