Practice Exam 2 – 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 is a subspace of the space of all matrices of order n.
2.
a) We show the set is a subspace. Let us call V the set in the problem. Consider
( x1 , y1 , z1 ),( x2 , y2 , z2 ) : x1  y1  z1  0, x2  y2  z2  0 .
( x1 , y1 , z1 )  ( x2 , y2 , z2 )  ( x1  x2 , y1  y2 , z1  z2 ) V since x1  x2  y1  y2  z1  z2  0 .
Similarly, if c  R, ( x, y, z ) V then c( x, y, z )  V since cx  cy  cz  0 .
Therefore the set is a subspace with the given operations. Notice in particular that the
equation represents 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 two 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.
3.
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 .
4.
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' .
5.
If the vectors span R 3 then any vector in R 3 can be written as a linear
combination of the three vectors. This means that for any vector (b1 , b2 , b3 ) , there exist
scalars c1 , c2 , c3 such that (b1 , b2 , b3 )  c1 (1, 2,1)  c2 (1,3,0)  c3 (0,5,1) . This equation
1 1 0 
leads to a linear system whose coefficient matrix is A   2 3 5  . The system will


1 0 1 
always have solutions if and only if the row-echelon form of the matrix has three leading
ones, or equivalently, det( A)  0 . However, computation shows det( A)  0 . Now, in
order to determine the subspace that is spanned by the given set we look at the augmented
b1
1 1 0

b1  
1 1 0 b 1  1 1 0

1





(b 2 2b 1 )  .
matrix:  2 3 5 b 2  0 5 5 b 2 2b 1  0 1 1
5


1 0 1 b 3  0 1 1 b 3 b 1  
b 3 b 1 
0 1 1
It is now obvious that the system will have solutions if and only if
1
(b 2 2b 1 )  b 3 b 1 , or
5
equivalently 3b 1 b 2 5b 3  0 .
6.
The set is linearly independent if the vector zero cannot be written as a nontrivial
linear combination of the vectors in the set. Reasoning as in the problem above, this is
t 1 1
equivalent to requiring the matrix A  1 t 1 to be nonsingular. We will compute the


1 1 t 
determinant of the matrix, but before doing so we reduce the matrix by subtracting the
third row from the second and subtracting a multiple of t of the third row from the first
row. We have:
0 1 t 1 t2
det( A)  0 t  1 1  t  (1  t ) 2  (t  1)(1  t 2 )  (1  t ) 2 (2  t ) .
1 1
t
Therefore, the matrix is nonsingular for t  1, 2 . So these are the values for which the
three vectors are linearly independent.
7.
If the set spans P3 then any polynomial in P3 can be written as a linear
combination a3 x3  a2 x 2  a1 x  a0  c1 ( x 2  2 x)  c2 ( x 3  8)  c3 ( x 3  x 2 )  c4 ( x 2  4) .
Since the two represent the same polynomial we need to have:
c2  c3  c4  a3
c1  c3  a2
2c1  a1
8c2  4c4  a0
It is easy to observe that the above system has always a solution, and therefore the given
set of polynomials spans P3 .
8.
The two sets of vectors span the same subspace if each vector in S1 belongs to
span( S2 ) , and each vector in S 2 belongs to span( S1 ) .
Notice that (0,0,1)  (1,1, 2)  (1,1,1)  span( S2 ), (2,1,1)  S2  span( S2 ) . Let us write
(0,1,1)  c1 (1,1,1)  c2 (1,1, 2)  c3 (2,1,1) . If we can find the scalars c1 , c2 , c3  R then
(0,1,1)  span(S2 ) . The above vector equation leads to a linear system with augmented
1 1 2 0 1 1 2 0 1 1 2 0 
matrix A  1 1 1 1  ~ 0 0 1 1  ~ 0 1 1 1  . The rwo echelon form of
1 2 1 1  0 1 1 1  0 0 1 1 
the augmented matrix shows that we can write (0,1,1)  2(1,1,1)  (2,1,1)  span(S2 ) .
So far we proved that span(S1 )  span(S2 ) .
Conversely, we check whether span(S2 )  span(S1 ) . Notice that
1
1
(1,1,1)  (2,1,1)  (0,1,1)  span( S1 ),
2
2
1
1
(1,1, 2)  (2,1,1)  (0,1,1)  (0, 0,1)  span( S1 ), and (2,1,1)  span( S1 )
2
2
Therefore span(S2 )  span(S1 ) and together with span(S1 )  span(S2 ) it follows that
span( S1 )  span( S2 ) .
Observation: Notice that the proof does not find the geometric description of the
subspace spanned by the above sets. In particular, you can check that span( S1 )  R 3
implying span( S2 )  R3 as well.
9.
In order to prove that the set spans R 3 we need to show that for any
(u1 , u2 , u3 )  R 3 we can find scalars c1 , c2 , c3  R such that
c1 (1, 2,3)  c2 (3, 2,1)  c3 (0,0,1)  (u1 , u2 , u3 ) (1)
In order to prove linear independence we need to show that the only solution to
c1 (1, 2,3)  c2 (3, 2,1)  c3 (0,0,1)  (0,0,0)
(2)
is c1  c2  c3  0 .
Notice that both checking for linear independence and that the set spans R 3 reduces to
solving a linear system in which the coefficient matrix is
1 3 0 
A   2 2 0 
 3 1 1 
If we prove that the matrix A is nonsingular then (1) will always have a solution and (2)
will only have the trivial solution. We can compute det( A)  12  0 , so A is nonsingular
and this shows that the set of vectors B is linearly independent and spans R 3 .