Math 3130
Abrams
Spring 2016
PRACTICE EXAM for Exam 1 - SOLUTIONS
Give complete answers. Show all your work!! Write legibly. Write out your explanations in full detail.
Circle answers when appropriate. You may use the back of the sheet if necessary. Good luck.
1. SHORT ANSWER
(a) (2 pt) If V is a vector space and S = {v1 , v2 , ..., vn } is a finite set of vectors in V , then S is called a
basis for V in case
(i) S spans V
and
(ii) S is linearly independent
(b) (1 pt) Referring to question 1a, the number n is called the dimension of V .
(c) (1 pt) Give an example of a vector space V which has dim(V ) = 4, but where V is NOT R4 .
There are many, for instance P3 or M2,2 .
(d) (3 pt) Do the vectors (1, 2, 3), (1, 0, 0), and (4, 4, 6) form a basis of R3 ? Justify your answer appropriately.
ByThe Theorem
 we can use the determinant test: Form the matrix M using these vectors as columns:
1 1 4
M =  2 0 4  . Compute det(M ); it’s easiest to expand on the second column. We get
3 0 6
2 4
1+2
det(M ) = (−1)
· 1 · det
= (−1)(12 − 12) = 0. So these three vectors do NOT form a basis of R3 .
3 6
2. (3 pt) Is the vector u = (3, −1, 3, 2) in the span of the vectors v1 = (1, 0, 0, 0), v2 = (1, −1, 1, 0), and
v3 = (0, 0, 1, 1)? (Rephrased: Is u in span{v1 , v2 , v3 }?) If so, write u as a linear combination of v1 , v2 , and v3 .
If not, explain why not by an appropriate computation.
We are asked whether or not there are scalars k1 , k2 , k3 for which
(3, −1, 3, 2) = k1 (1, 0, 0, 0) + k2 (1, −1, 1, 0) + k3 (0, 0, 1, 1).
Equating the appropriate entries, this leads to a system
k1 +
k2
−k2
k2 + k3
k3
= 3
= −1
= 3
= 2
This system is easy to solve by inspection: we see that k3 = 2, and k2 = 1 (which is consistent with the second
equation), and then k1 = 2. So we have solved the system, with k1 = 2, k2 = 1, and k + 3 = 2. So
u = 2v1 + v2 + 2v3 .
1
3. (3 pt) Consider the set S of vectors in P2 of the form a + bx + cx2 where a ≥ 0. S is NOT a subspace of
P2 . Explain why not.
There are many reasons which can be given. For instance, this set is not closed under scalar multiplication.
As an example, 1 + x is in S, but −1 · (1 + x) = −1 − x is not in S.
4. (7 pt) Let W denote the subset of P3 consisting of all polynomials of the form a0 + a1 x + a2 x2 + a3 x3 for
which 2a0 − a1 + a3 = 0.
(a) Give three specific examples of polynomials in P3 which are in the subset W .
0 and 1 − 2x3 and x2 , for instance. Any polynomial for which the coefficients satisfy the given relation.
(b) Give three specific examples of polynomials in P3 which are NOT in the subset W .
1 and x and x3 , for instance.
(c) Prove that W is a subspace of P3 . (As in the homework, use The Subspace Theorem.) Make sure to
show all your work, and that your notation is clear and precise.
1) Show that W is closed under addition. So pick two elements a0 + a1 x + a2 x2 + a3 x3 and b0 + b1 x + b2 x2 + b3 x3
in W . We must show that a0 + a1 x + a2 x2 + a3 x3 + b0 + b1 x + b2 x2 + b3 x3 is in W , in other words, that
(a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 + (a3 + b2 )x3 is in W . We know by definition of W that 2a0 − a1 + a3 = 0, and
that 2b0 − b1 + b3 = 0. We must show that 2(a0 + b0 ) − (a1 + b1 ) + (a3 + b3 ) = 0.
But 2(a0 +b2 )−(a1 +b1 )+(a3 +b3 ) = (by algebra) (2a0 −a1 +a3 )+(2b0 −b1 +b3 ) = (by substitution) 0+0 = 0.
So part 1 of the Subspace Theorem has been established.
2) Show that W is closed under scalar multiplication. So pick an element a0 + a1 x + a2 x2 + a3 x3 in W , and
pick k ∈ K. We must show that k(a0 + a1 x + a2 x2 + a3 x3 ) is in W , in other words, that ka0 + ka1 x + ka2 x2 + ka3 x3
is in W . We know by definition of W that 2a0 − a1 + a3 = 0. We must show that 2(ka0 ) − (ka1 ) + (ka3 ) = 0.
But 2(ka0 ) − (ka1 ) + (ka3 ) = (by algebra) k(2a0 − a1 + a3 ) = (by substitution) k(0) = 0. So part 2 of the
Subspace Theorem has been established, and the problem is done.
5. (4 pt) It can be easily shown that the set of vectors v1 = (1, 0, 0), v2 = (−1, 1, 0), v3 = (0, 1, 2) forms a
basis for R3 . Find the coordinate vector of u = (1, 1, 3) relative to this basis.
We must solve the equation
(1, 1, 3) = k1 (1, 0, 0) + k2 (−1, 1, 0) + k3 (0, 1, 2).
This gives the system
k1 − k2
k2 +
k3
2k3
= 1
= 1
= 3
It is easy to solve this system by inspection; we get k3 = 3/2, and then k2 = −1/2, and then k1 = 1/2. So the
coordinate vector of u relative to the given basis is the vector (1/2, −1/2, 3/2).
2






2 0 −1
6. (11 pt total) Consider the matrix A =  1 1 0  .
4 0 −2
(a) Find a basis for the null space of A.
We must solve 
the homogeneous system whose coefficient matrix is A. So we setup
..
1
2 0 −1 . 0


..

1 1 0 . 0 
. Doing Gauss Jordan we get (after some tedious computation)  0
.
0
4 0 −2 .. 0
The corresponding system is then
x
− 1/2z
y + 1/2z
the augmented matrix

..
0 −1/2 . 0

.
1 1/2 .. 0 
.
..
0
0
. 0
= 0
= 0


x
We solve this system parametrically. Let z = t. Then the solutions to the system are of the form  y  =
z






1/2t
1/2
1/2
 −1/2t  = t  −1/2 . So the one-element set { −1/2 } is a basis for the null space of A.
t
1
1
(b) Find a basis for the column space of A.


1 0 −1/2
From part (a) of this question we see that the Gauss Jordan form of A is the matrix  0 1 1/2  . Since
0 0
0
the leading 1s appear in the first two columns of this matrix, this tells us that the first two columns of the original
matrix A form a basis of the column space. So the two element set
   
2
0



{ 1 , 1 }
4
0
is a basis of the column space of A.
(c) Use your answer to part (b) to find the dimension of the row space of A. (Do NOT find a basis of
the row space of A.)
The dimension of the row space equals the dimension of the column space, namely, 2 (by part (b)).
(d) For any m × n matrix M , what is the equation which arises in the Dimension Theorem for Matrices?
dimension of null space + dimension of column space = n. Rewritten, nullity(A) + rank(A) = n.
(e) For the specific matrix A of this question, give the specific values which arise in the Dimension
Theorem for Matrices.
nullity(A) = 1, rank(A) = 2, and 1 + 2 = 3, the number of columns of A.
3
7. (9 pt total) Let V be the vector space F(R),Ri.e., the set of functions from R to R. Let W denote the subset
1
of V consisting of all those functions f for which −1 f (x)dx = 0.
(a) Is the function f (x) = x5 in W ? Explain why or why not.
R1
6
6
6
Yes: −1 x5 dx = x6 |1−1 = 16 − (−1)
= 0.
6
(b) Prove that W is a subspace of V . (Use the Subspace Theorem.)
R1
1) Pick f (x) and g(x) in W . We must show that f (x) + g(x) is in W . So we have that −1 f (x)dx = 0 and
R1
R1
g(x)dx = 0. We must show −1 (f (x) + g(x))dx = 0. But
−1
Z 1
Z 1
Z 1
g(x)dx = 0 + 0 = 0.
f (x)dx +
(f (x) + g(x))dx = (property of definite integrals)
−1
−1
−1
2) Pick f (x) in W , and any scalar k. We must show that kf (x) is in W . So we have that
R1
we must show −1 kf (x)dx = 0. But
Z 1
Z 1
kf (x)dx = (property of definite integrals) k
f (x)dx = k · 0 = 0.
−1
R1
−1
f (x)dx = 0, and
−1
(c) Verify that the set consisting of the two functions {x, x3 } is a linearly independent subset of W .
We compute the Wronskian
det
x x3
1 3x2
= x · 3x2 − x3 · 1 = 3x3 − x3 = 2x3 6= 0.
So this set of two functions is linearly independent.
(d) What is the dimension of W ?
W has infinite dimension: The set {x, x3 , x5 , x7 , ...} is a linearly independent subset of W .
8.
(1 pt each) TRUE / FALSE.
(a)
FALSE
If S is a set of 3 vectors in R3 , then S must be a basis of R3 .
(b)
TRUE
If S is a set of 3 vectors in R4 , then S cannot be a basis of R4
(c) TRUE If you and your friend each find a basis for a vector space V , then you will each necessarily
have found the same number of vectors.
(d) TRUE
The function sin x is a vector.
(e) FALSE
Any set of 2 vectors in P4 is a linearly independent set.
(f)
Let S be a linearly independent set of 3 vectors in R3 . Then S is necessarily a basis for
TRUE
3
R.
(g) FALSE
A basis for a vector space is a subspace of the vector space.
(h) FALSE For any matrix A, the row space of A and the null space of A have the same dimension.
(i) FALSE If a set of vectors S in a vector space V has the property that no vector in S is a scalar
multiple of any other vector in S, then S is a linearly independent subset of V.
4