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Your Name:
Please circle your TA’s name: Jaeho Lee Andrew Bridy
Mathematics 340, Spring 2011 Lectures 1 & 2 (Wilson)
Second Midterm Exam April 14, 2011
There is a problem on the back of this sheet: Be sure not to skip over it by accident!
There is a sheet of scratch paper at the end of this exam. If you need more scratch paper, please ask for it.
You may refer to notes you have brought on part of a sheet of paper, as announced previously.
In proving something you may cite any theorems, lemmas, or corollaries from the text, so long as they came before what you are proving. I.e., don’t indirectly use something to prove itself!
The symbol ~x means
x 1 x 2 · · · xn
,^
~b means
b 1 b 2 · · · bn
, and^ ~ 0 means
, for whatever size is
appropriate.
You may assume that vector spaces Rn, Rn, Mmn, and Pn (with their usual operations), that we have frequently used, are vector spaces, without further proof. Do not assume other spaces are vector spaces unless specifically told to do so.
BE SURE TO SHOW YOUR WORK, AND EXPLAIN WHAT YOU DID. YOU MAY RECEIVE REDUCED OR ZERO CREDIT FOR UNSUBSTANTIATED ANSWERS.
Problem Points Score
TOTAL 100
Problem 1 (10 points)
(a) Consider the matrices
[
]
[
]
, and
[
]
, as vectors in the vector space
M 22 of all 2 × 2 matrices.
Do these matrices span M 22?
Be sure to give reasons for your answer!
(b) Now use the matrices
[
]
[
]
[
]
, and
[
]
. Are these four vectors
in M 22 linearly independent?
Again be sure to give reasons!
(Note that some of these matrices are the same as in part (a), if that is any help.)
Problem 3 (10 points)
Let v 1 =
, v 2 =
, v 3 =
, v 4 =
, and v 5 =
Find a basis for R^3 consisting of some of the vectors ~v 1 ,... , ~v 5.
(If you happen to do problem 6 before this one, you might find useful some calculations done
for that problem.)
Problem 4 (12 points)
Let V be any vector space. Suppose for some vector ~u in V and some number c, c ~u = ~0,
where is whatever the scalar multiplication is on V and ~0 is whatever is the zero vector for
the space V.
Prove that either c = 0 (the real number 0) or ~u = ~0.
(This is part (c) of Theorem 4.2: Parts (a) and (b) were that 0 ~u = ~0 and c ~0 = ~0 for any
vector ~u and for any number c. You may use those facts in your proof for this problem if you
find them helpful.)
Problem 6 (10 points)
Let A =
(You may find useful some work done in connection with problem 3.)
(a) Find a basis for the null space of A, the space of solutions of A~x = ~0.
(b) What is the rank of A?
Problem 7 (12 points)
For the vector space V = P 2 of polynomials with degree at most 2:
(1) The set B 1 = {1 + t, 1 + t^2 , t^2 } is a basis for V.
(2) The set B 2 = {t, 1 + t, 1 + t^2 } is also a basis for V.
(a) Find the “change of basis” matrix (the matrix that the book calls PB 1 ←B 2 ) that tells how
coordinates with respect to B 2 change to become coordinates with respect to B 1.
(b) Find the coordinates of 1 + t − t^2 with respect to B 2.
(c) Find the coordinates of 1 + t − t^2 with respect to B 1.
(d) As a check on your work, multiply the matrix you got in (a) on the left of the coordinate
vector resulting from (b) and show that you get the vector corresponding to (c).
Problem 9 (12 points)
Prove that the only subspaces of R^2 are
1. The subspace consisting of just
[
]
2. The subspace consisting of all multiples of some non-zero vector ~v, and
3. The whole space R^2.
Hint: Consider cases. If V is a subspace of R^2 , is there anything non-zero in it? If there is,
are there any two vectors in V that are linearly independent? Could there be more than two
vectors in V that were linearly independent?
Scratch Paper
(not a command!)
Mathematics 340, Spring 2011 Lectures 1 & 2 (Wilson)
Second Midterm Exam April 14, 2011 Answers
Problem 1 (10 points)
(a) Consider the matrices
[
]
[
]
, and
[
]
, as vectors in the vector space M 22 of all 2 × 2 matrices. Do these matrices span M 22? ANSWER: These three vectors cannot span M 22 : We have seen (e.g. in one of the homework problems) that this is a space of dimension 4, so no set of fewer than 4 vectors could span it.
(b) Now use the matrices
[
]
[
]
[
]
, and
[
]
. Are these four vectors in M 22 linearly independent?
ANSWER: You could write a lot solving equations, but it is also apparent if you happen to look at them “right” that the third matrix is the sum of the first and last. So one of the matrices is a linear combination of some of the others, so they cannot be linearly independent.
Problem 2 (12 points) Let S = {~v 1 , ~v 2 ,... , ~vk} be a set of k vectors in some vector space V. Prove that S is linearly dependent if and only if (at least) one of the vectors ~vj can be written as a linear combination of all the other vectors in S. ANSWER: Here we have to prove the fact that I used in answering problem 1(b)! Assume the vectors in S are linearly dependent. That means there is some combination a 1 ~v 1 + a 2 ~v 2 + · · · + ak~vk = ~0 giving the zero vector, without all of the coefficients aj being zero. Hence in particular aj 6 = 0 for some j. Rearranging the terms, −aj~vj = (the sum of all the terms ai~vi except for i = j). But since aj 6 = 0 (and hence −aj 6 = 0) we can multiply by 1/(−aj ) and get ~vj = − a a^1 j ~v 1 − a a^2 j ~v 2 + · · · − a akj ~vk, where the sum on the right includes terms for each vi except vj. So ~vj can be written as a linear combination of the other vectors. Now assume ~vj can be written as a linear combination of the others, and we have to show the vectors are linearly dependent. We have ~vj = a 1 ~v 1 + a 2 ~v 2 + · · · + ak~vk where the right-hand sum extends over all subscripts from 1 to k except j and the coefficients ai are some numbers. We can add −~vj to both sides and arrange the terms to get ~0 = a 1 ~v 1 + a 2 ~v 2 + · · · − ~vj + · · · + ak~vk, where the coefficients are the same as above except that we now have a term (−1)~vj. This is a linear combination of the vectors in S that gives ~0: Many of those coefficients might be zero, but the coefficient −1 on ~vj is definitely
not zero, so this is a linear combination giving ~0 with not all of the coefficients equal to zero, so the vectors are linearly dependent.
Problem 3 (10 points)
Let v 1 =
, v 2 =
, v 3 =
, v 4 =
, and v 5 =
Find a basis for R^3 consisting of some of the vectors ~v 1... ~v 5. ANSWER:
I will write the vectors as the columns of a matrix, A =
. Now row reduce
A and get either (in Row Echelon Form, where there might be variations in this matrix) AR =
(^) or (in Reduced Row Echelon Form) ARR =
. In either case
we see that the first, third, and fourth columns contain leading entries, so we use the first, third, and
fourth vectors v 1 =
, v 3 =
, and v 4 =
(^) from the original set for our basis.
(This is a procedure you could use mechanically. You could also have seen that v 2 is twice v 1 , so throw out v 2 , and proceeded like that discarding vectors that are linear combinations of the ones you don’t throw out.)
Problem 4 (12 points) Let V be any vector space. Suppose for some vector ~u in V and some number c, c ~u = ~0, where is whatever the scalar multiplication is on V and ~0 is whatever is the zero vector for the space V. Prove that either c = 0 (the real number 0) or ~u = ~0.
(This is part (c) of Theorem 4.2: Parts (a) and (b) were that 0 ~u = ~0 and c ~0 = ~0 for any vector ~u and for any number c. You may use those facts in your proof for this problem if you find them helpful.) ANSWER: Suppose c ~u = ~0 and c 6 = 0: We need to show that in this case ~u = ~0. Since c 6 = 0, there is a number (^1) c such that (^1) c c = 1. Multiply that number on both sides of our equation c ~u = ~0, getting 1 c (c^ ~u) =^
1 c ~0. By an earlier part of this same theorem,^
1 c ~0 =^ ~0. By the defining requirements for a vector space, (^1) c (c ~u) =
c c
~u, or (^1) c (c ~u) = 1 ~u, and by another of the defining requirements for a vector space 1 ~v = ~v for any vector ~v. Putting these pieces together, ~u = 1 ~u =
c c
~u = (^1) c (c ~u) = (^1) c ~0 = ~0 and we are through.
Problem 5 (10 points)
(a) Show that the set of vectors
[
a b c d
]
in M 22 which satisfy a + b + c + d = 0 is a subspace of M 22. ANSWER: It is clear from the way it is described that the set in question is a subset of M 22 , so we just have to show (i) the set is not empty, (ii) it is closed under addition, and (iii) it is closed under scalar multiplication.
Since
[
]
is in the set, the set is not empty. For (ii), suppose
[
a b c d
]
and
[
e f g h
]
are
both in the set, i.e. a + b + c + d = 0 and e + f + g + h = 0. Their sum is
[
a + e b + f c + g d + h
]
, so to determine whether that sum is in our set we ask whether (a + e) + (b + f ) + (c + g) + (d + h) is 0. Rearranging that as (a + b + c + d) + (e + f + g + h) = 0 + 0 = 0 we see the sum is in
(b) What is the rank of A? ANSWER: You could look at the REF and read off the number of non-zero rows, 3. Or you could use the fact that the rank plus the dimension of the solution space must equal the number of columns: From (a) we know the dimension of the solution space is 2, and there are 5 columns, so the rank must be 5 − 2 = 3.
Problem 7 (12 points) For the vector space V = P 2 of polynomials with degree at most 2: (1) The set B 1 =
1 + t, 1 + t^2 , t^2
is a basis for V. (2) The set B 2 =
t, 1 + t, 1 + t^2
is also a basis for V. (a) Find the “change of basis” matrix (what the book denotes PB 1 ←B 2 ) that tells how coordinates with respect to B 2 change to become coordinates with respect to B 1. ANSWER: We find the coordinates of each vector in B 2 with respect to B 1 , and write those as the columns of a matrix. For the first vector in B 2 , t, could either set up equations and solve them or just “think through it”. I see that I could get 1 as (1+t^2 )−t^2 , and then t as (1+t)−1 = (1+t)−[(1+t^2 )−t^2 )]. So t = 1 × (1 + t) + (−1) × (1 + t^2 ) + 1 × (t^2 ), i.e. the coordinates of t with respect to B 1 are 1,
−1, and 1 or as a column vector
Next we find the coordinates of 1 + t with respect to B 1 : Since the first vector in B 1 is exactly
1 + t, the coordinates are 1, 0, and 0 or
Similarly, 1 + t^2 is exactly the second vector in B 1 , so its coordinates are 0, 1, or 0 or
So the matrix we want is PB 1 ←B 2 =
(b) Find the coordinates of 1 + t − t^2 with respect to B 2. ANSWER: Reasoning as before (You could instead set this up as a system of equations to be solved): 1 = (1 + t) − t. t^2 = (1 + t^2 ) − 1 = (1 + t^2 ) − [(1 + t) − t] = (1 + t^2 ) − (1 + t) + t. So 1 + t − t^2 = (1 + t) − t^2 = (1 + t) − [(1 + t^2 ) − (1 + t) + t] = − 1 × (t) + 2 × (1 + t) − 1 × (1 + t^2 )
and hence the coordinates as a vector are
(c) Find the coordinates of 1 + t − t^2 with respect to B 1. ANSWER: This time the calculations are simpler: 1 + t and t^2 are themselves basis vectors in B 1 , so we can
see 1 + t − t^2 as 1 × (1 + t) + 0 × (1 + t^2 ) − 1 × (t^2 ), i.e. the coordinates as a vector are
(d) As a check on your work, multiply the matrix you got in (a) on the left of the coordinate vector resulting from (b) and show that you get the vector corresponding to (c). ANSWER:
If we multiply
(^) we do get
Problem 8 (12 points)
Let A =
(a) Find the adjoint adj(A). ANSWER:
Remember that the determinant of a 2 × 2 matrix
[
a b c d
]
is ad − bc. We need to compute nine determinants, each being just that easy, and affix ± signs. In the upper left of our 3 × 3 matrix we put, with a plus sign, the determinant of the matrix we get if we delete the first row and column from our matrix, i.e. det
([
])
. That is 2 × 1 − 2 × 1 = 0. For the entry in the i, j position we take the determinant of the 2 × 2 matrix we would get by deleting the jth^ row and the ith^ column (don’t forget the transpose here, i.e. the swap of i and j), and use for a sign − 1 i+j^. So for example in the middle of the top row, i = 1 and j = 2, we take the determinant of
[
]
and get 1 × 1 − 0 × 1, and − 1 1+2^ = − 13 = −1 so we use a minus sign. Filling in the calculations without the ± signs we have
2 × 1 − 2 × 1 1 × 1 − 0 × 1 1 × 2 − 0 × 2
2 × 1 − 2 × 0 2 × 1 − 0 × 0 2 × 2 − 0 × 2
1 × 2 − 0 × 2 2 × 1 − 1 × 0 2 × 2 − 1 × 2
(^) or
Now we put in the signs in their checkerboard pattern and get for the answer
(b) Find the determinant det(A) ANSWER: We could of course just calculate det(A) directly, but we also know that A × adj(A) should give us det(A) × I 3 , so we can both get the determinant and have a check on our answer to (a) if we just multiply A and our answer. We get
which is a scalar matrix, i.e. a scalar times I 3 (validating our computation of the adjoint) where the scalar is −2, so the determinant must be −2.
