Download Math 312 Syllabus

Math 312 – Homework 1 – Solutions
Last modified: July 15, 2012
This homework is due on Thursday, July 12th, 2012 at 12:10pm. Please turn it in
during class, or in my mailbox in the main math office (next to 4W1). Please staple your
pages before you turn in your homework and make sure the problems are labeled clearly and
are, preferably, in sequential order. Solutions are in blue.
1. Consider the following general system of two equations in the two variables x, y:
ax + cy = e
bx + dy = f
Find a condition on a, b, c, d so that the system above has a solution for all possible
e, f .
Solution: First assume that both a = 0 and b = 0. In this case, we see that the two
equations become cy = e and dy = f . If c = 0 or d = 0 then this system contains
the equation 0 = e or 0 = f , but we can choose e, f arbitrarily so this would be
an inconsistent system. On the other hand, if c, d 6= 0, then the system becomes
y = e/c = f /d but, again, we can vary e, f so that e/c 6= f /d and we obtain an
inconsistent system.
Thus, at least one of a or b is non-zero, and, by symmetry, we can assume without loss
of generality that a 6= 0. Now we look at the augmented matrix obtained from the
system:
a c e
b d f
which row reduces to:
1
c/a
e/a
0 d − bc/a f − be/a
We can choose f so that f − be/a 6= 0 so that this system is consistent if and only
if d − bc/a 6= 0 or equivalently ad − bc 6= 0. When this is the case, we see that
x = (de − cf )/(ad − cb) and y = (af − be)/(ad − cb) is a solution for an arbitrary e, f .
2. Let v1 , . . . , vm ∈ Fn be given and consider Fn as a vector space over F. Denote by
vi = (a1i , . . . , ani ), aij ∈ F, and consider the matrix A ∈ Mn×m (F) defined by Aij = aij .
Show that
1
(a) If Span(v1 , . . . , vm ) = Fn then the row reduced echelon form of A has n pivots.
(b) If the row reduced echelon form of A has n pivots then Span(v1 , · · · , vm ) = Fn .
(c) Deduce: if Span(v1 , . . . , vm ) = Fn then m ≥ n.
Solution: Say v = (b1 , . . . , bn ) ∈ Fn is given. Observe that v ∈ Span(v1 , . . . , vm ) if
and only if the system of linear equations Ax = (b1 , . . . , bn )t =: b has a solution.
(a) Suppose that the RREF form of A has less than n pivots (note that there can
be at most n). Then the RREF form of A must have a row of zeros. Denote by
B the RREF form of A, which is obtained from A by applying the elementary
row operations R1 , . . . , Rs . Denote by Ri0 the row operations which reverses Ri
– i.e. Ri0 is the row operation which is the inverse to Ri . In particular, we can
obtain A back from B by applying Rs0 , . . . , R10 in that order. By our assumption,
B contains a row of zeros and thus the augmented matrix:


0

.. 

. 
C :=  B


0 
1
corresponds to an inconsistent system. Applying the row operations Rs0 , . . . , R10
to C, we obtain the matrix:


b1

.. 
D :=  A
. 
bn
which is also inconsistent since it is obtained by elementary row operations from
an inconsistent system. In particular, (b1 , . . . , bn ) ∈
/ Span(v1 , . . . , vm ).
(b) Conversely, if B has n pivots, then it has a pivot in every row. Thus, the row
reduction of


b1

.. 
D :=  A
. 
bn
is precisely


C :=  B

c1
.. 
. 
cn
and, since B has a pivot in every row, both systems of linear equations are consistent.
2
(c) If Span(v1 , . . . , vm ) = Fn then B has a pivot in every row. However, B is a matrix
of size n × m. By the definition of “pivot” we see that there are precisely n pivot
columns. Since B has m columns, we deduce that n ≤ m.
3. Let F be a field (i.e. F = Q, R or C). Let A ∈ Mm×n (F) be given. Prove that the
following sets are subspaces in the corresponding vector space:
(a) The set {A · B : B ∈ Mn×k (F)}, as a subset of Mm×k (F).
(b) The set {B · A : B ∈ Mk×m (F)}, as a subset of Mk×n (F).
(c) The set {B ∈ Mn×k (F) : A · B = 0}, as a subset of Mn×k (F).
(d) The set {B ∈ Mk×m (F) : B · A = 0}, as a subset of Mk×m (F).
Solution: Recall that a set S ⊂ V is a subspace if the following two conditions hold:
• S 6= ∅ and
• For all v, v 0 ∈ S and a, a0 ∈ F one has av + a0 v 0 ∈ S.
To show that the sets above are subspaces, we check these two conditions.
(a) An arbitrary element in this set is of the form A · B for some B. As 0 = A · 0
is contained in the set, it is non-empty. Let A · B and A · B 0 be two arbitrary
elements of the set and a, a0 be two arbitrary scalars. Then a(A · B) + a0 (A · B 0 ) =
A · (aB + a0 B 0 ) is contained in the set. Thus, this set is closed under matrix
addition and scalar multiplication, as required.
(b) An arbitrary element in this set is of the form B · A for some B. As 0 = 0 · A
is contained in the set, it is non-empty. Let B · A and B 0 · A be two arbitrary
elements of the set and a, a0 be two arbitrary scalars. Then a(B · A) + a0 (B 0 · A) =
(aB + a0 B 0 ) · A is contained in the set. Thus, this set is closed under matrix
addition and scalar multiplication, as required.
(c) Observe that A · 0 = 0 so that 0 is an element of this set. Suppose B, B 0 are
contained in the set – equivalently, AB = AB 0 = 0 – and let a, a0 be two arbitrary
scalars. Then A · (aB + a0 B 0 ) = aAB + a0 AB 0 = a0 + a0 0 = 0 so that aB + a0 B 0
is contained in the set as well.
(d) Observe that 0 · A = 0 so that 0 is an element of this set. Suppose B, B 0 are
contained in the set – equivalently, BA = B 0 A = 0 – and let a, a0 be two arbitrary
scalars. Then (aB + a0 B 0 ) · A = aBA + a0 B 0 A = a0 + a0 0 = 0 so that aB + a0 B 0
is contained in the set as well.
4. Exercise 1.3.8 from the text, reproduced here for your convenience: Are the following
sets subspaces of R3 under the usual vector addition and scalar multiplication? Explain.
(a) W1 = {(a1 , a2 , a3 ) : a1 = 3a2 , and a3 = −a2 }.
3
(b) W2 = {(a1 , a2 , a3 ) : a1 = a3 + 2}.
(c) W3 = {(a1 , a2 , a3 ) : 2a1 − 7a2 + a3 = 0}.
(d) W4 = {(a1 , a2 , a3 ) : a1 − 4a2 − a3 = 0}.
(e) W5 = {(a1 , a2 , a3 ) : a1 + 2a2 − 3a3 = 1}.
(f) W6 = {(a1 , a2 , a3 ) : 5a21 − 3a22 + 6a23 = 0}.
Solution:
(a) Observe that (0, 0, 0) ∈ W1 so that W1 is non-empty. Suppose (a1 , a2 , a3 ), (b1 , b2 , b3 ) ∈
W1 are arbitrary. Let a, b ∈ F be scalars and consider the linear combination:
a(a1 , a2 , a3 ) + b(b1 , b2 , b3 ) = (aa1 + bb1 , aa2 + bb2 , aa3 + bb3 ) =: w
Observe that aa1 + bb1 = a3a2 + b3b2 = 3(aa2 + bb2 ) and aa3 + bb3 = a(−a2 ) +
b(−b3 ) = −(aa2 +bb3 ) so that w ∈ W1 . Thus, we see that W1 is indeed a subspace.
(b) This is not a subspace as (0, 0, 0) ∈
/ W2 .
(c) This is a subspace, as follows. First, we observe that (0, 0, 0) ∈ W3 . Suppose
v = (a1 , a2 , a3 ) and v 0 = (b1 , b2 , b3 ) are elements of W3 and take an arbitrary
linear combination av + bv 0 = (aa1 + bb1 , aa2 + bb2 , aa3 + bb3 ). We have:
2(aa1 + bb1 ) − 7(aa2 + bb2 ) + (aa3 + bb3 ) = a(2a1 − 7a2 + a3 ) + b(2b1 − 7b2 + b3 ) = 0.
Thus, we see that av + bv 0 ∈ W3 as well, proving that W3 is indeed a subspace.
(d) In a similar way to part (3), we see that W4 is a subspace.
(e) The vector (0, 0, 0) ∈
/ W5 so that W5 is not a subspace.
(f) This is √
not a subspace as it is not p
closed under vector addition. For example,
√
0
vp= (0, 2, 1) ∈ W6 and v = (1, 0, 5/6) ∈ W6 . However, v + v 0 = (1, 2, 1 +
5/6) ∈
/ W6 since (check this!):
p
√
5 · (1)2 − 3 · ( 2)2 + 6 · (1 + 5/6)2 6= 0.
5. Find examples of the following finite sets of vectors in the corresponding vector space:
(a) A collection of mn matrices of dimension m × n which span Mm×n (F).
(b) A collection of n·(n+1)
symmetric n × n square matrices which span the subspace
2
Symn := {A ∈ Mn×n (F) : At = A} of symmetric matrices.
(c) A collection of n·(n−1)
skew-symmetric n × n square matrices which span the
2
subspace Skewn := {A ∈ Mn×n (F) : At = −A} of skew-symmetric matrices.
Solution:
4
(a) Consider the set G = {Eij : 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Observe that:
X
A=
Aij · Eij
i,j
for an arbitrary m × n matrix A, where Aij denotes the i, j component of A.
Thus, indeed, Span G = Mm×n , as required. Moreover, we see immediately that
#G = m · n.
(b) Consider the set G = {Eij + Eji : 1 ≤ i ≤ j ≤ n}. Observe that (Eij )t = Eji
so that (Eij + Eji )t = Eij + Eji and so G indeed contains symmetric matrices.
Moreover, we observe that, for an arbitrary symmetric matrix A, one has:
A=
X Aii
i
2
· (Eii + Eii ) +
X
Aij (Eij + Eji ).
i<j
Thus, Span G = Symn , as required. Moreover, we see that:
#G = 1 + 2 + 3 + · · · + n =
n(n + 1)
.
2
(c) Consider the set G = {Eij − Eji : 1 ≤ i < j ≤ n}. Observe that (Eij − Eji )t =
Eji − Eij = −(Eij − Eji ) so that, indeed, G contains skew-symmetric matrices.
For any skew symmetric matrix A, one has:
X
A=
Aij (Eij − Eji )
i<j
so that Span G = Skewn . Moreover,
#G = 1 + 2 + 3 + · · · + (n − 1) =
n(n − 1)
.
2
6. Let V be a vector space over F and let W1 , W2 be two subspaces of V . Prove or disprove
the following statements:
(a) The set W1 ∩ W2 = {w ∈ V : w ∈ W1 and w ∈ W2 } is a subspace of V .
(b) The set W1 ∪ W2 = {w ∈ V : w ∈ W1 or w ∈ W2 } is a subspace of V .
(c) The set W1 + W2 = {w1 + w2 : wi ∈ Wi } is a subspace of V .
Solution:
(a) This is a subspace. First note that 0 ∈ W1 ∩ W2 as 0 ∈ W1 and 0 ∈ W2 . Suppose
w, w0 ∈ W1 ∩ W2 and a, a0 are scalars. Consider aw + a0 w0 . Since w, w0 ∈ W1 and
W1 is a subspace, we see that aw + a0 w0 ∈ W1 and, since w, w0 ∈ W2 with W2 a
subspace we see that aw + a0 w0 ∈ W2 as well. Thus aw + a0 w0 ∈ W1 ∩ W2 .
5
(b) This is not a subspace. Consider, e.g., V = R2 , let W1 denote the x-axis and
W2 denote the y-axis. Then W1 ∪ W2 is the union of the two coordinate axes, in
particular:
W1 ∪ W2 = {(x, y) : x = 0 or y = 0}
but then (1, 0) ∈ W1 ∪W2 , (0, 1) ∈ W1 ∪W2 while (1, 0)+(0, 1) = (1, 1) ∈
/ W1 ∪W2 .
(c) This is a subspace. First note that 0 = 0 + 0 ∈ W1 + W2 as 0 ∈ W1 ∩ W2 . Now
suppose w, w0 ∈ W1 + W2 are arbitrary. Then w = w1 + w2 and w0 = w10 + w20
for some wi , wi0 ∈ Wi , i = 1, 2. Let a, a0 be arbitrary scalars. Then aw + a0 w0 =
(aw1 +a0 w10 )+(aw2 +a0 w20 ) ∈ W1 +W2 since aw1 +a0 w10 ∈ W1 and aw2 +a0 w20 ∈ W2 .
7. Let V be a vector space over F and let W1 , W2 be two subspaces of V . We say that V
is the direct sum of W1 and W2 , and write V = W1 ⊕ W2 provided that:
• W1 ∩ W2 = {0} is the zero subspace.
• V = W1 + W2 ; i.e. any element of V can be written as a sum w1 + w2 where
wi ∈ Wi .
Prove the following:
(a) Mn×n (R) = Symn ⊕Skewn ; here Symn denotes the subspace of symmetric matrices
and Skewn denotes the subspace of anti-symmetric matrices.
(b) C 0 (R, R) = O⊕E; here C 0 (R, R) denotes the vector space of continuous functions
f : R → R, O denotes the subspace of odd functions (f such that f (−x) = −f (x))
and E denotes the subspace of even functions (f such that f (−x) = f (x)).
Hint: If A ∈ Mn×n (F), consider A + At and A − At .
Solution: In both situations, we need to show that an arbitrary vector v ∈ V can be
written as a sum w1 + w2 where wi ∈ Wi , then show that W1 ∩ W2 = {0}.
1. Let A be an n × n matrix. Observe that:
t
t
1
1
1
1
t
t
t
· (A + A ) = · (A + A ), and
· (A − A ) = − · (A − At ).
2
2
2
2
Thus,
1
2
· (A + At ) is symmetric while
A=
1
2
· (A − At ) is skew-symmetric. Moreover,
1
1
· (A + At ) + · (A − At ) ∈ Symn + Skewn .
2
2
Lastly, we must show that Symn ∩ Skewn = {0}. For this, suppose that A is both
symmetric and skew-symmetric. Namely, At = A = −A so that 2A = 0 but then
A = 0. Thus, we deduce that Mn×n = Symn ⊕ Skewn , by the definition of ⊕.
6
2. This is similar to the proof above as follows. Let f be a continuous function and observe
that the function fe (x) = 12 · (f (x) + f (−x)) is even while fo (x) = 12 · (f (x) − f (−x))
is odd; moreover, f = fe + fo so that, indeed, C 0 (R, R) = O + E. On the other
hand, if a function f ∈ O ∩ E, we see that f (−x) = −f (x) = f (x) so that 2f (x) = 0
which implies that f (x) = 0 for all xl thus O ∩ E = {0}. From this we deduce that
C 0 (R, R) = O ⊕ E as required.
7