QUIZ
MATH 307
Consider the map T : R3 M22 (R) defined by
uv
0
T (u, v, w) =
.
0
uw
Show T is linear, compute and describe N (T ) and R(T ), and verify the dimension
theorem for T .
Solution. Pick x = (x1 , x2 , x3 ) and y = (y1 , y2 , y3 ) R3 . Here
x + y
HOMEWORK 8
MATH 307
(1) Let (V, F, h, i) denote an inner product space over R. Prove the polar
identity;
hx, yi = 14 kx + yk2 14 kx yk2
(2) Let V be a vector space over F, and let (W, F, h, i) denote an inner product
space over F. If T : V W is linear, p
HOMEWORK 9
MATH 307
Let V be an inner product space, and T : V V be a linear operator.
(1) Show R(T ) = N (T ).
(2) If V is finite dimensional, show that R(T ) = N (T ) .
(3) Suppose W V is a T -invariant subspace. If T is self-adjoint, prove the
followin
Section 5.1:
2
3
0
2
3
1 5 ; F = R.
Problem 3-b) A = 4 1 1
2
2
5
Solution:
i) The characteristic polynomial is f (t) = t3 6t2 +11t 6 = (t 3) (t 1) (t 2) :
Therefore the eigenvalues are 1 = 1; 2 = 2; and 3 = 3 (note that since we
have three distinct eigenv
Section 4.2
Problem 25) Prove that if A is an n
det ( A) = n det (A).
Solution. Let B = I: Then det (B) =
det ( A)
n matrix and
n
is a scalar then
: Then we have
= det ( IA)
= det (BA)
= det (B) det (A)
n
=
det (A) :
As a second method you can apply the e
Section 3.3
Problem 8) Let T : R3 ! R3 be dened by T (a; b; c) = (a + b; b
Is the vector v = (1; 3; 2) in R (T )?
2c; a + 2c) :
Solution. The vector v will be in the range of T (i.e., v will be an output)
if there exists a vector (x1 ; x2 ; x3 ) 2 R3 such
Section 6.1:
Problem 8)
(a) Let a = c = 0; and b = d = 1: Then h(1; 0) ; (1; 0)i = 1 which is not
positive. (Note that we should have hx; xi > 0 )
(b) Let A = I2 : Then hA; Ai = h I2 ; I2 i = tr ( I2 I2 ) = tr ( 2I2 ) =
4 which not positive.
R1
(c) Let f
Section 3.2
Problem 6-b) Suppose T : P2 (R) ! P2 (R) is dened by T (f (x) =
(x + 1) f 0 (x) : Determine whether T is invertible, and compute T 1 if it exists.
Solution. Let 3= 1; x; x2 be the standard ordered basis for P2 (R) : Then
2
0 1 0
[T ] = 4 0 1 2
Section 2.1
Problem 17-a: We use the "contradiction method". In other words, we
assume that T is onto and then we will arrive to a nonvalid statement. If T is
onto, then rank (T ) = dim (R (T ) = dim (W ) : Then by the dimension theorem
we have:
nullity (
Section 1.6
Problem 5) Is f(1; 4; 6) ; (1; 5; 8) ; (2; 1; 1) ; (0; 1; 0)g a linearly independent
subset of R3 ?
Solution:
No, because there are more vectors in the set than the dimension of R3 :
(Note that by theorem, any linearly independent subset of R3
Section 2.1
Related to Problem 5) For T : P2 (R) ! P3 (R) dened by T (f (x) =
x2 f 0 (x) + f (x), show that T is a linear map, and nd bases for N (T ) and
R (T ) :
Solution:
T is linear:
We need to show that T (f (x) + g (x) = T (f (x) + T (g (x) for ever
Section 1.4
Problem 4-d): The rst vector could be expressed as a linear combination
of the other two vectors if we could nd scalars a and b such that:
x3 + x2 + 2x + 13
= a 2x3 3x2 + 4x + 1 + b x3 x2 + 2x + 3
or
x3 + x2 + 2x + 13 = (2a + b) x3 + ( 3a b) x
Section 2.3
Problem 9) We can let U (a; b) = (b; 0) and T (a; b) = (a; 0) : Then
U (T (a; b)
= U (a; 0)
= (0; 0)
But
T (U (a; b) = T (b; 0)
= (b; 0)
6= (0; 0) (if b 6= 0)
If we let
be the standard basis for F 2 and we let A = [U ] =
0
0
1
0
1 0
; then you
HOMEWORK 7
MATH 307
(1) Equip CR [0, 1] with the inner product from lectures. Let t(t) = t and
g(t) = et . Compute hf, gi, kf k, kgk and kf + gk. Then verify both the
Cauchy-Schwarz inequality and the triangle inequality.
R1
This is routine. For example,
HW5
MATH 307
(1) Compute your expected grade in the class, taking the midterm into account.
(2) Let T1 : M22 (R) M22 (R) be the projection onto W1 along W2 . (For
the definition of this, the previous homework). Let T2 : M22 (R)
M22 (R) be the projection
HOMEWORK 7
MATH 307
(1) Equip CR [0, 1] with the inner product from lectures. Let t(t) = t and
g(t) = et . Compute hf, gi, kf k, kgk and kf + gk. Then verify both the
Cauchy-Schwarz inequality and the triangle inequality.
(2) Let hv1 , . . . , vk i be an
MATH 307 HW1 SOLUTIONS
DR. T. MURPHY
(1) (a) No. If (6, 4, 2) = t (3, 1, 2) for some t 6= 0, then from the first two
coordinates, we would need 6 = 3t, or t = 2, as well as 4 = 1t, or
t = 4. As 2 6= 4, this is impossible.
(b) Yes, (9, 3, 21) = 3 (3, 1, 7)
HOMEWORK 4
MATH 307
(1) Study and understand the proof of Theorem 1.11 in the text.
(2) In M22 (Q), define the following two subspaces:
W1 = N (tr), W2 = span(cfw_I2 ),
where tr : M22 (Q) Q denotes the trace function, so that
tr(A) = a11 + a22 ,
the sum o
Homework 5 Solutions
Math 307: Linear Algebra
April 10, 2016
1. Check again the posted grades.
2. From the definition of T2 in the last homework, we have
a b
(a + d)/2
0
T2
=
.
c d
0
(a + d)/2
Hence if we apply T2 to the basis vectors in , we find
1
1
HW6
MATH 307
(1) In the following problems, determine which of the indicated column vector
are eigenvectors of the matrix A. Give the corresponding eigenvalue.
(a)
4 2
5
2
2
A=
K1 =
K2 =
K3 =
.
5 1
2
5
5
The answer is K3 , with eigenvalue 1.
(b)
1 2 2
A =
HOMEWORK 3 SOLUTIONS.
MATH 307
(1) Study Theorem 1.5 in the textbook, and use it to prove that if S1 and S2 are
subsets of V with S1 S2 , then Span(S1 ) Span(S2 ). In particular, note
that if we also know Span(S1 ) = V , then we can deduce that Span(S2 )
HOMEWORK 3
MATH 307
Throughout, let V be a vector space.
(1) Study Theorem 1.5 in the textbook, and use it to prove that if S1 and S2 are
subsets of V with S1 S2 , then Span(S1 ) Span(S2 ). In particular, note
that if we also know Span(S1 ) = V , then we
HOMEWORK 4 ROUGH SOLUTIONS
MATH 307
These are rough solutions, and do not constitute full answers.
(1) Study and understand the proof of Theorem 1.11 in the text. Up to you.
(2) Here we prove that M22 is
W1 and W2 are subspaces (I
a b
=
c d
a sum of W1
MATH 307 HW2
DR. T. MURPHY
(1) Show that if U is a subspace of V , and V is a subspace of W , that U is a
subspace of W .
(2) Let U1 and U2 be two subspaces of a vector space V . Is U1 U2 a subspace?
If yes, prove it. If not, explain why not.
(3) Show tha
HOMEWORK 8
MATH 307
(1) Let (V, h, i) denote an inner product space over R. Prove the polar identity;
hx, yi = 41 kx + yk2 14 kx yk2
This is straightforward:
kx + yk2 = hx + y, x + yi
= kxk2 + kyk2 + 2hx, yi
Similarly
kx yk2 = hx + y, x + yi
= kxk2 + kyk
MATH 307 HW2
DR. T. MURPHY
(1) We have to show U is a subspace of W . We apply the subspace theorem.
Firstly, it is clear 0W , the zero element of W , is in U . This is because
OW = OV (since V is a subspace of W , by a theorem from class) and
similarly O
MATH 307 HW1
DR. T. MURPHY
(1) Determine whether the vectors emenating from the origin and terminating
at the following pairs of points are parallel.
(a) (3, 1, 2) and (6, 4, 2).
(b) (3, 1, 7) and (9, 3, 21).
(c) (5, 6, 7) and (5, 6, 7).
(d) (2, 0, 5) and
HW6
MATH 307
(1) In the following problems, determine which of the indicated column vector
are eigenvectors of the matrix A. Give the corresponding eigenvalue.
(a)
4 2
5
2
2
A=
K1 =
K2 =
K3 =
.
5 1
2
5
5
(b)
1 2 2
A = 2 1 2
2
2
1
4
1
0
K1 = 1 K2 = 4 K3 =
Section 2.1
Problem 10) Suppose T : R2 ! R2 is a linear and T (1; 0) = (1; 4) and
T (1; 1) = (2; 5) : Find T (a; b) for every a; b 2 R and then nd T (2; 3) : Is T is
1-1?
Solution) Note that (a; b) = (a
T (a; b)
=
=
=
=
=
b) (1; 0) + b (1; 1) : Therefore
January 31, 2008
Section 1.2
Problem 15: V is not a vector space over the eld of complex numbers. Because, for example,
(1; 0; 0; :; 0) 2 V but i (1; 0; 0; :; 0) = (i; 0; 0; :; 0) does not belong to V anymore.
Problem 18: V is not a vector space because i