Math 2135 - Linear Algebra
Homework #2 Solutions
1. Find a basis for and the dimension of the subspaces dened by the following conditions:
(a) (x1 , x2 , x3 , x4 ) R4 such that
x1 + x4 = 0 ,
3x1 + x2 + x4 = 0 .
We solve the equations. First we have x1 = x
Math 2135 - Linear Algebra
Practise Mid-Term
1. For each of the following subsets W of a vector space V , determine if W is a subspace of V .
Say why or why not in each case:
(a) V = R3 and W = cfw_(a1 , a2 , a3 )|a1 3a2 + 4a3 = 0 , and a1 = a2
Let a = (
CMNS 2850 First Full Draft (25%)
Due November 13th, (Week 10) (2015 4pm)
The purpose of this assignment is to produce a FULL DRAFT of your project.
The completed elements of the project will include:
Completed Sections: Introduction, Background/Literature
Math 2135 - Linear Algebra
Practise Final Solutions
Note: In addition the exam will have a short answer section for the rst question.
1. For each of the following subsets W of a vector space V , determine if W is a subspace. Justify
your answer.
(a) V = R
Math 2135 - Linear Algebra
Homework #1 Solutions
1. Find the inverse of the matrix
0 1 1
M = 1 1 0
0 0 1
using Z2 as the set of scalars. Hint: follow the usual steps of Gaussian elimination, but use
modulo 2 operatios. Compare this to the inverse of M wh
Math 2135
Mid Term Test
Tues Feb 25
Answer the questions in the space provided on the question sheet. If you
run out of room for an answer, continue on the back of the page.
Make sure to note on the page that the answer is continued on the back.
g
Calc
Math 2135 - Linear Algebra
Homework #3 Due Feb. 20th
1. Show that equality holds in the Cauchy-Schwarz inequality if and only if u and v are linearly
dependent.
First we assume that u and v are linearly dependant. So u = cv for some scalar c. Then,
< u, v
Math 2135 - Linear Algebra
Homework # 4 Solutions
1. Find the determinant of the following matrices.
(a)
1 0 2
0 2 0
1 0 3
Using the formula, we get
1(6) 0(0) + 2(2) = 2
(b)
1 1 2
1 0 1
1 1 0
Again we get
1(1) 1(1) + 2(1) = 2
2. What is the determinan
Math 2135 - Linear Algebra
Homework #5 Due April 3
1. A linear transformation T : V V is called skew-adjoint or skew-Hermitian if T = T
(Recall T = T T the conjugate transpose). Prove
(a) The eigenvalues of a skew-adjoint mapping are purely imaginary.
Le
Math 2135 - Linear Algebra
Practise Final
Note: In addition the exam will have a short answer section for the rst question.
1. For each of the following subsets W of a vector space V , determine if W is a subspace. Justify
your answer.
(a) V = R3 and W =
Math 2135 - Linear Algebra
Homework #5 Due April 3
1. A linear transformation T : V V is called skew-adjoint or skew-Hermitian if T = T
(Recall T = T T the conjugate transpose). Prove
(a) The eigenvalues of a skew-adjoint mapping are purely imaginary.
(b
> restart;
> # First I will construct a set of orthonormal polynomials on [0,
L]
> # I start with cfw_1,x,x^2,x^3 as my basis and apply Gram-Shmidt
> L:=2*Pi;
(1)
> v1:=1/sqrt(int(1^2,x=0.L);
(2)
> v2:=x-int(x*v1,x=0.L)*v1;
(3)
> v2:=v2/sqrt(int(v2^2,x=0.
General Journal
Date
1/2/2015
101 Bank
GJ_1
F
Dr
J1
22000.00
123 Furniture & Equipment
Cr
8000.00
301 Capital Henri Matin
30000.00
Initial Investment
1/2/2015
101 Bank
J2
125 Vehicles
15000.00
15000.00
303 Capital Wes Corbett
30000.00
Initial Investment
1