SOLUTION TO ASSIGNMENT 1 - MATH 251, WINTER 2007
1. Consider the following vector spaces (you do not need to prove those are vector spaces, unless specified): (1) The vector space V1 of continuous fu
McGill University
MATH 251: Algebra 2
Assignment 5 Solutions
1. (a) Since a(u, 0) + b(v, 0) = (au + bv, 0) and a(0, u) + b(0, v) = (0, au + bv) we see that U1 , V1 are subspaces
since (0, 0) U1 , V1 .
McGill University
MATH 251: Algebra 2
Assignment 6 Solutions
1. (a) Since A2 = nA, we have A(A nI) = 0 so that the possible eigenvalues of A are 0, n.
If the characteristic of F does not divide n, the
Least Squares and the Generalized Inverse
An important problem is to nd a polynomial curve y = f (x) which best ts a given set of m
data points (x1 , y1 ), (x2 , y2 ), . . . , (xm , ym ). If f (x) = a
The Real Jordan Canonical Form and the Rational Canonical Form
Not all matrices over a given eld have a Jordan canonical form as not all polynomials split
completely into linear factors. For example,
McGill University
MATH 251: Algebra 2
Assignment 4 Solutions
1. The function y = f (xf is a solution of the given dierential equation f Ker(D4 6D3 + 13D2 12D + 4),
where D is the dierentiation operato
McGill University
MATH 251: Algebra 2
Assignment 3 Solutions
1. (a) If aex + be2x + ce3x = 0 for all x then on dierentiating twice we get
aex + be2x + ce3x = 0
aex + 2be2x + 3ce3x = 0
aex + 4be2x + 9c
Notes on the Dual Space
Let V be a vector space over a eld F . The dual space of V is the vector space V = Lin(V, F ) consisting
of the linear mappings : V F . The elements of V are called linear form
McGill University
MATH 251: Algebra 2
Solutions to Assignment 8
1. Since P 2 P = 0 we have V = Ker(P2 P) = Ker(P(P1) = Ker(P)Ker(P1) = Ker(P)Ker(Q).
Now v Ker(Q) = v = P(v) = v Im(P) and v Im(P) = v =
MATH 251, Winter 2009
Honours Algebra 2
Solutions to the Midterm Exam
1. Let V = R5 , a real vector space, and W1 , W2 be
of V .
80 1 0
10
3
>1
>
>B C B
>2
<B C B 7 C B
CB
W1 = Span B 0 C , B 2 C , B
SOLUTION TO ASSIGNMENT 2 - MATH 251, WINTER 2007
1. Let B = {(1, 1), (1, 5)} and C = {(2, 1), (1, -1)} be bases of R2 . Find the change of basis matrices B MC 8 and C MB between the bases B and C. Le
SOLUTION TO ASSIGNMENT 3 - MATH 251, WINTER 2007
1. Let T : V V be a nilpotent linear operator. Prove that if n = dim(V ) then T n 0. Show that for every n 2 there exists a vector space V of dimen
SOLUTION TO ASSIGNMENT 4 - MATH 251, WINTER 2007
1. Deduce from the theorems on determinants the following: (1) If a column is zero, the determinant is zero. (2) det(A) = det(At ), where At is the tr
SOLUTION TO ASSIGNMENT 5 - MATH 251, WINTER 2007
1. (A) Let W be a k-dimensional subspace of Fn . Prove that there are n - k linear equations such that W is the solutions to that homogenous system. (
SOLUTION TO ASSIGNMENT 10 - MATH 251, WINTER 2007
1. Let A be a matrix in block form: A1 0 A= 0 Prove that A = A 1 A 2 A r , and mA = lcm{mA1 , mA2 , , mAr }. You may use the formula Ab 1 0
SOLUTION TO ASSIGNMENT 11 - MATH 251, WINTER 2007
1. It is known that a differentiable function f : R2 R has a maximum at a point P if f /x = f /y = 0 at P and the 2 2 symmetric matrix -
2f x2 2f
SOLUTION TO ASSIGNMENT 9 - MATH 251, WINTER 2007
1. Calculate the characteristic and minimal polynomial of the following matrices with real entries. In each case determine the algebraic and geometric
SOLUTION TO ASSIGNMENT 8 - MATH 251, WINTER 2007
In this assignment F = R or C. 1. Consider results of an experiment given by a series of points: (x1 , y1 ), (x2 , y2 ), . . . , (xn , yn ), where x1
SOLUTION TO ASSIGNMENT 6 - MATH 251, WINTER 2007
1. Let T : V W be a linear map and define T : W V by (T (g)(v) := g(T v). Prove the following lemma: LEMMA 1. (1) T is a well-defined linear ma
189-251B: Honors Algebra 2
Midterm Exam: Corrections and
comments
The fol lowing was the grade dsitribution in this midterm:
>= 100 5 95 99 5 90 94 13
80 89 13 70 79 6 60 69 7
50 59 1
So overal l the
McGill University
MATH 251: Algebra 2
Assignment 1 Solutions
1. If V is a vector space and f is an isomorphism then f (u + v) = f (u) + f (v), f (cv) = cf (v). Setting
u = f (u), v = f (v), we see tha