This WeekÕs Homework: (On Tuesday I cut this assignment down, so you you want to look at it agian.) Make sure you can do 2.1:1,2,3,7,10,20,21,24,25 and 2.2 : 1,2(c,e,f,g),5,11

Turn in 2.1:4,5,6,9,11,12,14,15,16,17,18,19,22,23 and 2.2: 2(a,b,d),3,4,7,8,15

Proposed Topics For this Week: We will begin chapter 2 and our study of linear maps. In particular we will explore the notions of rank and nullity and the representation of a linear map between finite dimensional spaces by a matrix. The X-session Topic We will continue our geometric excursion and look at several interesting mappings in th presence of a bilinear form. While no homework will be assigned out of this x-session several of Monday's exam problems will involve exploring these issues. Also, depending on the pace, it is possible that during the x-session we will be discussing chapter two and that this material will show up in Friday's lecture instead.

Homework Number 3 Many people had a diffeicult time with 1.6 number 21. This problem is not easy to do with out using some of the theorems form section 1.6, and many people tried to do it from scratch and ran into problems. In one form or another the worst and most common problem was to assume that a vector space has a unique basis.

In the problem W₁ and W₂ are subspaces of W with $dim(W_1) =m \geq n =dim(W_2)$.

Show $dim(W_1 \bigcap W_2) \leq n$.

Proof: $W_1 \bigcap W_2$ is subset of W₂ and in fact from theorem 1.3 we have that that $W_1 \bigcap W_2$ is subcpace of W₂. So by corollary 2 (page 45) $dim(W_1 \bigcap W_2) \leq dim(W_2) = n$.

done

Show $dim(W_1 + W_2) \leq n + m$.

proof: Well we know frpm problem 23 page 21, that W₁ + W₂ is subspace. Letting $\beta_1$ and $\beta_2$ be bases of W₁ and W₂ respectively, by the defintion of + we know that $\beta_1 \bigcup \beta_2$ gerates W₁ + W₂. By the defintion of dimension $\beta_1$ contains m elements and $\beta_2$ contians n elements, so $\beta_1 \bigcup \beta_2$ contains fewer than or exacly to n +m elements. So by the replacemnt theorem any basis (in fact any linearly independent set) of W₁ + W₂ must contain fewer than n+m elemtns. Hence by the defintion of dimension $dim(W_1 + W_2) \leq n + m$. done

Some Maple Code From Lecture:

Maple Code.

Homewor number 2 homework. The Grader informed me that number 6 in 1.5 did not go well, so I thought I'd wirite a solution. I recomend looking at your solution and asking yourself "where should I have been more carefull?" so that any problem you suffered will not occur again.

The problem was: Let u and v be distinct vectors in a vector space V. Show that {u,v} is a linearly dependent if and only if u or v is a multiple of the other.

(First this is an if and only if statement so we know we have to prove two things indicated by the arrows below.)

$\rightarrow$(the "only if" case)

We'll suppose {u,v} is linearly dependent and prove u = cv or v=cu with $c \in F$. Well by the definition of linearly dependence we know

au+ bv = 0

with either $a \neq 0$ or $b \neq 0$ both elements of F.

By subtracting bv form the left and right hand sides of the above equation we see that

au = -bv.

In the case where $a \neq 0$ we may divide by a and arrive at

$$u = \frac{-b}{a}v.$$

(notice we are not done yet...)

In the case where $b \neq 0$ we may devide by -b and arrive at

$$v = \frac{-a}{b} u.$$

So in either of these cases v is a multiple of u or u is a multiple of v as needed.

$\leftarrow$ (the "if" case)

Let us now assume either v is a multiple of u or u is a multiple of v, and prove linear dependence.

If v = cu with $c \in F$ then by subtracting cu from both sides we have v-cu =0. Since 1 is the coeeficient of v and $1 \neq 0$, we may conclude that by the very defintion of linear dependence {u,v} is linearly dependent.

(Notice we are not done yet...)

Similarly, if u = cv then by subtracting cv from both sides we have u-cv =0, and once again $1 \neq 0$ and {u,v} is linearly dependent.

So ineither case we have that {u,v} is linearly dependent as need.

Done