Paul's Homework 1

Comments:

1.1 Question 2: Good! If you wrote only one point I would have been mucho upset. I am pretty sure you got them all.

1.1 Question 9: Also good: you noticed that it was not a complete circle but it had 1/4 of a circle for positive s and t.

1.1 Question 11: Ok, so you are right, but I want you to write out the expansion of ||x+y||^2 (i.e ||x+y||^2 = (x+y)*(x+y) = x*x + 2x*y+y*y = ||x||^2+2x*y+||y||^2 where * denotes dot product.

1.2 Question 8: Dude, I gave you this one? Beautiful problem. Especially after number theory. But dude. Dude. SO CLOSE. You set it up perfectly but the sum of the first n squares is n * (n+1)* (2N+1)/6. (Your final answer is good though, but the miracle of limits).

1.2 Question 1.3: a.(duh), b.good, c. (duh), good, d. correct, but simplier argument is..when does the sum of square components equal 0? When they are all zero. e. Good, you used the reasoning I gave in d to contradict f. g. good. h. BAD (please don't get offended like the kid), just because the vectors are linearly depended does not mean it does not SATISFY the properties of a subspace. YOU MUST check the properties directly hold!

1.2 Question 4: Good. This question is a standard one and I will guarentee it will be on a future exam.

3.1 Question 1: Yawn, computation. But yeah, you get the point.

I cannot find the book questions for 3, 8, 6, and 22: you write at the top Ch. 4 Lin alg...so I am not sure.

There are two GLARINGLY BAD statements though (sorry, please don't get offended), inverting a nx1 matrix does not make sense (you can talk about pseudoinverses, but that is not only not guarenteed to exist, but not useful in the context you are using them, which is to remove x from Ax=0).

Secondly, incredibly incredibly bad statement: MATRICES NEED NOT COMMUTE. (A+B)*(A-B) = A^2 - AB + BA -B^2. But AB need not equal BA.

Also, you are correct that A^2 = A can only be A = 0 or A = I. But your proof is bad in that an inverse need not exist. The correct proof would be to rely on distributivity and substraction: A^2 = A iff A^2 - A = 0 iff A(A-I) = 0 iff A = 0 or A = I.
 * UPDATE****

The above statement is not true (sorry!). Consider the counterexample A= [0 0'; 0 1]. A^2 = A but A is neither the identity or the zero matrix.

NOTE: My apologies for the crumpledness of the homework: Paul handed his work in mint condition, but they got crumpled up while I was moving out of the dorm. Sorry-Philip