06-240/Final Exam Preparation Forum

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If you have questions, ask them here and hopefully someone else will know the answer. (Answering questions will probably help you understand it more).

Since many of us (including I) don't really know how to use Wiki's, I suggest that we keep the formatting simple: I will post a template at the top of this page, and if you want to add something just click on the "edit", copy the template, and insert your question. Order the questions according to section (solved/unsolved, as judge by whoever created the question), with the newest at the top, except for the template question. In general, I wouldn't retype the question if it's from the book because that's tedious and we all have the book.

(By the way, I think you leave a space between lines in the code to make a new line; that is, simply pressing enter once will not make a new line. Also, you can press a button at the top of the editing textbox that lets you put in simple equations.)

Contents

Unsolved Questions

Question Template

Q: Can someone help me prove: "If an integer n is greater than 2, then a^n + b^n = c^n has no solutions in non-zero integers a, b, and c."? I had the answer in my head at one point, but the margins of the piece of paper I was working with was too small to fit it.

Exam April/May 2006 #3(b)

Q: Let T : M3x2(C) -> M2x3(C) be defined as follows. Given A Є M3x2(C), let B be the matrix obtained from A by adding i times the second row of A to the third row of A. Let T(A) = Bt, where Bt is the transpose of B. (Note: Here, i is a complex number such that i2 = -1.) Determine whether the linear transformation T is invertible.

Totally lost on this question :/ Please show some example matrix and how it is transformed as the question asks if possible. I want to see what actually happens to the elements in the matrix rather than the answer (think that would be more important)

A(Matrix Elements): This is my interpretation:

A = \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{pmatrix} where a_{ij} \in C, then B = \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\ia_{21}+a_{31}&ia_{22}+a_{32}\end{pmatrix}.

Therefore, T(A) = Bt is T\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\\a_{31}&a_{32}\end{pmatrix}=\begin{pmatrix}a_{11}&a_{21}&ia_{21}+a_{31}\\ia_{12}&a_{22}&ia_{22}+a_{32}\end{pmatrix}

Exam April/May 2006 #7

Q: Let T : V -> V and U : V -> V be linear operators on a finite-dimensional vector space V. Assume that U is invertible and T is diagonalizable. Prove that the linear operator UTU-1 = U o T o U-1.

I dont know where or how to start this question ><.

Sec. 2.4 Lemma p. 101

Q: In the proof of the lemma, the second line, we have "T(beta) spans R(T) = W". How do we know that R(T) = W? This would be true if dim V = dim W, because then T would be onto, but we can't assume what we're trying to prove.

Exam April/May 2006 #4

Q: Suppose that A, B Є Mmxn(F), and rank(A) = rank (B). Prove that there exist invertible matrices P Є Mmxm(F) and Q Є Mnxn(F) such that B = PAQ.

A(partial): Here is a sketch. If you rref A and B by applying a series of elementary row operation matricies, they will both look similar. That is, they will have a section of 1's and 0's (each 1 is the only number in its column) and then a section of "remaining stuff", and these sections will be the same "size" because their ranks are the same. Then, using the elementary column matrix operations, you can essentially modify the "remaining stuff" as much as you like, by adding multiples of the "nice" columns (with single-1's). These row and column operations can then be grouped nicely and set to be equal to P and Q, which are invertible because products of elementary matricies are invertible.

I know this is very rough, but even if I did have a full answer I wouldn't now how to typeset it.

Complex Numbers

Q: If 'C' is used in the context of a vector space (as in "define T:C->C"), then should we consider C to be the vector space of C over the field C, or instead C over the field R?

Solved Questions

Question Template

Q: How many ways are there to get to the nth stair, if at each step you can move either one or two squares up?

A: This question can be easily modeled by the Fibonacci numbers, with the nth number being the ways to get to the nth stair. This is because, to get to the nth stair, you can come only from the n-1th or the n-2th. This is exactly how the Fibonacci numbers are defined; the proof is simple by induction.

Sec. 1.3 Thm 1.3 Proof

Q: In the first paragraph of the proof, it says "But also x + 0 = x , and thus 0'=0." How do we know 0 (that is 0 of V) even exists in W? I understand that we know some zero exists (0'), but not why the zero (0) exists.

A: x is in W as well in V. Thus, x + 0 = x (VS 3).

Reply: Oh I see... now it looks so obvious =/. Thanks.