Linear Algebra Proofs and Problems

[Dustinsfl]

We used to have a bunch of problems and proofs that were in a pdf could be downloaded by anyone. Since we aren't able to upload pdf files of a certain size, I provided a link to google docs. If there is an error, typo, or something is just drastic wrong let me know.

Undgraduate Final Review Practice problems with solutions
Theorems/Proofs Undergraduate level

However, with this first link, I can't edit this document. It was created with Maple which I no longer have. So errors have to just be corrected in the thread and then consolidated for readability.

This pdf has more advanced proofs in it.

Linear Alg Workbook

I have completed the second set. The only ones that need solutions are $A5$ part 2, $B7$ part2, $C4$ part 2 and 3, $C5$, $D4$, $F7$ needs to be checked, $H3$, $H10$, $I4$ part 3, $I5$, $I10$ part 2, $J7$, and $J10$.
The rest of the problems I believe to be right but they should still be checked out.

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-linear-algebra-proofs-problems-4230.html

 

[Dustinsfl]

For the first document, here is what Ackbeet$\equiv $Ackbach suggested on MHF that needed to be adjusted

Very nice review! I just had a few comments:


1. From Test 5, Problem 4, on page 4. I would say more than eigenvectors must be nonzero, by definition. It's not that the zero eigenvector case is trivial: it's that it's not allowed.


2. Page 6, Problem 8: typo in problem statement. Change "I of -I" to "I or -I".


3. Page 8, Problem 21: the answer is correct, but the reasoning is incorrect. It is not true that $\mathbf{x}$ and $\mathbf{y}$ are linearly independent if and only if $|\mathbf{x}^{T}\mathbf{y}|=0.$ That is the condition for orthogonality, which is a stronger condition than linear independence. Counterexample: $\mathbf{x}=(\sqrt{2}/2)(1,1),$ and $\mathbf{y}=(1,0).$ Both are unit vectors, as stipulated. We have that $|\mathbf{x}^{T}\mathbf{y}|=\sqrt{2}/2\not=0,$ and yet
$a\mathbf{x}+b\mathbf{y}=\mathbf{0}$ requires$a=b=0,$ which implies linear independence.


Instead, the argument should just produce a simple counterexample, such as $\mathbf{x}=\mathbf{y}=(1,0)$.


Good work, though!