Page 1 of 2 12 LastLast
Results 1 to 15 of 24

Math Help - Lin Alg Proofs and Counterexamples

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Lin Alg Proofs and Counterexamples

    I have compiled 57 prove or disprove lin alg questions for my final; however, these may be useful to all.

    Contributors to some of the solutions are HallsofIvy, Failure, Tikoloshe, jakncoke, tonio, and Defunkt.

    If you discover any errors in one of the solutions, then feel free to reply with the number and correction.
    Attached Files Attached Files
    Last edited by dwsmith; May 6th 2010 at 04:06 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Test 2, Q1:
    You are supposed to prove here that det(AB) = det(BA) for any matrices A,B. Why, then, do you assume that B is singular?


    Test 2, Q2:
    You are meant to give a counter example here, like you did in Q3.


    Test 2, Q4:
    Your proof is incorrect. You are saying, essentially, that if A,B are row equivalent then A=E_1E_2...E_k where E_1E_2...E_k=B but then A=E_1E_2...E_k=B \Rightarrow A=B and clearly not any two row equivalent matrices are equal. The correct formulation is that A=E_1E_2...E_kB with E_i elementary matrices.
    You should also note that this proposition is not true: Take A=I_2 and do the row operation R_1 \leftrightarrow R_2 to get B, then: det(A) = 1, det(B) = -1 \neq det(A)


    Test 2, Q5:
    This is also incorrect. I don't see how \mathbb{R}^2 supposedly not having a third coordinate has anything to do with closure under addition. You should note that in \mathbb{R}^3, \mathbb{R}^2 would be \{(x,y,z) \in \mathbb{R}^3 : z = 0\}. It should now be obvious that \mathbb{R}^2 is a subspace of \mathbb{R}^3 (and this makes much more sense, does it not?)


    Test 4, Q3:
    As already mentioned, if you think a proposition is false you need to provide a counter example, not just say that it may not always be true.


    Test 4, Q5:
    0 is not a positive number.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Test 2, Q1:
    You are supposed to prove here that det(AB) = det(BA) for any matrices A,B. Why, then, do you assume that B is singular?
    By doing so, we can show that det(AB)=det(BA).

    det(A)det(B)=det(A)0=0
    det(B)det(A)=0det(A)=0
    det(AB)=det(BA)
    Last edited by dwsmith; May 5th 2010 at 08:19 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Test 4, Q3:
    As already mentioned, if you think a proposition is false you need to provide a counter example, not just say that it may not always be true.
    This answer comes directly from the def of an orthonormal matrices.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    But you're still only showing it for a singular B... What if both A and B are non-singular? Your proof doesn't cover that case. det(AB) doesn't always equal 0.

    This answer comes directly from the def of an orthonormal matrices.
    Yes, but I believe the purpose of those questions is giving a counter-example. I don't think you would get a full grade for only explaining why it is not always true.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Yes, but I believe the purpose of those questions is giving a counter-example. I don't think you would get a full grade for only explaining why it is not always true.

    Giving a counterexample was not a requirement for this file I created. If it is easier to use a counterexample, I did. If it is easy to use a definition to show it doesn't work, I did that too.

    This is just something I created. It doesn't necessarily mean a counterexample is mandatory.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Test 2, Q5:
    This is also incorrect. I don't see how \mathbb{R}^2 supposedly not having a third coordinate has anything to do with closure under addition. You should note that in \mathbb{R}^3, \mathbb{R}^2 would be \{(x,y,z) \in \mathbb{R}^3 : z = 0\}. It should now be obvious that \mathbb{R}^2 is a subspace of \mathbb{R}^3 (and this makes much more sense, does it not?)
    It probably could be worded better but check out this link.
    Why R2 is not a subspace of R3? [Archive] - Physics Forums
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Test 4, Q5:
    0 is not a positive number.
    We discussed this in class and decided to disregard the semantic issue about 0 to instead answer the question.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    But you're still only showing it for a singular B... What if both A and B are non-singular? Your proof doesn't cover that case. det(AB) doesn't always equal 0.
    If you provide the proof you deem necessary, I will update the file.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Test 2, Q1:
    Test 2, Q4:
    Your proof is incorrect. You are saying, essentially, that if A,B are row equivalent then A=E_1E_2...E_k where E_1E_2...E_k=B but then A=E_1E_2...E_k=B \Rightarrow A=B and clearly not any two row equivalent matrices are equal. The correct formulation is that A=E_1E_2...E_kB with E_i elementary matrices.
    You should also note that this proposition is not true: Take A=I_2 and do the row operation R_1 \leftrightarrow R_2 to get B, then: det(A) = 1, det(B) = -1 \neq det(A)
    You are correct here and I will update.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by dwsmith View Post
    If you provide the proof you deem necessary, I will update the file.
    Of course, det(AB)=det(A)det(B) is only correct when A,B are square matrices. Otherwise, this is not true. I'll leave it to you to find a counter example in this case.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    The row equivalent question has been change and new file uploaded.

    Test 2 Q4
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by Defunkt View Post
    Of course, det(AB)=det(A)det(B) is only correct when A,B are square matrices. Otherwise, this is not true. I'll leave it to you to find a counter example in this case.
    This probably important, and not mentioned, assume all matrices are nxn for test 2.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    The file has been update again for test 2 with the relevant info.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by dwsmith View Post
    The file has been update again for test 2 with the relevant info.
    Once a final draft has been established I will stick this thread as a useful resource.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Lin Alg Proofs and Counterexamples
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: December 15th 2011, 06:58 PM
  2. Proofs/ counterexamples
    Posted in the Geometry Forum
    Replies: 3
    Last Post: October 1st 2011, 11:29 AM
  3. Proofs and Counterexamples
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: August 1st 2011, 04:53 AM
  4. Ask for two counterexamples regarding convergence in L^p
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: June 19th 2010, 08:37 AM
  5. convergence: proofs and counterexamples
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 1st 2007, 03:25 PM

Search Tags


/mathhelpforum @mathhelpforum