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anonymous

  • 4 years ago

Can someonehelp me prove the left cancellation property with matrices?

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  1. Akshay_Budhkar
    • 4 years ago
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    Not me.. sorry

  2. anonymous
    • 4 years ago
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    LOL I know that by now

  3. Akshay_Budhkar
    • 4 years ago
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    oh give me a week i will learn matrices before you! :P

  4. anonymous
    • 4 years ago
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    Sound like hero! r u guys related

  5. anonymous
    • 4 years ago
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    I can show u how they proved right cancellation Give me a sec

  6. LollyLau
    • 4 years ago
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    not something worth lolling much.

  7. Akshay_Budhkar
    • 4 years ago
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    yea if u show me maybe i can explain

  8. anonymous
    • 4 years ago
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  9. Akshay_Budhkar
    • 4 years ago
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    that seems so simple!

  10. Akshay_Budhkar
    • 4 years ago
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    which step do u dont get?

  11. anonymous
    • 4 years ago
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    Ya but i dont get what they did

  12. anonymous
    • 4 years ago
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    This is right cancellation how do i change it for left cancellation?

  13. anonymous
    • 4 years ago
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    What they are doing is saying that if C is invertible, then by definition a matrix C^-1 must exist. Thus, they multiply both sides on the right by C^-1, then use the associative property and kill it. The proof for left is basically the same.

  14. anonymous
    • 4 years ago
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    If C is invertible, then \[CA=CB \implies C^{-1}(CA)=C^{-1}(CB)\] \[\implies (C^{-1}C)A=(C^{-1}C)B \implies IA=IB\] \[\implies A=B\] QED :)

  15. anonymous
    • 4 years ago
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    LOL Its that simple???????

  16. Akshay_Budhkar
    • 4 years ago
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    LOL

  17. anonymous
    • 4 years ago
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    Thanks imperialist :P

  18. anonymous
    • 4 years ago
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    No problem :)

  19. Akshay_Budhkar
    • 4 years ago
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    imperialist is cool hehe

  20. anonymous
    • 4 years ago
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    hehe

  21. anonymous
    • 4 years ago
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    Haha, thank you Akshay. Just trying to combine two of my great loves, teaching and math. :D

  22. Akshay_Budhkar
    • 4 years ago
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    i like the way you tackle tough problems :D won't be surprised to see u a legend soon. Yea this is the best place to do that!

  23. Akshay_Budhkar
    • 4 years ago
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    rld613 is typing a reply…

  24. anonymous
    • 4 years ago
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    Or not

  25. Akshay_Budhkar
    • 4 years ago
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    Lol

  26. anonymous
    • 4 years ago
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    LOL Called it first to be ur student

  27. anonymous
    • 4 years ago
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    Sorry lost connection for a second

  28. Akshay_Budhkar
    • 4 years ago
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    1 second o_O ?

  29. Akshay_Budhkar
    • 4 years ago
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    now imperialist is typing a reply…

  30. Akshay_Budhkar
    • 4 years ago
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    imperialist is "STILL" typing a reply…

  31. anonymous
    • 4 years ago
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    Akshay u r impatient LOL

  32. anonymous
    • 4 years ago
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    Haha, if you really want to be my student, I guess I'll take you on board. I have just been doing math for a long time Akshay. I don't particularly see problems as tough or not, I just like to see them as new ways of looking at things. For example, if I had never done this problem before and saw a proof of right cancellation, my natural inclination would be to wonder if left also works. The solution just grows naturally out of the problem. And yes, I think a lot when I type, so I type slow, haha :)

  33. Akshay_Budhkar
    • 4 years ago
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    i like things to be "FAST" :P

  34. Akshay_Budhkar
    • 4 years ago
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    oh cool!

  35. Akshay_Budhkar
    • 4 years ago
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    u got a great teacher rld.. go for it

  36. Akshay_Budhkar
    • 4 years ago
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    as far as calc or geography or weather is concerned i may help u :P

  37. anonymous
    • 4 years ago
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    Yay!!!!! Today is my lucky today, Its not as if imperialist ahs been helping me b4 LOL

  38. anonymous
    • 4 years ago
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    Lol

  39. anonymous
    • 4 years ago
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    Ok i guess its official. ya dont worry akshay i will still be posting questions

  40. anonymous
    • 4 years ago
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    I am a slow student so like I hope u dont regret it :D

  41. Akshay_Budhkar
    • 4 years ago
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    i want to help u as well ;) :P anyways gotta help that poor guy there so u guys can discuss the terms and conditions :P bye bye

  42. anonymous
    • 4 years ago
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    Haha, there are no slow students. People just learn in different ways; it's the responsibility of the teacher to find the best way for each individual student to learn. My way of teaching is often very rigorous and methodical, and that is very helpful for some people. For instance, I could have just said "the proof for the left is the same of the right by symmetry" but I instead chose to actually write out the whole proof. I'll also often show my entire thought process as I'm solving a problem, like saying why I think of things in certain ways, so that might be of some help too. And rld, if I ever explain something and you don't understand, feel free to bug me until you do. :)

  43. Akshay_Budhkar
    • 4 years ago
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    can i bug you as well?

  44. Akshay_Budhkar
    • 4 years ago
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    say like why is 1+1=2?

  45. anonymous
    • 4 years ago
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    AKSHAY!!!!!!! BEHAVE

  46. anonymous
    • 4 years ago
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    Lol. Russell has a 347 page proof of that, I think he's covered all the bases. Go look it up there. :p

  47. Akshay_Budhkar
    • 4 years ago
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    ok ok sorry :P

  48. anonymous
    • 4 years ago
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    LOL I dont like to bug :D I usually say thank you andhead over to another person LOL

  49. anonymous
    • 4 years ago
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    neways I have another question LOL

  50. anonymous
    • 4 years ago
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    Haha, sorry, 379 http://en.wikipedia.org/wiki/Principia_Mathematica#Quotations

  51. anonymous
    • 4 years ago
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    LOL akshay go read it now

  52. anonymous
    • 4 years ago
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    I gotta prove another thing

  53. Akshay_Budhkar
    • 4 years ago
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    LOL, i agree! 1+1 is indeed 2

  54. anonymous
    • 4 years ago
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    "The above proposition is occasionally useful."

  55. Akshay_Budhkar
    • 4 years ago
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    lol

  56. anonymous
    • 4 years ago
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    that A^2=A when A is singular and when A=matrix

  57. anonymous
    • 4 years ago
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    *identity matrix

  58. Akshay_Budhkar
    • 4 years ago
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    neways its late here, 00;40, i need to go to sleep, what is the time there @rld :P

  59. anonymous
    • 4 years ago
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    LOL the same as u

  60. anonymous
    • 4 years ago
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    Same time here

  61. Akshay_Budhkar
    • 4 years ago
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    oh wow! where r u from guys?

  62. anonymous
    • 4 years ago
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    I guess in the same time zone. We are from toronto

  63. Akshay_Budhkar
    • 4 years ago
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    Lol yea we r from toronto :D

  64. anonymous
    • 4 years ago
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    Duke university in Durham, North Carolina

  65. Akshay_Budhkar
    • 4 years ago
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    lemme google it, my geography is very bad

  66. anonymous
    • 4 years ago
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    Lol, the US

  67. Akshay_Budhkar
    • 4 years ago
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    oh ok :P

  68. anonymous
    • 4 years ago
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    I guess just a couple of hours away

  69. Akshay_Budhkar
    • 4 years ago
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    really?

  70. anonymous
    • 4 years ago
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    I mean more than a couple if u r driving LOL

  71. razor99
    • 4 years ago
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    good morning rld

  72. Akshay_Budhkar
    • 4 years ago
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    LOL,, i am off now.. @razor are we gonna eat u if u tell us good morning?

  73. anonymous
    • 4 years ago
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    Lol, later Akshay. Anyway, could you repeat your question rld, I didn't really understand what you want to show.

  74. razor99
    • 4 years ago
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    good morning rld

  75. anonymous
    • 4 years ago
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    |dw:1327211029340:dw| This can occur if A=identity matrix or singular y is this so?

  76. anonymous
    • 4 years ago
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    Gmorning razor lol Awful username uve got there LOL

  77. razor99
    • 4 years ago
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    is it mornig to you

  78. anonymous
    • 4 years ago
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    quarter to one here so i guess u can call it morning

  79. razor99
    • 4 years ago
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    so whats the question

  80. anonymous
    • 4 years ago
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    I posted it like scroll up a bit

  81. anonymous
    • 4 years ago
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    Oh, this one isn't so bad. Suppose that A is invertible. Then we have \[A^2=A \implies A^2A^{-1}=AA^{-1}\] \[\implies A(AA^{-1})=AA^{-1} \implies AI=I\] \[\implies A=I\] Thus, if A is invertible and satisfies A^2=A, then A MUST be the identity matrix.

  82. anonymous
    • 4 years ago
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    Now just find an example, such as the zero matrix, that is singular and satisfies A^2=A, and you have shown that A^2=A implies that A is either singular or the identity matrix and you are done.

  83. anonymous
    • 4 years ago
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    what is teh zero matrix?

  84. anonymous
    • 4 years ago
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    A matrix (of whatever size you want) where all of its entries are zero

  85. anonymous
    • 4 years ago
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    LOL oh ok will do:D

  86. anonymous
    • 4 years ago
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    ok that is it Thanks

  87. anonymous
    • 4 years ago
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    No problem :)

  88. LollyLau
    • 4 years ago
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    LOL count: 22

  89. razor99
    • 4 years ago
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    http://en.wikipedia.org/wiki/Singular_value think this link would be a help rld

  90. anonymous
    • 4 years ago
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    No, that is not the same thing. There are many many uses of the word singular in mathematics and not all of them are really related to each other. This question uses singular to mean noninvertible, much simpler than dealing with operators on Hilbert spaces!

  91. anonymous
    • 4 years ago
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    Thanks @razor

  92. razor99
    • 4 years ago
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    i think singular y

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