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UsukiDoll

Can anyone help me with this proof? Am I on the right track? Let A and B be equivalent square matrices. Prove that A is nonsingular if and only if B is nonsingular

  • one year ago
  • one year ago

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  1. UsukiDoll
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    Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

    • one year ago
  2. UsukiDoll
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    Let's see.... Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

    • one year ago
  3. UsukiDoll
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    We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

    • one year ago
  4. UsukiDoll
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    |dw:1361430166380:dw|

    • one year ago
  5. UsukiDoll
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    Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.

    • one year ago
  6. UsukiDoll
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    |dw:1361430248917:dw|

    • one year ago
  7. UsukiDoll
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    |dw:1361430265828:dw|

    • one year ago
  8. UsukiDoll
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    @UnkleRhaukus

    • one year ago
  9. UsukiDoll
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    ?????????????

    • one year ago
  10. UnkleRhaukus
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    it looks like your on the right track, but i can not remember this stuff

    • one year ago
  11. UsukiDoll
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    @joemath314159 and @hartnn

    • one year ago
  12. UsukiDoll
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    @zepdrix

    • one year ago
  13. UsukiDoll
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    thank goodness! I was searching everywhere for someone to help me see if I'm on the right track.

    • one year ago
  14. zepdrix
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    I'm afraid this might be over my head :c

    • one year ago
  15. UsukiDoll
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    AWWW ffdsfkadjfksldsjfkadsjfoiweuro

    • one year ago
  16. UsukiDoll
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    : (

    • one year ago
  17. UsukiDoll
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    have you done proof writing before or no?

    • one year ago
  18. zepdrix
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    mmmm not so much... and i don't have a very good grasp on matrices yet i'm afraid :P darn.

    • one year ago
  19. UsukiDoll
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    @Risadinha

    • one year ago
  20. zepdrix
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    Poor lil pony :o

    • one year ago
  21. UsukiDoll
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    crud. This thing is like due in 11-12 days! I like to plan ahead and such. wahhh

    • one year ago
  22. UsukiDoll
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    @KingGeorge do you know how to write proofs in Linear Algebra? I've done some and I was wondering if you could tell me if I'm on the right track.

    • one year ago
  23. KingGeorge
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    This seems to be on the right track to me.

    • one year ago
  24. UsukiDoll
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    ok so what do I need to write next for this proof or is that it?

    • one year ago
  25. KingGeorge
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    have you learned that a matrix A is nonsingular if and only if it's invertible?

    • one year ago
  26. UsukiDoll
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    invertible?

    • one year ago
  27. UsukiDoll
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    oh that! yes square matrix and the AB=Ba=In. yeah I dealt with that before.

    • one year ago
  28. UsukiDoll
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    if a matrix is singular, there's no inverse, so it's impossible for AB=BA=In to work

    • one year ago
  29. KingGeorge
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    Alright. Suppose A is nonsingular, and \(A=PBQ\) for some non-singular matrices P,Q. This means that A, P, Q are invertible. Then \[B=P^{-1}AQ^{-1}\]But since \(A^{-1}\) exists, look at the product\[P^{-1}AQ^{-1}QA^{-1}P =BQA^{-1}P=\text{Id}\]So B has an inverse and must be non-singular. This is how I would do it. Just switch A and B to prove the other statement you need.

    • one year ago
  30. UsukiDoll
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    ok the switching is a bit nerve wrecking. omg... that's a lot of letters.

    • one year ago
  31. KingGeorge
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    Just work through it slowly. You had most of it yourself. You're only missing the last couple lines.

    • one year ago
  32. UsukiDoll
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    ok let me try ummm... B is nonsingular and B =PAQ for some non-singular matrices P,Q. This means that B, P, and Q are invertiable

    • one year ago
  33. KingGeorge
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    right.

    • one year ago
  34. UsukiDoll
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    ok so that means that |dw:1361502760161:dw|

    • one year ago
  35. UsukiDoll
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    ok so B inverse must exist

    • one year ago
  36. KingGeorge
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    You already know that B inverse exists. You need to show that A inverse exists.

    • one year ago
  37. KingGeorge
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    What can you multiply \[P^{-1}BQ^{-1}\]by to get the identity?

    • one year ago
  38. UsukiDoll
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    to get if B is nonsingular A is nonsingular. |dw:1361503034980:dw|

    • one year ago
  39. KingGeorge
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    I'm not understanding what you're trying to do there. Would you mind elaborating?

    • one year ago
  40. UsukiDoll
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    I'm trying to prove the other statement. I know there's two statements If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

    • one year ago
  41. UsukiDoll
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    Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

    • one year ago
  42. UsukiDoll
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    so much letters going oN!

    • one year ago
  43. KingGeorge
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    So can you find some product of matrices such that \(P^{-1}BQ^{-1}\) multiplied by that product of matrices gives you the identity?

    • one year ago
  44. UsukiDoll
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    hmmm... B = PAQ A = P^-1BQ^-1

    • one year ago
  45. UsukiDoll
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    (PAQ)(P^-1BQB^-1) = (P^-1BQB^-1)(PAQ)

    • one year ago
  46. UsukiDoll
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    :/

    • one year ago
  47. KingGeorge
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    I think you're on the right track. You want \[P^{-1}BQ^{-1}X=\text{Id}\]though. So \(BQ^{-1}X=P\) and \(Q^{-1}X=B^{-1}P\). If you continue this one more step, what do you get to be for \(X\)?

    • one year ago
  48. KingGeorge
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    Did that make sense?

    • one year ago
  49. UsukiDoll
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    omg no. I'm lost... :/

    • one year ago
  50. UsukiDoll
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    ok wait so A = P^-1BQ^-1 that must be Ax = Id

    • one year ago
  51. UsukiDoll
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    o.o wahhh this is torture.

    • one year ago
  52. KingGeorge
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    That's the right idea. We're finding an \(X\) such that \(AX=\text{Id}\). So if \[AX=P^{-1}BQ^{-1}X=\text{Id},\]you need to find \(X\) in terms of \(P,Q,\) and \(B\) (and/or their inverses). Make more sense?

    • one year ago
  53. UsukiDoll
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    yeah more work and torture D:

    • one year ago
  54. UsukiDoll
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    but how do I find x in terms of P? Won't that be P = whatever it is X.

    • one year ago
  55. UsukiDoll
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    |dw:1361504529420:dw|

    • one year ago
  56. KingGeorge
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    That's what I was going to suggest next actually.

    • one year ago
  57. UsukiDoll
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    |dw:1361504588920:dw|

    • one year ago
  58. UsukiDoll
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    |dw:1361504640692:dw|

    • one year ago
  59. UsukiDoll
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    |dw:1361504663129:dw|

    • one year ago
  60. UsukiDoll
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    |dw:1361504710718:dw|

    • one year ago
  61. KingGeorge
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    Hold up. Look at it as an algebraic equation like this where you solve for x.\[\frac{1}{p}b\frac{1}{q}x=1\]

    • one year ago
  62. UsukiDoll
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    |dw:1361504883640:dw|

    • one year ago
  63. UsukiDoll
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    |dw:1361504957573:dw|

    • one year ago
  64. KingGeorge
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    Sorry, internet glitched out on me there. But yes. That right. Converting that back into matrix form, you get \[X=PB^{-1}Q\]That make sense?

    • one year ago
  65. UsukiDoll
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    yeah sorry...I feel a bit weak...need to eat dinner

    • one year ago
  66. KingGeorge
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    No worries. I'll probably be gone when you get back though. Good luck!

    • one year ago
  67. UsukiDoll
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    @KingGeorge are you going to be on openstudy tomorrow? I had a busy day today and feel tired.

    • one year ago
  68. UsukiDoll
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    Oh well can I email it to you tomorrow?

    • one year ago
  69. UsukiDoll
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    @ArkGoLucky

    • one year ago
  70. UsukiDoll
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    |dw:1361519596240:dw|

    • one year ago
  71. UsukiDoll
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    |dw:1361519710516:dw|

    • one year ago
  72. UsukiDoll
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    |dw:1361519769721:dw|

    • one year ago
  73. UsukiDoll
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    @AriPotta

    • one year ago
  74. UsukiDoll
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    ughhhh so close!!! X( I'm trying to prove the second statement

    • one year ago
  75. ArkGoLucky
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    Sorry don't know this stuff

    • one year ago
  76. UsukiDoll
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    Ok. I may have got it. Suppose B is nonsingular and B=PAQ for some non-singular matrices P and Q. That means B, P, and Q are invertible. |dw:1361526378870:dw| |dw:1361526405075:dw| |dw:1361526433824:dw| |dw:1361526465398:dw| Since B^-1 exist we look at the product of nonsingular matrices. |dw:1361526502682:dw| Therefore A has an inverse and it must be nonsingular

    • one year ago
  77. UsukiDoll
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    @KingGeorge I may have gotten it

    • one year ago
  78. UsukiDoll
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    @wio

    • one year ago
  79. UsukiDoll
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    can you please check if I got this right?

    • one year ago
  80. wio
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    Hold on.

    • one year ago
  81. experimentX
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    don't divide by matrices, division is not defined. Inverses and division are different.

    • one year ago
  82. UsukiDoll
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    ok so how do I prove the converse of this

    • one year ago
  83. UsukiDoll
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    Given: Matrix A and Matrix B are equivalent square matrices. Since we are given a biconditional statement, we need to prove two situations: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

    • one year ago
  84. wio
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    The converse of what statement?

    • one year ago
  85. UsukiDoll
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    If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

    • one year ago
  86. UsukiDoll
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    these are the two situations

    • one year ago
  87. UsukiDoll
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    King George and I did the first one yesterday

    • one year ago
  88. wio
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    Okay, what is meant by A and B are equivalent?

    • one year ago
  89. UsukiDoll
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    just the second one. I attempted through the drawings last night

    • one year ago
  90. experimentX
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    use the same argument as you used on the first one, P and Q are invertible by definition.http://en.wikipedia.org/wiki/Matrix_equivalence

    • one year ago
  91. toxicsugar22
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    Hi can you help me after you help this person please

    • one year ago
  92. UsukiDoll
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    but the letters are different on the second

    • one year ago
  93. UsukiDoll
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    I drew the attempt on it last night

    • one year ago
  94. UsukiDoll
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    I'm not sure if that's it or something

    • one year ago
  95. UsukiDoll
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    no and I can't use fancy stuff that my class didn't learn. I already got chewed once by the professor

    • one year ago
  96. UsukiDoll
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    like he will give points but taunt you in the comments

    • one year ago
  97. UsukiDoll
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    Theorem 2.13 states that two m x n matrices A and B are equivalent of and only if B = PAQ for some nonsingular matrices P and Q.

    • one year ago
  98. UsukiDoll
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    We have a product of nonsingular matrices P and Q. If B is nonsingular, B=PAQ, then A is nonsingular A=P^-1BQ^-1

    • one year ago
  99. UsukiDoll
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    Conversely, if A is nonsingular, A = PBQ, then B is nonsingular, B=P^-1AQ^-1 because P and Q are products of nonsingular matrices.

    • one year ago
  100. experimentX
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    I think you earlier asked the question AB is invertible iff both A and B are invertible. Isn't this same as proving it?

    • one year ago
  101. UsukiDoll
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    the second part. the one with conversely..... I don't know if I got it right even though it says to flip the letters A and B

    • one year ago
  102. UsukiDoll
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    ? ? ? I think that's a different question that involves something different.

    • one year ago
  103. experimentX
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    this is same, B = PAQ what did you prove above and what do you want to prove next? write in short.

    • one year ago
  104. UsukiDoll
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    THere's two situations to this: If Matrix A is nonsingular then Matrix B is nonsingular If Matrix B is nonsingular then Matrix A is nonsingular

    • one year ago
  105. UsukiDoll
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    I want to prove through the usage of Theorem 2. 13 whatever I posted previously

    • one year ago
  106. UsukiDoll
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    I already have one done and I attempted the second one (it's drawn)

    • one year ago
  107. experimentX
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    \[ B = PAQ => P^{-1}BQ^{-1}=A \] Let \( P^{-1}=P'\) and \(Q^{-1}=Q'\) isn't this same as before?

    • one year ago
  108. UsukiDoll
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    THat's if B is nonsingular then A is nonsingular since we have the product of nonsingular matrices P and Q

    • one year ago
  109. experimentX
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    sorry, I didn't get you ... where do you have problem understanding it?

    • one year ago
  110. wio
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    If matrix equivalence is an equivalence relation, then it follows the transitive property.

    • one year ago
  111. experimentX
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    you don't have to go through the rigorous way like you did the first one, Both P^-1 and Q^-1 are invertible matrices. If you take P and Q on the side of B, you will get same expression as you did on the first one.

    • one year ago
  112. UsukiDoll
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    really? so A=PBQ then B=P^-1AQ^-1

    • one year ago
  113. experimentX
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    yes P and Q are invertible matrices by definition of Equivalent Matrices. So their inverses are also invertible matrices making the the relation symmetric.

    • one year ago
  114. UsukiDoll
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    omg I've been freaking out and stressing for nothing then! -________-

    • one year ago
  115. UsukiDoll
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    but ok I should put that for if A is nonsingular then B is nonsingular

    • one year ago
  116. experimentX
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    No ... use symmetry argument. A = PBQ B = P'AQ' <-- this is (similar) to first.

    • one year ago
  117. zzr0ck3r
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    you can do this with elementry matricies.

    • one year ago
  118. UsukiDoll
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    symmetry argument?

    • one year ago
  119. experimentX
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    Theorem 1: A = PBQ if B, P, and Q is non invertible then A is non invertible. B = P'AQ',P', Q' is non invertible, then B is non invertible from theorem 1. Guess not symmetry argument.

    • one year ago
  120. experimentX
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    write it up like that.

    • one year ago
  121. UsukiDoll
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    what about... A = PBQ. IF B, P, and Q are non invertible, then A is non invertible. B = P^-1AQ^-1. If P^-1 and Q^-1 are non invertible then B is non invertible.

    • one year ago
  122. experimentX
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    sure that works too.

    • one year ago
  123. UsukiDoll
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    so. If B=PAQ. If A, P, and Q are non invertible then B is non invertible A = P^-1BQ^-1. If P^-1 and Q^-1 are non invertible then A is non invertible

    • one year ago
  124. experimentX
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    yes

    • one year ago
  125. UsukiDoll
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    :D thank you

    • one year ago
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