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brinethery Group Title

Find a basis for the space V of all skew-symmetric 3x3 matrices. What is the dimension of V?

  • 2 years ago
  • 2 years ago

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  1. brinethery Group Title
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    I'll be back in a sec. Brewing some coffee. Would you like some?

    • 2 years ago
  2. No-data Group Title
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    Yeah, I will in maybe two hours =P

    • 2 years ago
  3. No-data Group Title
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    Well you need to remember that a basis for any vectorial space generates it and it's vectors are linear independent.

    • 2 years ago
  4. No-data Group Title
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    And 3x3 skew-symmetric matrix is something like this:

    • 2 years ago
  5. No-data Group Title
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    \[A=-A^T\]

    • 2 years ago
  6. brinethery Group Title
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    :-)

    • 2 years ago
  7. No-data Group Title
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    Hmm thinking...

    • 2 years ago
  8. No-data Group Title
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    Do you have any idea to begin with?

    • 2 years ago
  9. No-data Group Title
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    Mm you need three matrices am I right?

    • 2 years ago
  10. No-data Group Title
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    Well the dimension is 3 haha

    • 2 years ago
  11. brinethery Group Title
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    I am stupid haha

    • 2 years ago
  12. brinethery Group Title
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    and I need a longer explanation b/c A=-A^T isn't gonna cut it

    • 2 years ago
  13. brinethery Group Title
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    soy estupido haha

    • 2 years ago
  14. No-data Group Title
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    uff I almost burn my brain hehe

    • 2 years ago
  15. brinethery Group Title
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    I told you linear algebra's no good!

    • 2 years ago
  16. No-data Group Title
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    Well I think that the three vectors are \[\left[\begin{matrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0\end{matrix}\right]\]

    • 2 years ago
  17. No-data Group Title
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    Hmmm you're right

    • 2 years ago
  18. No-data Group Title
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    Is the idea of a basis clear to you?

    • 2 years ago
  19. No-data Group Title
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    Like you know why (1,0,0), (0,1,0),(0,0,1) is the basis of R^3?

    • 2 years ago
  20. brinethery Group Title
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    yes.

    • 2 years ago
  21. No-data Group Title
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    Well then you know that the basic idea behind a basis is that you can get any vector of the space by forming a linear combination.

    • 2 years ago
  22. No-data Group Title
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    like \[\vec{v}=a\hat{i} + b\hat{j} + c\hat{k}\]

    • 2 years ago
  23. No-data Group Title
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    |dw:1337293630890:dw|

    • 2 years ago
  24. No-data Group Title
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    well that is the idea in a graphic form.

    • 2 years ago
  25. No-data Group Title
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    If you want to obtain a basis for the 3x3 skew symmet

    • 2 years ago
  26. No-data Group Title
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    ric matrices you need to know that is the general form of matrix of that kind.

    • 2 years ago
  27. brinethery Group Title
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    okay, I'm following what you're saying

    • 2 years ago
  28. No-data Group Title
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    the definition says that the elements of an skew symmetric matrix are such as: \[a_{ij}=-a_{ji}\]

    • 2 years ago
  29. No-data Group Title
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    It seems that that doesn't tell us a lot about those matrices, but you can figure out what you have on the diagonal.

    • 2 years ago
  30. No-data Group Title
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    if i = j, then \[a_{ii}=-a_{ji}\]

    • 2 years ago
  31. No-data Group Title
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    which is true if the diagonal is made up zeros.

    • 2 years ago
  32. No-data Group Title
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    and I think that's all. An 3x3 skew symmetrical matrix has this form:

    • 2 years ago
  33. brinethery Group Title
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    Here's something... http://www.euclideanspace.com/maths/algebra/matrix/functions/skew/index.htm

    • 2 years ago
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    \[\left[\begin{matrix}0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23}\\ -a_{13} & -a_{23} & a_{33} \end{matrix}\right]\]

    • 2 years ago
  35. brinethery Group Title
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    Looks like it's all zeros down the diagonal.

    • 2 years ago
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    shame on my haha

    • 2 years ago
  37. No-data Group Title
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    yeah a_33 = 0

    • 2 years ago
  38. brinethery Group Title
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    you're a bad math teacher haha! :-p

    • 2 years ago
  39. No-data Group Title
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    yeah! I'm terrible =(

    • 2 years ago
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    but it's fun!

    • 2 years ago
  41. No-data Group Title
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    well now you can split that matrix on three parts:

    • 2 years ago
  42. brinethery Group Title
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    Yep this part is where I need explaining...

    • 2 years ago
  43. No-data Group Title
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    \left[\begin{matrix}0 & a_{12} & 0 & \\ -a_{12} & 0 & 0\\ 0&0&0\end{matrix}\right]

    • 2 years ago
  44. No-data Group Title
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    \[\left[\begin{matrix}0 & 0 & a_{13} & \\ 0 & 0 & 0\\ -a_{13}&0&0\end{matrix}\right]\]

    • 2 years ago
  45. No-data Group Title
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    \[\left[\begin{matrix}0 & 0 & 0 & \\ 0 & 0 & a_{23}\\ 0&-a_{23}&0\end{matrix}\right]\]

    • 2 years ago
  46. No-data Group Title
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    Is it clear why I did that?

    • 2 years ago
  47. brinethery Group Title
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    Oh I see. So each part of the basis will be a 3x3 matrix?

    • 2 years ago
  48. brinethery Group Title
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    And there will be (3) 3x3 matrices

    • 2 years ago
  49. No-data Group Title
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    yeah! it has to bee a 3x3 matrix

    • 2 years ago
  50. brinethery Group Title
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    I asked on cramster and some (puta) was really insulting to me AND she gave me the wrong answer.

    • 2 years ago
  51. No-data Group Title
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    what is cramster? and puta?

    • 2 years ago
  52. No-data Group Title
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    we are not finished yet.

    • 2 years ago
  53. No-data Group Title
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    you can do this:

    • 2 years ago
  54. brinethery Group Title
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    I thought it meant b)it(ch but I guess it's mexican slang for prostitute or whore

    • 2 years ago
  55. brinethery Group Title
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    but you get the idea.

    • 2 years ago
  56. No-data Group Title
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    \[\left[\begin{matrix}0& 1 & 0 \\ -1 & 0 &0\\ 0&0&0\end{matrix}\right]\]

    • 2 years ago
  57. No-data Group Title
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    ahh I thought I did haha.

    • 2 years ago
  58. No-data Group Title
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    \[\left[\begin{matrix}0& 0 & 1 \\ 0 & 0 &0\\ -1&0&0\end{matrix}\right]\]

    • 2 years ago
  59. No-data Group Title
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    and \left[\begin{matrix}0& 0 & 0 \\ 0 & 0 &1\\ 0&-1&0\end{matrix}\right]

    • 2 years ago
  60. No-data Group Title
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    You multiply by the scalars a,b,c each of those matrices and form a linear combination.

    • 2 years ago
  61. No-data Group Title
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    and the result is a skew symmetrical matrix

    • 2 years ago
  62. No-data Group Title
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    I'm sorry If my explanation was awful, and my english worst =P

    • 2 years ago
  63. brinethery Group Title
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    So the dimension of all skew-symmetric matrices is 2?

    • 2 years ago
  64. brinethery Group Title
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    or that's only the dimension of the basis?

    • 2 years ago
  65. No-data Group Title
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    the dimesion is the number of vectors of the basis. so it's 3.

    • 2 years ago
  66. brinethery Group Title
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    hahaha I am so stupid!

    • 2 years ago
  67. brinethery Group Title
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    WOW. I can't believe I asked that.

    • 2 years ago
  68. No-data Group Title
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    if you have an 4x4 skew symmetrical matrix the dimension of the basis is 4.

    • 2 years ago
  69. brinethery Group Title
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    :-) yeah yeah yeah

    • 2 years ago
  70. No-data Group Title
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    Don't worry, don't judge yourself so severely.

    • 2 years ago
  71. No-data Group Title
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    Do you have any doubt regarding this brinethery?

    • 2 years ago
  72. brinethery Group Title
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    I have another question for you! No I don't have any doubt. You did a great job explaining.

    • 2 years ago
  73. No-data Group Title
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    Ok

    • 2 years ago
  74. No-data Group Title
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    Linear Algebra?

    • 2 years ago
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