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Properties of Determinants

  1. Jul 11, 2005 #1
    Prove that, if [tex]AA^T = A^TA = I_n[/tex], then [tex]\det{A} = \pm 1[/tex].

    This is daunting.
     
    Last edited: Jul 11, 2005
  2. jcsd
  3. Jul 11, 2005 #2

    quasar987

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    try w/o the spaces btw the brackets and the "tex". you don't need to rewrite [ tex ] on both sides of the "=" u know.

    [tex]AA^T = A^TA = I_n[/tex]

    then

    [tex]\det{A} = \pm 1[/tex]
     
    Last edited: Jul 11, 2005
  4. Jul 11, 2005 #3

    quasar987

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    As for the solution:

    [tex]\det{AA^T} = \det{A}\det{A^T} = \det{A}^2 = \det{I_n} = 1 \Leftrightarrow \det{A} = \pm 1[/tex]
     
  5. Jul 12, 2005 #4
    omg that's it. i hate this course
     
  6. Jul 12, 2005 #5

    matt grime

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    don't hate the course because it is daunting; see it as a challenge. as you see the answer is straight forward (i presume that's what the 'omg that's it' means), and all teh answers are like that. just be calm and check your notes for things
     
  7. Jul 12, 2005 #6
    I just have to get used to the proving techniques
     
  8. Jul 14, 2005 #7
    Prove that any scalar multiple of a symmetric matrix is symmetric.

    Let [tex]A = (a_i_j)[/tex]

    Since [tex]A[/tex] is symmetric, [tex]A = A^T[/tex].

    Then [tex]A^T = (a_i_j)[/tex].

    Therefore, [tex](cA) = c(A) = c(A^T) = (cA^T)[/tex]

    Was it necessary to show that [tex]A = (a_i_j)[/tex] and [tex]A^T = (a_i_j)[/tex]? Is [tex]A^T = (a_i_j)[/tex] even right? I can't express myself mathematically
     
    Last edited: Jul 14, 2005
  9. Jul 14, 2005 #8
    Prove that, if A and B are two matrices such that A + B and AB are defined, then both A and B are square matrices.


    - Let A be an m x r matrix and B an r x n matrix such that,
    [tex]A_m_x_rB_r_x_n = (AB)_m_x_n[/tex]

    - We know that the sum A + B of the two matrices is the m x n matrix

    How do I express this mathematically to show that they are square?
     
    Last edited: Jul 14, 2005
  10. Jul 14, 2005 #9
    Just put the 'T' out of the brackets at the end (that is, over all of: cA).
     
  11. Jul 14, 2005 #10
    If A+B is defined, then their sizes are similar- i.e., they're both mxn.
    Since AB is defined, as you yourself wrote, we must have m=n. (Because the product is an mxn*mxn, which is only defined when m=n).
    Both matrices are therefor nxn, square matrices!
     
  12. Jul 14, 2005 #11
    Oops typo, thanks.
     
  13. Jul 14, 2005 #12
    The r in A is the jth column and in B it's the ith row. So when you say m=n, are you refering to the r's?

    And if I were to prove that any scalar multiple of a diagonal matrix is a diagonal matrix, how is that different from, say, letting [tex] A = (a_i_j) [/tex] be any m x n matrix and c any real number?

    A diagonal matrix is a square matrix that all of its nonzero entries are on the diagonal.

    Then [tex] cA = c(a_i_j) = (ca_i_j)[/tex] but it may not be diagonal.

    I guess we assume that [tex] A = (a_i_j) [/tex] is a diagonal matrix?
     
    Last edited: Jul 14, 2005
  14. Jul 14, 2005 #13
    Yes, I was referring to the 'r's.
    And I'm sure I got your question. It's true that for any matrix A=(aij) that keeps aij=0 for all i!=j, cA keeps the same thing for any scalar c. In the private case of square matrix, it shows that any scalar multiple of a diagonal matrix is, indeed, a diagonal matrix.
    Did I answer your question? :uhh:
     
  15. Jul 14, 2005 #14
    Yeah thanks, but I will be back with more
     
  16. Jul 14, 2005 #15
    Is that a threat?

    Anyway, I'm off to bed. Be back in about 12 hours... (It's the middle of the night here).
     
  17. Jul 14, 2005 #16
    Haha what?
     
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