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Proving #2

  1. Dec 12, 2003 #1
    Hello , it's me again.

    It's about proving linear aalgebra concepts.

    1. Show that if A = A^-1, then det (A) = 1 or -1.

    2. Show that if A^T = A^-1, then det (A) = 1 or -1.
     
  2. jcsd
  3. Dec 12, 2003 #2
    [tex]\begin{align}
    \det(AB)&=\det(A)\det(B) \\
    \det(A^t)&=\det(A)
    \end{align}[/tex]

    These are all you need.
     
  4. Dec 12, 2003 #3
    Hmmmmm

    Hello again,

    So the two are enough to prove the two theorems.

    But, I got only +1, how about -1?
     
  5. Dec 12, 2003 #4

    Hurkyl

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    How did you do it?
     
  6. Dec 12, 2003 #5
    Here's mine

    For "If A = A^-1, then det A = +1 or -1."

    1. det A = det (A^-1) \\ determinating both sides
    2. det A = 1 / (det A) \\ proeprty: det (A^-1) = 1 / (det A)
    3. (det A)(det A) = 1 \\ Multiplying det A both sides
    4. (det A^2) = 1 \\ simplifying
    5. det A = + or 1 (square root of 1 ) \\ extracting sq.root
    6. Thus, det A = +1 or -1.

    For "If A^t = A^-1, then det A = 1 or -1."

    1. (A^T) A = (A^-1) A \\ If the inverse of A is said to be existing, then A itself must exist and the product of the two must be an identity matrix. I multiplied A both sides of equation.

    2. (A^T) A = I \\ Identity matrix.
    3. det [ (A^T) A ] = det I \\ determinating both sides.
    4. det(A^T) det A = det I \\ Prop: det AB = det A det B
    5. det A det A = det I \\ Prop: det (A^T) = det A.
    6. det A^2 = 1 \\ determinant of identity matrix is 1.
    \\ simplifying
    7. det A = + or - square root of 1 \\ extracting a square root.
    8. Thus det A = 1 or -1.
     
  7. Dec 13, 2003 #6
    The first proof can be done using only property (1) that master_coda mentioned.

    1 = det(I) = det(A * A^-1) = det(A)det(A^-1) = det(A)det(A) = det(A)^2

    det(A)^2 = 1
    det(A) = +/- 1
     
  8. Dec 13, 2003 #7
    Thanks!

    Hi!

    Hmmm, that's correct. =) Thank you. =)

    Well, was my proving to both of the two in the previous "replies" correct? (I want to know if I am doing fine in my proving).
     
  9. Dec 13, 2003 #8
    Hey Muzza...

    Hello Muzza,

    I have read your answer. It seems correct but I doubt on one point.

    The det A^-1 is equal to [1 / (det A)], how come it turns out to be det A?
     
  10. Dec 13, 2003 #9
    Your proofs seem correct to me.

    det(A) = 1/det(A) does hold, for det(A) = 1, or det(A) = -1 (1 = 1/1 and -1 = 1/(-1) are both true statements), which is well... what you showed that the possible values for det(A) could be :P
     
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