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Math Help - Eigenvalues - Determinant proof

  1. #1
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    Eigenvalues (edit - completed)

    Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

    Question:

    If A is a square matrix, show that:

    (i) The determinant of A is equal to the product of its eigenvalues.

    (ii) The trace of A is equal to the sum of its eigenvalues


    Please help. Thanks.

    edit - i have done it now. thanks.
    Last edited by WWTL@WHL; February 14th 2009 at 05:50 PM.
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  2. #2
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    Quote Originally Posted by WWTL@WHL View Post
    Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

    Question:

    If A is a square matrix, show that:

    (i) The determinant of A is equal to the product of its eigenvalues.

    (ii) The trace of A is equal to the sum of its eigenvalues


    Please help. Thanks.
    \left| \begin{array}{ccc}<br />
a & b  \\<br />
c & d   \end{array} \right| = ad-bc

    Now for the characteristic equation:


    \left( \begin{array}{ccc}<br />
\lambda & 0  \\<br />
0 & \lambda   \end{array} \right)-<br />
\left( \begin{array}{ccc}<br />
a & b  \\<br />
c & d   \end{array} \right) = \left( \begin{array}{ccc}<br />
  \lambda -a &  - b  \\<br />
\lambda - c & \lambda - d   \end{array} \right)

    And eigen values given by:

      \left| \begin{array}{ccc}<br />
  \lambda -a &  - b  \\<br />
- c & \lambda - d   \end{array} \right| = 0

    So

      (\lambda - a)(\lambda - d) - (-c)(-b) = 0

      \lambda^2 - a\lambda - d \lambda +ad - bc = 0

      \lambda^2 +\lambda(-a - d) +ad - bc = 0

    You see where I'm going?
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  3. #3
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    Question done.
    Last edited by WWTL@WHL; February 14th 2009 at 05:49 PM.
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  4. #4
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    Ok, I have figured it out.

    No help needed now. Thanks.
    Last edited by WWTL@WHL; February 14th 2009 at 05:49 PM.
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  5. #5
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    Quote Originally Posted by WWTL@WHL View Post
    Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

    Question:

    If A is a square matrix, show that:

    (i) The determinant of A is equal to the product of its eigenvalues.

    (ii) The trace of A is equal to the sum of its eigenvalues
    Let A be a square n\times n matrix. Define f(x) = \det (xI - A). This is a polynomial f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0. If x=0 then a_0 = \det (-A) \implies a_0 = (-1)^n \det (A). However, a_0 = (-1)^n k_1k_2...k_n where k_j are eigenvalues (repeated if necessary). Therefore, \det (A) = k_1 ... k_n. To prove this result for the trace that takes more work, see this.
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