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Math Help - Diagonal Matrices from Characteristic Poly

  1. #1
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    Diagonal Matrices from Characteristic Poly

    A 3 x 3 symmetric matrix A has characteristic polynomial (~ -1)^2(~ - 2). Find all diagonal matrices similar to A. Any ideas? Never seen a question like this before. When multiplied out the poly is (~^3 - 4~^2 + 5~ - 2).
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    Quote Originally Posted by chadlyter View Post
    A 3 x 3 symmetric matrix A has characteristic polynomial (~ -1)^2(~ - 2). Find all diagonal matrices similar to A. Any ideas? Never seen a question like this before. When multiplied out the poly is (~^3 - 4~^2 + 5~ - 2).
    The charachteristic polynomial of two similar matrices are the same. Thus, if A has the charachteristic polynomial X^3 - 4X^2 + 5X - 2 then A^3 - 4A^2 + 5A - 2I = \bold{0} by the Cayley-Hamilton theorem. Of course, this theorem is very advanced and you probably never seen it before. Therefore, there is a weaker version for this theorem which states that a diagnolizable matrix satisfies its charachteristic polynomial. Since A is a symettric matrix it means it is diagnolizable and the rest follows.
    Thus, you need to find \left[ \begin{array}{ccc}a&0&0\\0&b&0\\0&0&c \end{array} \right] which solve this polynomial equation.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    The charachteristic polynomial of two similar matrices are the same. Thus, if A has the charachteristic polynomial X^3 - 4X^2 + 5X - 2 then A^3 - 4A^2 + 5A - 2I = \bold{0} by the Cayley-Hamilton theorem. Of course, this theorem is very advanced and you probably never seen it before. Therefore, there is a weaker version for this theorem which states that a diagnolizable matrix satisfies its charachteristic polynomial. Since A is a symettric matrix it means it is diagnolizable and the rest follows.
    Thus, you need to find \left[ \begin{array}{ccc}a&0&0\\0&b&0\\0&0&c \end{array} \right] which solve this polynomial equation.
    Hence a, b and c are roots of x^3 - 4x^2 + 5x - 2 = {0}, and any diagonal matrix with each diagonal element equal to a root (not necessarily distinct) satisfies the charateristic equation.

    RonL
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