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Thread: Cayley-Hamilton Theorem for matrices

  1. #1
    Member Last_Singularity's Avatar
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    Cayley-Hamilton Theorem for matrices

    The Cayley Hamilton Theorem states that if $\displaystyle T$ is a linear operation on vector space $\displaystyle V$ and $\displaystyle f(t)$ is the characteristic polynomial of $\displaystyle T$, then $\displaystyle f(T)$ is the zero transformation.

    How do I extend this to matrices? As in, how do I show that: if $\displaystyle A$ is $\displaystyle n \times n$ and $\displaystyle f(t)$ is the characteristic of $\displaystyle A$, then $\displaystyle f(A)$ is the zero matrix?

    I attempted to prove it using $\displaystyle f(A) = det(A - AI) = det(O) = 0$ but this reasoning seems faulty...
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  2. #2
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    Remember that linear transformations between finite dimensional vector spaces are in 1-1 correspondence with matrices (and they do the same thing with the exception that the matrices act on $\displaystyle \mathbb{F} ^n$ ) so translate everything from matrices to lin.tranf. and see what $\displaystyle f(A)$ does in $\displaystyle \mathbb{F} ^n$
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