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Math Help - [SOLVED] Diagonalization

  1. #1
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    [SOLVED] Diagonalization

    if there is a 3x3 matrix which has exactly 3 distinct eigenvalues why must it be diagonalizable?
    Last edited by jkeatin; December 18th 2008 at 04:13 PM.
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  2. #2
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    Quote Originally Posted by jkeatin View Post
    Suppose now that A is some other 3x3 matrix which has exactly 3 distinct eigenvalues. why A must be diagonalizable.
    well, it depends how much you know about diagonalizable matrices. i'll assume you know nothing but this definition that an n \times n matrix A with entries in a field \mathbb{F} is diagonalizable iff \mathbb{F}^n has a

    basis \mathfrak{B} such that [T]_{\mathfrak{B}} is a diagonal matrix, where as usual T: \mathbb{F}^n \longrightarrow \mathbb{F}^n is defined by T(\bold{x})=A\bold{x}.


    so suppose an n \times n matrix A with entries in a field \mathbb{F} has n distinct eigenvalues in \mathbb{F}, say \lambda_i, \ 1 \leq i \leq n. let A\bold{x}_i=\lambda_i \bold{x}_i with \bold{0} \neq \bold{x}_i \in \mathbb{F}^n. then \mathfrak{B}=\{\bold{x}_1, \cdots, \bold{x}_n \} is a basis for \mathbb{F}^n. \ \ \ \ \ \ \color{red}(*)

    now consider the linear map T: \mathbb{F}^n \longrightarrow \mathbb{F}^n defined by T(\bold{x})=A \bold{x}. then T(\bold{x_i})=\lambda_i \bold{x}_i and thus: [T]_{\mathfrak{B}}=\begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}. \ \Box



    \color{red}(*) to see this we only need to show that \bold{x}_1, \cdots, \bold{x}_n are linearly independent. so suppose c_{i_1}\bold{x}_{i_1} + \cdots + c_{i_k}\bold{x}_{i_k}=\bold{0}, where not all of the coefficients are 0 and k is minimal. call it (1). taking T

    of both sides of (1) gives us: c_{i_1}T(\bold{x}_{i_1}) + \cdots + c_{i_k}T(\bold{x}_{i_k})=\bold{0} and hence c_{i_1}\lambda_{i_1}\bold{x}_{i_1} + \cdots + c_{i_k}\lambda_{i_k}\bold{x}_{i_k}=\bold{0}. call this one (2). now multiply (1) by \lambda_{i_1} and subtract the result from (2) to get:

    c_{i_2}(\lambda_{i_2}-\lambda_{i_1})\bold{x}_{i_2} + \cdots + c_{i_k}(\lambda_{i_k}-\lambda_{i_1})\bold{x}_{i_k}=\bold{0}, which contradicts the minimality of k unless c_{i_2}(\lambda_{i_2}-\lambda_{i_1}) = \cdots = c_{i_k}(\lambda_{i_k}-\lambda_{i_1})=0. since \lambda_i are distinct, we must have c_{i_2}= \cdots = c_{i_k}=0,

    and thus c_{i_1}=0 by (1). contradiction! thus the elements of \mathfrak{B} are linearly independent. Q.E.D.
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  3. #3
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    so it basically has to be linearly independent? meaning 3 independent eigenvectors?
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