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Thread: [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; Dec 18th 2008 at 03:13 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 $\displaystyle n \times n$ matrix $\displaystyle A$ with entries in a field $\displaystyle \mathbb{F}$ is diagonalizable iff $\displaystyle \mathbb{F}^n$ has a

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


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

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



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

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

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

    and thus $\displaystyle c_{i_1}=0$ by (1). contradiction! thus the elements of $\displaystyle \mathfrak{B}$ 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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