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Thread: Eigenvalues

  1. #1
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    Eigenvalues

    Let $\displaystyle A$ be an $\displaystyle n \times n$ real matrix such that $\displaystyle A^{2}=I, A \neq \pm{I}$
    (where I denotes the n n-identity matrix). Show that
    (i) A has two eigenvalues $\displaystyle \lambda_{1},\lambda_{2}$.
    (ii) Every element $\displaystyle x \in \mathbb{R}^{n}$ can be expressed uniquely as $\displaystyle x_{1}+x_{2}$,
    where $\displaystyle Ax_{1}=\lambda_{1}x_{1}$ and $\displaystyle Ax_{2}=\lambda_{2}x_{2}$
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  2. #2
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    Quote Originally Posted by Chandru1 View Post
    Let $\displaystyle A$ be an $\displaystyle n \times n$ real matrix such that $\displaystyle A^{2}=I, A \neq \pm{I}$
    (where I denotes the n n-identity matrix). Show that
    (i) A has two eigenvalues $\displaystyle \lambda_{1},\lambda_{2}$.
    (ii) Every element $\displaystyle x \in \mathbb{R}^{n}$ can be expressed uniquely as $\displaystyle x_{1}+x_{2}$,
    where $\displaystyle Ax_{1}=\lambda_{1}x_{1}$ and $\displaystyle Ax_{2}=\lambda_{2}x_{2}$
    Take two endomorphisms of $\displaystyle \mathbb{R}^{n}$ :
    $\displaystyle u = A - I$
    $\displaystyle v = A + I$

    You have

    • $\displaystyle u \circ v = v \circ u = 0$
    • $\displaystyle u \neq 0$
    • $\displaystyle v \neq 0$

    according to the problem.

    You see that

    • $\displaystyle Ker (u) \cap Ker (v) = \oslash$
    • $\displaystyle \mathbb{R}^{n} = Ker (u \circ v) = Ker (v) \cup v^{-1}(Ker (u) \cap Im (v))$

    and, after a little manipulation and comaprison of dimensions,
    dim Ker (u) + dim Ker (v) = n

    Which is sufficient to say that $\displaystyle Ker(u) \oplus Ker(v) = \mathbb{R}^{n}$

    $\displaystyle dim Ker (u) \neq 0$, cause it would mean v = 0
    $\displaystyle dim Ker (v) \neq 0$, cause it would mean u = 0

    So there are effectively 2 eigenvalues for A ( 1 and -1),
    and the associated eigenspaces are in direct sum (Ker u and Ker v),
    which guarantees the existance and unicity if the decomposition.
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