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Math Help - [SOLVED] Prove that the only vector common is the zero vector

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    [SOLVED] Prove that the only vector common is the zero vector

    Let \lambda and \mu be distinct eigenvalues for T:R^n \rightarrow R^n and let the corresponding eigenspaces be V_\lambda and V_\mu respectively.

    Prove that the only vector common to both V_\lambda and V_\mu is the zero vector.
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    Quote Originally Posted by Runty View Post
    Let \lambda and \mu be distinct eigenvalues for T:R^n \rightarrow R^n and let the corresponding eigenspaces be V_\lambda and V_\mu respectively.

    Prove that the only vector common to both V_\lambda and V_\mu is the zero vector.
    If the vector x is in both V_\lambda and V_\mu then Tx = \lambda x and Tx = \mu x. What can you deduce from that?
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    Quote Originally Posted by Opalg View Post
    If the vector x is in both V_\lambda and V_\mu then Tx = \lambda x and Tx = \mu x. What can you deduce from that?
    That certainly helps a lot, but I suspect something is missing. Am I just to show that the two do not equal unless the zero vector is used, or is there something else I'd need to show?
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    Quote Originally Posted by Runty View Post
    That certainly helps a lot, but I suspect something is missing. Am I just to show that the two do not equal unless the zero vector is used, or is there something else I'd need to show?
    If \lambda x = \mu x then (\lambda-\mu)x = 0, and if \lambda\ne\mu then \lambda-\mu is a nonzero real number, so you can divide by it. That's all that is needed.
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    Quote Originally Posted by Opalg View Post
    If \lambda x = \mu x then (\lambda-\mu)x = 0, and if \lambda\ne\mu then \lambda-\mu is a nonzero real number, so you can divide by it. That's all that is needed.
    Thanks, I had a feeling that's what I needed to do.
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