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Math Help - [SOLVED] Diagonalizable

  1. #1
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    [SOLVED] Diagonalizable

    If A is nxn diagonalizable matrix, then each vector in \mathbb{R}^n can be written as a linear comination of eigenvectors of A.

    This is true but how do I prove this?
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  2. #2
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    Since A is diagonalizable, there exist a diagonal matrix D and an invertible matrix P such that A= PDP^{-1}, D= P^{-1}AP

    Suppose \lambda is an eigenvalue of A. That means that there exist a non-zero vector v such that Av= \lambda v. Let u= P^{-1}v. Since P is invertible, u is non-zero also and v= Pu. But now, Du= P^{-1}APu= P^{-1}Av= \lambda P^{-1}v= \lambda u.

    That is, \lambda is also an eigenvalue of the diagonal matrix D and, since a diagonal matix has its eigenvalues on its diagonal, is one of the numbers on the diagonal of D. Further the eigenvectors of a diagonal matrix are simply the standard basis vectors <1, 0, 0...>, etc.

    That means that set of all eigenvectors u, corresponding to the eigenvalues of D are independent and their number is equal to the dimension of the space.. Now, v= Pu and since P is invertible, it maps independent sets of vectors into independent sets of vectors. Since their number is equal to the dimension of the space, it follow that they are a basis for the space.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    Let u= P^{-1}v
    How is P^{-1} being multiplied by v?
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