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  • 5 years ago

prove that if a linear operator T on a space V of dimension n has n distict eigenvalues then it can be represented by a diagonal matrix.

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  1. watchmath
    • 5 years ago
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    Hint: if T has n disticnt eigenvalues then T is diagonalizable.

  2. anonymous
    • 5 years ago
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    I'll add a little to this. How to think about this problem depends on your definition of diagonalizability (the defs are all equivalent, but the explanation will depend on which one we choose), I will take diagonalizable as meaning "there is a basis for our vector space consisting of eigenvectors for T". In this case, it comes down to showing that the fact that T has n (=dimension of the vector space) distinct eigenvalues means that it has n linearly independent eigenvectors, which would then form a basis. Try showing for starters that the eigenvectors for distinct eigenvalues are linearly independent. Let me know if you can't do this.

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