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Consider the following problem. (The question and the reasoning remains identical if negative is replaced by positive everywhere.)

Suppose A is an $n\times n$ symmetric matrix with distinct eigenvalues. Suppose further that A is non-singular and negative semidefinite. Is A a negative definite matrix?

  1. A is symmetric, so all its Eigenvalues are real.
  2. A is non-singular, so all of its Eigenvalues are non-zero.
  3. A is negative semidefinite, so all of its Eigenvalues are non-positive.

Combining 2 and 3, A is negative definite.

My question is, what is the significance of the fact that all Eigenvalues are distinct in the context of this question? Am I missing something?

Thanks in advance.

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    $\begingroup$ You're not missing anything; your answer is correct. $\endgroup$ – Omnomnomnom Aug 25 '16 at 14:42
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There is no significance of the fact that all eigenvalues are distinct in the context of this question; it is a superfluous hypothesis. A symmetric real matrix is always diagonalisable, whether or not its eigenvalues are distinct.

Maybe a failed attempt to put the student at ease by giving a supposedly reassuring hypothesis?

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