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?
- A is symmetric, so all its Eigenvalues are real.
- A is non-singular, so all of its Eigenvalues are non-zero.
- 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.
Best Answer
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?