Conditions for distinct eigenvalues of a matrix Given a matrix $A \in \mathbb{R}^{n\times n}$,

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*Under what conditions (e.g. symmetric, positive definite, etc.) will it have $n$ distinct eigenvalues?

*Under what conditions will it have $n$ distinct linearly independent eigenvectors? (eigenvalues corresponding to these eigenvectors need not be distinct)

*Is $\text{rank}(A) = n$ always equivalent to having $n$ distinct eigenvalues or eigenvectors?

 A: *

*A matrix has distinct eigenvalues iff the discriminant of its characteristic polynomial is nonzero. This is pretty tedious to compute away from the $2 \times 2$ or possibly the $3 \times 3$ case, though. But it can sometimes be theoretically useful to know that "almost all" matrices have this property (more precisely, that the set of matrices which do not have this property has measure zero).


*A matrix has a basis of eigenvectors iff it's diagonalizable (this is more or less just a restatement of definitions). By the spectral theorem, a sufficient but not necessary condition for this to be true is that the matrix is normal. A different sufficient but not necessary condition is that the eigenvalues are distinct.


*No, this is neither necessary nor sufficient. A zero matrix is diagonalizable but has rank zero, a Jordan block with nonzero eigenvalue is full rank but not diagonalizable, the matrix $\text{diag}(0, 1)$ has distinct eigenvalues but does not have full rank, and again a Jordan block with nonzero eigenvalue has full rank but does not have distinct eigenvalues.
