# $n$ linearly independent eigenvectors $\implies$ $n$ distinct eigenvalues?

Suppose $$A$$ is an $$n\times n$$ matrix. Is it true that $$n$$ linearly independent eigenvectors $$\implies$$ $$n$$ distinct eigenvalues? I know the other direction is true since the algebraic multiplicity of each eigenvalue is $$1$$.

Consider the identity matrix of which we can easily find $$n$$ linearly independent eigenvectors but the only eigenvalue is $$1$$.