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

Question

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 $1$.

Answer 1

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