It is not always the case that an $n\times n$ matrix has $n$ distinct eigenvectors. For example, consider the matrix \begin{equation} B = \begin{pmatrix} 3 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5\\ \end{pmatrix}, \end{equation} whose eigenvectors are again clearly the standard basis. But what are the eigenvalues of $B$? Again, the answer is obvious: $3$ and $5$. In such cases, the eigenvalue $3$ is a degenerate eigenvalue of $B$, since there are two independent eigenvectors of $B$ with eigenvalue $3$. Degenerate eigenvalues are also referred to as repeated eigenvalues. In this case, one also says that $3$ is a repeated eigenvalue of multiplicity $2$.

However, that's not the whole story. Setting \begin{equation} v = \begin{pmatrix}1\\0\\0\\\end{pmatrix}, \qquad w = \begin{pmatrix}0\\1\\0\\\end{pmatrix}, \end{equation} we of course have \begin{equation} Bv = 3v, \qquad Bw = 3w . \end{equation} But, by the linearity of matrix operations, we also have \begin{equation} B(av+bw) = a\,Bv+b\,Bw = 3(av+bw), \end{equation} so that any linear combination of $v$ and $w$ is also an eigenvector of $B$ with eigenvalue $3$. The eigenspace of $B$ corresponding to eigenvalue $3$ is therefore a 2-dimensional vector space (a plane), rather than the 1-dimensional eigenspaces (lines) that occur when the eigenvalues are distinct.