Let $n \ge 2 $ be an integer. Suppose that A is an $n \times n$ matrix and that $\lambda_1$, $\lambda_2$ are eigenvalues of A with corresponding eigenvectors $v_1$, $v_2$ respectively. Prove that if $v_1$, $v_2$ are linearly dependent then $\lambda_1 = \lambda_2$.
I have an intuition as to why this is true, but am having difficulty formalizing a proof. What I have doesn't seem tight enough.
If $v_1$ and $v_2$ are linearly dependent then $v_1$ lies in the span of $v_2$. If two eigenvectors lie in the span of one another then only one of them is required in order to form a basis of the eigenspace.
All eigenvalues correspond to a single $n\times 1$ eigenvector or a set of $n\times 1$ linearly independent vectors.
Since $v_1$ and $v_2$ are linearly dependent, we know that there can only be one eigenvalue that corresponds to the single eigenvector.
Thus $\lambda_1$ must equal $\lambda_2$.
Any thoughts or criticism are welcome.