if A is a matrix of size nxn, and if it has n linearly independent eigenvectors associated to a single eigenvalue, A is a diagonal matrix

Question

I need help on proving the following affirmation:
if A is a matrix of size nxn, and if it has n linearly independent eigenvectors associated to a single eigenvalue, A is a diagonal matrix

I have gotten to the result that a linear combination $v$ of $v_1, \ldots, v_n$ must be also an eigenvector associated with $\lambda$
being $v_1, \ldots , v_n$ the eigenvectors mentioned above and $\lambda$ the eigenvalue
but I don't know how to take that A must be diagonal from this

Answer 1

Hint: let that single eigenvalue of $A$ be $\lambda$ and let $v$ be any vector in $\Bbb{R}^n$. $v$ must be a linear combination of the eigenvectors of $A$ (do you see why?), so we must have $v = \lambda v$. What is the matrix representation of scalar multiplication by $\lambda$?