Let $\mathbf{A}$ be an $n \times n$ matrix with two distinct eigenvalues $\lambda_1$ and $\lambda_2$. If the dimension of the eigenspace $E(\lambda_1)$ is $n-1$, then $\mathbf{A}$ is diagonizable.
Prove or disprove.
I know that any matrix with no repeated eigenvalues can be diagonalized. So my intuition leads me to believe this is a true statement, but I am not sure how to use the dimensionality of the eigenspace to justify my answer, or how I could go about proving it.
Best Answer
If $A$ had just one eigenvalue, it wouldn't be diagonalizable because its algebraic multiplicity would be equal to $\,n\,$ (dimension of the whole vector space), but its geometric multiplicity: $\dim E_{\lambda_1}=n-1<n$
However, since $A$ has two different eigenvalues, their algebraic multiplicity must be at least $1$. So, if $\dim\left(E_{\lambda_1}\right)=n-1\,$ then $\,\dim\left(E_{\lambda_2}\right)=1$.
This means all the vectors in both $E_{\lambda_1}\,$ and $\,E_{\lambda_2}$ are eigenvectors so $A$ is diagonalizable.