Let $A$ be a diagonalizable $n \times n$ matrix, and suppose that $\lambda$ is an eigenvalue of $A$ with multiplicity $n$. I want to show that $A= \lambda I_n$.
I started by saying that the determinant $\det (A- \lambda I_n)$ yields a characteristic polynomial with root $\lambda$. Thus, the eigenvalues of A are just $\lambda$ with multiplicity $n$. To find the eigenvectors now, we need to find the null space of $A- \lambda I_n$.
This is where I get confused. In order for $A = \lambda I_n$ to be true, $A – \lambda I_n = 0$ must also be true. Moreover, let $\vec v$ be an eigenvector of $A$ belonging to an eigenvalue $\lambda$. Then $v ∈ N(A − λI)$. But I'm not sure how to tie these concepts together or solve for the eigenvectors of A.
Any help is greatly appreciated!
Best Answer
You have to use the diagonalizability somewhere, otherwise a matrix like $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda\end{pmatrix}$ provides a counterexample, since it has $\lambda$ as all its eigenvalues but is not equal to $\lambda I_n$.
Suppose $A$ is diagonalizable, say $A = P^{-1}DP$. Then we know that $D$ has all the eigenvalues of $A$ on its diagonal, but every eigenvalue is $\lambda$ so $D = \lambda I_n$, and therefore $A = P^{-1}(\lambda I_n) P = \lambda I_n (P^{-1}P) = \lambda I_n$ since the identity matrix commutes with every matrix.