matrix is not diagonazible if algebraic multiplicity = number of rows of matrix except for diagonal matrix

Question

Is it possible to say for real matrices, that if the algebraic multiplicity is equal to the number of rows of a matrix, that the matrix is never diagonazible?

But this should only be valid for non-diagonal matrices.

Example:

$ A=\begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}$

$\lambda_1=\lambda_2 = \lambda_3 = 1$

algebraic multiplicity = 3

but geometric multiplicity will never be 3, because $rank(A-\lambda E)$ is at least 1:

$geometric~multiplicity= 3-rank(A-\lambda E)< 3$

Answer 1

Yes. You can also prove it as follows.

Suppose that $A \in M_n(\mathbb{R})$ is diagonalizable with eigenvalue $\lambda$ having algebraic multiplicity $n$. Then there exists an invertible matrix $P$ such that $P^{-1} A P = \lambda I$, where $I$ is the identity matrix. But then $$A = P \lambda I P^{-1} = \lambda P P^{-1} = \lambda I. $$ Therefore $A$ itself is diagonal.