Not diagonalizable matrix

Question

I have the following matrix of dimension $n\times n$:

$A = \begin{pmatrix} a_{1} & 1 & \\ & a_1 & 1 & \\ & & \ddots & 1\\ & && a_{1} \\ \end{pmatrix}$

Where $a_1 \in \mathbb{C}$. I need to show that $A$ is not diagonalizable. And here is what I have done:

$\textbf{det }(A-I\lambda) = (a_1-\lambda)^n$

Now, what this means is that there is only one eigenvalue for this matrix. This, being $\lambda = a_1$. Therefore, there is only one eigenspace associated, meaning that the matrix cannot be further decomposed. Since the matrix cannot be further decomposed, it cannot be diagonalized.

Is this enough to prove that $A$ is not diagonalizable?

Thanks for the help.

Answer 1

Knowing that a matrix $A$ is diagonalizable if and only if the minimal polynomial factors out completely with $m(\lambda) = 1$ where $m(\lambda)$ denote the algebraic multiplicity of an eigenvalue,

To conclude that $A$ is not diagonalizable we just have to note that the minimal polynomial $m_{A}(t)$ of $A$ is $m_{A}(t) = (1-t)^{n}$

(Same as the characteristic polynomial)

Answer 2

Assuming that $n>1,A$ is not diagonal. There is exactly one eigenvalue, $a_1,$ which has algebraic multiplicity $n.$ A basis for the corresponding eigenspace is the vector with 1 as the first entry and all other entries 0. Thus the dimension of the eigenspace is 1. Since $1 \ne n$ the matrix $A$ is not diagonalizable.

Answer 3

I think you can show it directly that $A$ is not diagonazible. If $A$ is diagonazible then you can find an invertible matrix $T$ s.t. $T^{-1}AT=a_1I$ where $a_1$ is the only eigenvalue as you said. So, $$AT=T(a_1I)=a_1T\Rightarrow A=a_1TT^{-1}=a_1I$$ but clearly that $A\neq a_1I$. Contradiction.