# A demonstration on diagonalizable matrix

Determine the truth or falsity of the following proposition.

If $A$ is a square matrix of order $n$ and is invertible, then $A$ is diagonalizable.

This proposition is false. For example, the matrix

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

is invertible, but not diagonalizable. Since it has only a single eigenspace $E_{\lambda =1}=\textrm{span}\left \{ \left ( 0,1 \right ) \right \}$.

How to demonstrate that falsehood in general?

But I do not find to demonstrate that falsehood in general.

I don't think there is a way to demonstrate this in general. Such statements are considered false, so long as you can provide a case that the given condition is wrong.

If A is a square matrix of order n is invertible, then A is diagonalizable

The above statement is false, because there are square matrices that don't match that condition such as:

$\left ( \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right )$, $\left ( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right )$ etc...

Although your example is enough to disprove the statement, here is an example for any $n\in \mathbb N$ $$\begin{pmatrix}\begin{array}{cc|c|c|c|c}1 & 1 & 0 & 0 & \dots & 0\\0 & 1 &0 & 0 & \dots & 0\\ \hline 0 & 0 & 1 & 0 & \ddots & \vdots\\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ 0 & 0 & 0 & 0 &0 & 1\end{array}\end{pmatrix}$$

The matrix is an identity matrix except that $a_{12}$ is equal to $1$, while not being diagonalizable.