Classifing the (square) real matrices by invertibilty and diagonalizability Consider square matrices of real entries.
They can be classified into two categories by invertibility (invertible / not invertible), and they can also be classified into three by diagonalizabilty (not diagonalizable / diagonalizable with distinct eigenvalues / diagonalizable with repeated eigenvalues).
So, they can be classified into six categories.
I've made a Venn diagram and tried to examplify each cases for clear understanding of this topic.
(Note that the red rectangle is included in the blue one.)
I've filled all the cases but the fifth case when it is not invertible, diagonalizable with repeated eigenvalues.
Can anyone give me an example for (5)?

Below is the calculations for each cases;
(1) $A=\begin{bmatrix}3&0\\0&2\end{bmatrix}$
$$|A-\lambda I|=(3-\lambda)(2-\lambda)$$
$$\lambda_1=2$$
$$A-\lambda_1I=\begin{bmatrix}1&0\\0&0\end{bmatrix}$$
$$x_1=\begin{bmatrix}0\\1\end{bmatrix}$$
$$\lambda_2=3$$
$$A-\lambda_2I=\begin{bmatrix}0&0\\0&-1\end{bmatrix}$$
$$x_2=\begin{bmatrix}1\\0\end{bmatrix}$$

*

*$A$ has no zero eigenvalues ; invertible.

*$A$ has distinct eigenvalues ; diagonalizable

(2) $A=\begin{bmatrix}4&0&-2\\2&5&4\\0&0&5\end{bmatrix}$
$$|A-\lambda I|=(4-\lambda)(5-\lambda)^2$$
$$\lambda_1=4$$
$$A-\lambda_1I=\begin{bmatrix}0&0&-2\\2&1&4\\0&0&1\end{bmatrix}$$
$$x_1=\begin{bmatrix}1\\-2\\0\end{bmatrix}$$
$$\lambda_2=\lambda_3=5$$
$$A-\lambda_2I=\begin{bmatrix}-1&0&-2\\2&0&4\\0&0&0\end{bmatrix}$$
$$x_2=\begin{bmatrix}0\\1\\0\end{bmatrix},\quad x_3=\begin{bmatrix}2\\0\\-1\end{bmatrix}$$

*

*$A$ has no zero eigenvalues ; invertible.

*$A$ has repeated eigenvalues but it has sufficient eigenvectors ; diagonalizable

(3) $A=\begin{bmatrix}3&1\\0&3\end{bmatrix}$, $B=\begin{bmatrix}2&-1\\1&0\end{bmatrix}$
$$|A-\lambda I|=(3-\lambda)^2$$
$$\lambda_1=\lambda_2=3$$
$$A-\lambda_1I=A-\lambda_2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$
$$x_1=\begin{bmatrix}1\\0\end{bmatrix}$$
$$|B-\lambda I|=(2-\lambda)(-\lambda)+1=(\lambda-1)^2$$
$$\lambda_1=\lambda_2=1$$
$$B-\lambda_1I=B-\lambda_2I=\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$$
$$y_1=\begin{bmatrix}1\\1\end{bmatrix}$$

*

*$A$ and $B$ have no zero eigenvalues ; invertible.

*$A$ and $B$ have repeated eigenvalues and it has insufficient eigenvectors ; not diagonalizable

(4) $A=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $B=\begin{bmatrix}1&1\\1&1\end{bmatrix}$, $C=\begin{bmatrix}2&3\\4&6\end{bmatrix}$
$$|A-\lambda I|=(1-\lambda)(-\lambda)$$
$$\lambda_1=0$$
$$A-\lambda_1I=\begin{bmatrix}1&0\\0&0\end{bmatrix}$$
$$x_1=\begin{bmatrix}0\\1\end{bmatrix}$$
$$\lambda_2=1$$
$$A-\lambda_2I=\begin{bmatrix}0&0\\0&-1\end{bmatrix}$$
$$x_2=\begin{bmatrix}1\\0\end{bmatrix}$$
$$|B-\lambda I|=(1-\lambda)^2-1=\lambda(\lambda-2)$$
$$\lambda_1=0$$
$$B-\lambda_1I=\begin{bmatrix}1&1\\1&1\end{bmatrix}$$
$$y_1=\begin{bmatrix}1\\-1\end{bmatrix}$$
$$\lambda_2=2$$
$$B-\lambda_2I=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}$$
$$y_2=\begin{bmatrix}1\\1\end{bmatrix}$$

*

*$A$ and $B$ has zero eigenvalue ; not invertible.

*$A$ and $B$ has distinct eigenvalues ; diagonalizable

(5) Any examples?
(6) $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$$|A-\lambda I|=\lambda^2$$
$$\lambda_1=\lambda_2=0$$
$$A-\lambda_1I=A-\lambda_2I=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$
$$x_1=\begin{bmatrix}1\\0\end{bmatrix}$$

*

*$A$ has zero eigenvalue ; not invertible.

*$A$ has repeated eigenvalues and it has insufficient eigenvectors ; not diagonalizable.

 A: The $2\times2$ matrix $$B=\begin{bmatrix}
2 &3\\
4&6
\end{bmatrix},$$ is diagonalizable with distinct eigenvalues. The diagonal matrix is
$$D_B=\begin{bmatrix}
0 & 0\\
0 & 8
\end{bmatrix}.$$ The eigenvalues are then $$\lambda_1=0,\quad\lambda_2=8.$$ The eigenvalues are thus distinct and $B$ is in your category-$(4)$.

EDIT:  For category-$(5)$ we have a problem, for the $2\times2$ case. Note that if a $2\times2$ matrix is diagonalizable but NOT invertible, then one of its eigenvalues must be $0$. Since you also want repeated eigenvalue, both eigenvalues must be $0$ and the diagonal form of our desired matrix, say $A$ is
$$D_A=O_2=\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix},$$ the $2\times2$ matrix of zeros. Now, recall that to diagonalize any matrix, we take a suitable invertible matrix $P$ such that
$$A=PD_A P^{-1}.$$ But since $D_A=O_2$, $A=O_2$ for any $P$.
Thus, the only matrix in your category-$(5)$ is $O_2$.
A: 
not invertible, diagonalizable with distinct eigenvalues

Example : $A=\begin{pmatrix}0 &0\\0&1\end{pmatrix}$
Not invertible as $\det(A) =0$
Diagonalizable as already diagonal.
Eigenvalues are distinct : $0$ and $1$
Edit : For $5$ we can take $O=\begin{pmatrix}0 &0\\0&0\end{pmatrix}$
Diagonalizable, not nvertible but eigenvalue is $0$ with multiplicity $2$
Non Invertible, repeated eigenvalues, diagonalizable
Claim: Only $2×2$  matrix satisfying the above $3$ properties must be a null matrix.
Proof: $A$ is not Invertible means $0$ is an eigenvalue of $A$ .
Since $A$ has repeated eigenvalues hence $0$ is the only eigenvalue  of $A$ with multiplicity $2$.
Then $p(x) =x^2$ is characteristics polynomial. Hence $p(A) =A^2=0$ .
Hence $ A$ is nilpotent matrix . We know a nilpotent matrix is diagonalizable iff it is the $0$ matrix.

Conclusion : Only possible example for $5$ is the null matrix.
