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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)?

Venn diagram

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.
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2 Answers 2

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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.

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  • $\begingroup$ Sorry my question was wrong ; Can you give me an example of (5) instead of (4)? $\endgroup$
    – govin
    Jul 8, 2022 at 15:27
  • $\begingroup$ The OP asked example of a matrix that is NOT invertible but diagonalizable with repeated eigenvalues. $\endgroup$ Jul 8, 2022 at 15:36
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    $\begingroup$ $A=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$ $\endgroup$ Jul 9, 2022 at 5:23
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    $\begingroup$ @govin : Please try to be specific as to what you are asking for. You cannot keep changing your question. Please specify ONE question properly and try to stick with it. $\endgroup$ Jul 9, 2022 at 5:24
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    $\begingroup$ @sadman-ncc Sorry I think I was no ready when I post this question. I will be careful next time. $\endgroup$
    – govin
    Jul 9, 2022 at 22:29
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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 $O_2$.

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    $\begingroup$ I'm really appreciate your nontrivial answer. $\endgroup$
    – govin
    Jul 8, 2022 at 15:10
  • $\begingroup$ But I'm sorry to notify you that my question was wrong ; Can you give me an example of (5) instead of (4)? $\endgroup$
    – govin
    Jul 8, 2022 at 15:28
  • $\begingroup$ Thanks for your modified answer. So in 2 by 2 case, the matrix of (5) should be the zero matrix. I think it's too trivial case. If we move to, say, 3 by 3, can we get some nontrivial example like (2)? $\endgroup$
    – govin
    Jul 9, 2022 at 4:58

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