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Let $A$ $\in M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases when $A$ is diagonalizable over $\mathbb R$:

a. when $A^2 = A$;

b. when $(A - 3I)^2 = 0$;

c. when $A^2 + I = 0$.

I could eliminate c. by using the equation $\lambda^2+1$ and showing if a matrix has to satisfy this then it has to be diagonal.

I am also in doubt whether all the eigen values of $A$ satisfy this or not.

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How do you argue that if $A^2 + I = 0$ then $A$ has to be diagonal? To the contrary, in this case it cannot be diagonal. –  Marc van Leeuwen Jan 8 at 9:50
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3 Answers

up vote 5 down vote accepted

Let $A\in \mathcal M_3(\mathbb R)$.

  • If $A^2=A$, then $A^2-A=0_{\mathcal M_3(\mathbb R)}$, therefore the minimal polynomial $m_A$ of $A$ is in $\{x, x-1, x(x-1)\}$. Since any matrix is diagonalizable if, and only if, its minimal polynomial has only linear factors, it follows that $A$ is diagonalizable.
  • Consider $A=\begin{bmatrix} 3 & 1 & 0\\0 & 3 & 0\\ 0 & 0 & 3\end{bmatrix}.$
  • If $A^2+I=0$, then the minimal polynomial of $A$ is $\,\_\_\_\_\_\_$ which has degree $\_\_$, however $A$'s characteristic polynomial has degree $\_\_$ and therefore has a real root. This means such a real matrix cannot exist.
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Not sure what the question is, so I will take a stab at the answer. I am assuming you want to know what additional condition is needed to ensure that it is diagonalizable.

a) $A^2=A$

In this case the eigenvalues of $A$ have to be zero or one. Since we need three eigenvalues, at least one of them is repeated. Now Jordon block of the repeated eigenvalue has to be diagonal for (a) to hold. Hence $A$ is diagonalizable.

b) $(A-3I)^2=0$

This need not be diagonalizable, unless $A-3I=0$. If this is not true then A has to be similar to $$A = \begin{pmatrix}3 & 1 & 0\cr 0 & 3 & 0\cr 0 & 0 & 3\end{pmatrix}$$ which can't be diagonalized.

c) A^2+I = 0

This can't be diagonalized in the field of real numbers. This is because the eigenvalues of any diagonal matrix are the values in the diagonals. However, $A$ has eigenvalues $\pm \sqrt{-1}$.

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For b), note that $A$ is assumed not to be diagonal so the non-diagonalizable is the only possible one. –  Erick Wong Jan 8 at 8:16
    
Also, c) is impossible for the stated dimensions of $A$. –  Erick Wong Jan 8 at 17:50
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Yet another take on this is provided in my answer to this question's evil twin, Matrix diagonalizable or not!

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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This should really be pointed out in a comment. –  robjohn Jan 8 at 12:06
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