Find out how many invertible and diagonal solutions $X^2-2X=0 $ has when $ X \in\Bbb{R}^{3\times 3}$ I want to find out how many invertible solutions and how many diagonal solutions matrix equation 
has when 
$ X \in\Bbb{R}^{3\times 3}.$ 
I have researched everywhere and can't seem to find the solution to this problem. I would be very thankful for a lead in the right way.
 A: If X is diagonal and invertible then none of the diagonal elements is zero.
Let us call the diagonal elements, $a$,$b$,and $c$. 
On the other hand $X^2 -2X$ is a diagonal matrix whose diagonal elements are $a^2-2a$,$b^2-2b$, and $c^2-2c$.
The only non-zero solutions for $a$,$b$ and $c$ are $$a=b=c=2$$ 
Therefore there is only one such matrix $$ X = \begin{bmatrix} 2&0&0\\0&2&0\\0&0&2\end{bmatrix}$$
A: $X^2-2X=X(X-2)=0$ hence $f$ is diagonalizable and if $\lambda$ is an eigenvalue of $X$, $\lambda=0$ or $\lambda=2$.
There are $2*2*2$ diagonal solutions (each diagonal coefficent is $0$ or $2$).
Now, if $X$ should be invertible, $0$ is not an eigenvalue, hence, since $X$ is diagonalizable, $X=2I$.
A: If $$0 = X^2 - 2X = X(X - 2I)$$ then the minimal polynomial $\mu_X(\lambda)$ divides $\lambda(\lambda - 2)$.
There are three possibilities:


*

*If $\mu_X(\lambda) = \lambda$ then $X = 0$.

*If $\mu_X(\lambda) = \lambda - 2$ then $X = 2I$.

*The last possibility is $\mu_X(\lambda) = \lambda(\lambda - 2)$.
Since $\mu_X$ is split into linear factors, $X$ is diagonalizable and $\sigma(X) = \{0,2\}$ so the only two options are
$$\pmatrix{0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}, \pmatrix{0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 2}$$
i.e. $X$ has to be similar to one of the above two matrices.


Since you are only looking for invertible solutions, the only solution is 
$$X =2I= \pmatrix{2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2}$$
A: If $X$ is invertible, then multiplying each side of
$$
  X^2 - 2X = 0
$$
by the inverse of $X$ gives us that
$$
  X - 2I = 0
$$
and so
$$
  X = 2I.
$$
For the case where $X$ is diagonal, let
$$
  X = \begin{pmatrix} a & 0 & 0 \\
                  0 & b & 0 \\
                  0 & 0 & c \end{pmatrix}.
$$
Then we have that
$$
  0 = X^2 - 2x = \begin{pmatrix} a^2 & 0 & 0 \\
                                 0 & b^2 & 0 \\
                                 0 & 0 & c^2 \end{pmatrix}
 - 2 \begin{pmatrix} a & 0 & 0 \\
                     0 & b & 0 \\
                     0 & 0 & c \end{pmatrix}
 = \begin{pmatrix} a^2 - 2a & 0 & 0 \\
                   0 & b^2 - 2b & 0 \\
                   0 & 0 & c^2 - 2c \end{pmatrix}
$$
and so we have a solution if and only if $a^2 - 2a = b^2 - 2b = c^2 - 2c = 0$. This is equivalent to each of $a$, $b$, and $c$ being either $0$ or $2$, and so we see that there are $8$ solutions.
