<s>[INST] Find the determinant and possible sizes of a matrix $A$, which satisfies the equation $A^2 - 2A + 4E = 0$ Let $A \in M_n(\mathbb{R})$ satisfy the equation:
$$A^2 - 2A + 4I = 0$$
(a) What is $\det A$? (3 points)
(b) What are possible sizes of a matrix $A$? (6 points)
My solution
We can derive 2 useful equations from the given one:
1) $A^2 = -2(2I - A)\Rightarrow (\det A)^2 = \det (-2(2I - A)) = (-2)^n \cdot \det (2I - A)$
2) $A(2I - A) = 4E \Rightarrow \det A \cdot \det (2I - A) = 4^n = 2^{2n}$
From the first we see that the size $n$ must be even. This is the answer to the question (b).
Now, let's build the system of equations, where $x = \det A$, $y = \det (2I - A)$. Also, we can replace $(-2)^n$ with $2^n$, because we know, that $n$ is even:
$$\begin{cases} x^2 = 2^n \cdot y, \\ xy = 2^{2n} \end{cases} \Rightarrow \begin{cases} y = x^2\cdot2^{-n}, \\ x\cdot(x^2\cdot2^{-n}) = 2^{2n} \end{cases} \Rightarrow x = (2^{2n}\cdot2^n)^{1/3} = 2^n$$
So, $\det A = 2^n$ and it depends on the size of the matrix. This is the answer to the question (a).
The question
Is this solution correct? I have big doubts about it, because the question (b) is worth twice more points than (a), but happened to be much easier and also order of tasks suggests, that we should solve (a) first.
 [/INST] Using eigenvalues: Let $x$ be an eigenvector of $A$ associated with $\lambda$.  Note that
$$
0 = 0x = (A^2 - 2A + 4I)x = (\lambda^2 - 2\lambda + 4)x
$$
so, each $\lambda$ must be a root of $x^2 - 2x + 4$, which is to say that each eigenvalue is either $1 + \sqrt{3}i$ or $1 - \sqrt{3} i$.  Since $A$ is real, its eigenvalues come in conjugate pairs.  So, $n$ must be even, and the eigenvalues $1+\sqrt{3}i$ and $1 - \sqrt{3}i$ must have the equal multiplicities $n/2$.
The determinant of $A$ is the product of its eigenvalues, so 
$$
\det(A) = [(1 + \sqrt{3}i)(1 - \sqrt{3} i)]^{n/2} = 4^{n/2} = 2^n
$$
</s>