Matrix Algebra Question (Linear Algebra) Find all values of $a$ such that $A^3 = 2A$, where
$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$
The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.
 A: Hint:
$$A^3 = \begin{bmatrix} -4 + 2 (2 - a) & -2 (-4 + 2 a) + 2 (-2 + a^2) \\-2 + (2 - a) a &  4 - 2 a + a (-2 + a^2) \end{bmatrix}$$
$$2A = \begin{bmatrix} -4 & 4 \\-2 &  2 a \end{bmatrix}$$
Can you take it from here?
Spoiler
Hover over the blue area for the result.

 $a=2$ (Throw out the other solutions from equating the lower right items to solve for $a$).

A: Here's a different approach that avoids matrix multiplication in favor of determinants:
We want to find $a$ such that $A^3=2A$. This is equivalent to $A^3-2A=0$, and to $(A^2-2I)A=0$.
Thus either $A=0$, which is impossible, or $A^2-2I$ is a zero divisor.
Thus \begin{align}0&=\det (A^2-2I) \\ &= \det(A+\sqrt2I)\det(A-\sqrt2 I),\end{align} so one of these determinants must be $0$. That is, $(-2+\sqrt 2)(a+\sqrt 2)+2=0$ or $(-2-\sqrt 2)(a-\sqrt 2)+2=0$. These have the same sole solution: $a=2$.
Note that we get the value of $A^2$ too: plugging in $a=2$ gives $\det A = -2\ne 0$, so $A$ is not a zero divisor, so $A^2-2I=0$, so $A^2=2I$.
A: The matrix $A$ is clearly never a multiple of the identity, so its minimal polynomial is of degree$~2$, and equal to its characteristic polynomial, which is $P=X^2-(a-2)X+2-2a$. Now the polynomial $X^3-2X$ annihilates $A$ if and only if it is divisible by the minimal polynomial $P$. Euclidean division of $X^3-2X$ by $P$ gives quotient $X+a-2$ and remainder $(a^2-2a)X-(a-2)(2-2a)$. This remainder is clearly$~0$ if $a=2$ and for no other values of $a$, so $A^3=2A$ holds precisely for that value.
You can in fact avoid doing the Euclidean division by observing, since $X$ divides $X^3-2X$, that for $P$ to divide $X^3-2X$, unless $P$ itself has a factor $X$ (which happens for $a=1$ but divisibility then fails anyway), $P$ must divide (and in fact be equal to) $(X^3-2X)/X=X^2-2$; this happens for $a=2$.
A: Any element of the set $\left\{\lambda\right\}$ of $A$ eigenvalues satisfy the equations
$$
\lambda^{2} = \left(a - 2\right)\lambda + 2\left(a - 1\right)\,,
\quad
\lambda^{3} = 2\lambda\,,
\quad
\sum_{\lambda}\lambda = a - 2
$$
Then, $\sum_{\lambda}\lambda^{3} = 2\left(a - 2\right)$. Also,
\begin{align}
\sum_{\lambda}\lambda^{3}
&=
\sum_{\lambda}\lambda\left[\left(a - 2\right)\lambda + 2\left(a - 1\right)\right]
=
\left(a - 2\right)\sum_{\lambda}\lambda^{2}
+
2\left(a - 1\right)\left(a - 2\right)
\\[3mm]&=
\left(a - 2\right)^{3} + 6\left(a - 1\right)\left(a - 2\right)
\end{align}
We get
$$
\left(a - 2\right)\left[\left(a - 2\right)^{2} + 6a - 8\right] = 0
$$
$$
\color{#ff0000}{\large%
a = 2\,,\qquad\qquad a = -1 \pm \sqrt{5\,}}
$$
