I was asked the following question:
Let matrix $X \in \Bbb R^{3 \times 3}$ be such that $$X^6+X^5-2X^4-X^2-X+2=0$$ Which of these is not a possible value of the trace of $X$?
a. $-4$
b. $-2$
c. $0$
d. $2$
Now, all I know about Trace$(X)$ is its definition, and a few places where it pops up. But I don't have any idea how to connect a matrix equation with its trace.
I tried to factorise the given polynomial and found that $$x^6+x^5-2x^4-x^2-x+2=(x-1)^2 (x+1) (x+2) (x^2+1)$$
So, $X$ can be $I$, $-I$ and $-2I$ where $I$ is the identity matrix. The trace corresponding to these solutions are $3$, $-3$ and $-6$ respectively, none of which are in the options provided.
I don't have any other ideas about this question. Any help, even a hint, would be appreciated. Also, feel free to edit the tags if necessary. Thanks in advance.
$M_3(\mathbb{R})$ is not an integral domain (the product of two non-zero matrices can be zero) and $X$ can be more matrices than you thought.
However, we have a polynomial that cancels $X$, with roots $1, 1, -1, -2, i, -i$, we know that $\operatorname{Sp}(X) \subset \{1, -1, -2, i, -i\}$, with $1$ that can have a multiplicity of $2$. Since the sum of the eigenvalues is the trace, and we have $3$ eigenvalues (because $X \in M_3(\mathbb{R})$), the possible values for the trace are the possible sums you can make by choosing $3$ eigenvalues (you can choose $1$ twice, and if you choose $i$ you must choose $-i$ because complex eigenvalues are conjugate). Hence, $2$ and $-4$ are not possible values for $\operatorname{tr}(X)$.