Existence of matrix which satisfies the equation

Question

Does there exist a $3\times 3$ real matrix $A$ which satisfies the following equation: $$ A^2 + A +7I = 0$$ Where $I$ is $3 \times 3 $ unit matrix. I suppose that there doesn't exist such matrix, but i don't now how to find contraddiction. From this equation we can find eigenvalues. Suppose $Av = \lambda x$, then $$ (A^2 + A +7I )(v) = (\lambda^2 + \lambda + 7)(v) = 0$$ And finally we see that eigenvalues are: $$\lambda_{1, 2} = \dfrac{-1 \pm 3i\sqrt{3}}{2} $$ But is it usefull that they are complex? And more generally, when there exists $p \times p $ matrix $A$ whiche is ''zero'' of $$ Q(x) = x^k + a_{k -1}x^{k-1} + \ldots + a_0$$ By this i mean the following equation $$ A^k + a_{k-1}A^{k-1}+ \ldots + a_0I = 0$$ What conditions do we need on $p$, and coeffitients of polynomial?

Answer 1

If there was such a matrix $\;A\;$ then the polynomial $\;x^2+x+7\;$ is its minimal polynomial, as this is an irreducible quadratic. This means that all the matrix's eigenvalues are among the roots of this polynomial, which is impossible as any real polynomial of odd degree has a real root...can you now complete the argument and prove the impossibility of existence of such a real $\;3\times3\;$ matrix?