a) There exists a square matrix $A$ with rational entries and dimension $2019 \times 2019$ such that $A^3 + 6A^2 - 2I = 0$?
b) And a square matrix $A$ with rational entries and dimension $2019 \times 2019$ such that $A^4 + 6A^3 - 2I = 0$?
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Im trying to answer this questions. For the first one, i thought the following:
$A^2\cdot\dfrac{(A + 6I)}{2} = I \Longrightarrow det(A)^2 = \dfrac{2^{2019}}{det(A+6I)}$
Then, because the matrix has rational entries, the determinant of $A$ must be rational, and then:
$det(A + 6I) = k^2\cdot(2^{2n + 1})$, where $k \in \mathbb{Q^*}$, $n \in \mathbb{Z}$.
And we obtain $det(A) = \dfrac{2^{1009-n}}{k}$.
I was hopping i could get a contradiction out of this but i'm not able to find one. Any hints? Thanks! (On the question b) i tried the same thing but didn't work).
THIS IS WRONG, PLEASE CHECK EDIT
(a) Let, $$ A = \begin{pmatrix} 0 & 0 & 2 \\ 1 & 0 & 0 \\ 0 & 1 & -6 \end{pmatrix} $$ Then a matrix that'll work is, $$ \begin{pmatrix} A & 0 \\ 0 & I_{2016} \end{pmatrix} $$ (b) Let, $$ B = \begin{pmatrix} 0 & 0 & 0 & 2 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -6 \end{pmatrix} $$ Then a matrix that'll work is, $$ \begin{pmatrix} B & 0 \\ 0 & I_{2015} \end{pmatrix} $$ The $A$ and $B$ are called companion matrices of the corresponding polynomials.
EDIT : As pointed out in the comments, the above solution is wrong. I sort of typed it in a hurry, and didn't check if it actually worked or not. I apologise for that. I'm keeping this for context of the comments.
Having said that, as pointed out by @greg in the comments, $I \otimes A$ does work for (a) and (b) is impossible, as pointed out by @JyrkiLahtonen.