Let $A\in M_3[\mathbb{R}]$ such that $A^{2019}+A=$

Question

Let $A\in M_3[\mathbb{R}]$ such that $A^{2019}+A=\begin{bmatrix} 2 & 2 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 2\end{bmatrix}.$ Find $tr(A^{2019})$ and $\det(A).$

My work:

Let characteristic equation of $A$ be $x^3-tx^2+sx-d=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)=0$

$\implies x^{2019}=Q(x)(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)+ax^2+bx+c,\text{ where }a\lambda_{i}^2+b\lambda_i+c=\lambda_i^{2019}\,\,\forall\,\, i=1,2,3.$

$\implies A^{2019}+A=aA^2+(b+1)A+cI_3=\begin{bmatrix} 2 & 2 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 2\end{bmatrix}$

$\implies a\times tr(A^2)+b\times tr(A)+3c=6$

$\implies a\sum\lambda_i^2+(b+1)\sum\lambda_i+3c=6$

$\implies \sum (a\lambda_i^2+(b+1)\lambda_i+c)=6$

Answer 1

Outline of a solution:

Let $p(x)$ denote the polynomial $p(x) = x^{2019} + x$. First, show the following:

• If $A$ has linearly independent eigenvectors $v,w$, then the same holds for $q(A)$ (for any polynomial $q(x)$, but particularly for $q(x) = p(x) = x^{2019} + x$)
• $p(A)$ does not have a linearly independent pair of eigenvectors (all eigenvectors are a multiple of $(1,0,0)$)
• By contrapositive, $A$ cannot have a linearly independent pair of eigenvectors
• If $A$ had two distinct eigenvalues, then it would also have a linearly independent pair of eigenvectors. By contrapositive, $A$ has only one eigenvalue.

Thus, $A$ has exactly one eigenvalue. Let $\lambda$ be this eigenvalue. $p(\lambda)$ must be an eigenvalue of $p(A)$, but the only eigenvalue of $p(A)$ is $2$. Thus, $\lambda$ satisfies $p(\lambda) = 2$, i.e. $$ \lambda^{2019} + \lambda = 2. \tag{*} $$ The non-real, complex eigenvalues of $A$ necessarily come in conjugate pairs, but $A$ has only one eigenvalue. So, this eigenvalue must be real. So, $\lambda$ must be a real solution to the equation (*) above. (Alternatively, $A$ is a $3 \times 3$ matrix, and every matrix with an odd size has at least one real eigenvalue).

The only real solution to $(*)$ is $\lambda = 1$, so this is the sole eigenvalue of $A$.

To find $\operatorname{tr}(A^{2019})$, note that the trace of a matrix is the sum of its eigenvalues, so that $$ \operatorname{tr}(A^{2019}) = \operatorname{tr}(p(A) - A) = \operatorname{tr}(p(A))- \operatorname{tr}(A) = 6 - (1 + 1 + 1) = 3. $$ To find $\det(A)$, note that the determinant of a matrix is the product of its eigenvalues, so that $\det(A) = 1 \cdot 1 \cdot 1 = 1$.

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The fact that $p(\lambda) = 2$ has exactly one real solution may be argued with the help of calculus. We find that the derivative of $p$ is given by $p'(x) = 2019 x^{2018} + 1$. However, $x^{2018}$ is an even power of a real number and hence non-negative, which means that $p'(x)$ is positive for all $x$, which means that $p(x)$ is strictly increasing over all $x \in \Bbb R$. Thus, $p(x) = 2$ has at most one solution.

On the other hand, plugging in $x = 1$, we easily see that $p(1) = 2$.

In order to determine that $x = 1$ is the solution to $p(x) = 2$, one could apply the rational root theorem to deduce that the only possible rational solutions to the equation are $\pm 1, \pm 2$, then check these possibilities.

Answer 2

You can use that if $\lambda$ is an eigenvalue of $A$ with eigenvector $u$ then: $$ (A^{2019} + A) u = (\lambda^{2019} + \lambda) u $$ so $\lambda^{2019} + \lambda$ is an eigenvalue of $A^{2019} + A$.

$A^{2019} + A$ has $2$ as a triple eigenvalue. These means the eigenvalues of A satisfy: $$ \eta^{2019} + \eta - 2 = 0 $$ so $\eta = 1$.

The trace of $A^{2019}$ is the sum of the eigenvalues so: $$ tr(A^{2019}) = 3 \eta^{2019} = 3 $$ and the determinant of $A$ is the product of eigenvalues, which is $1$.