# Find $\operatorname{Tr}(A^{2018})$ if $\det(A^2-2018I_2)=0$

Find $\operatorname{Tr}(A^{2018})$ if $\det(A^2-2018I_2)=0, A\in M_2(\mathbb{Q})$

My attempt:

Let $A^2=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ then $B=A^2-2018I_2=\begin{bmatrix}a-2018&b\\c&d-2018\end{bmatrix}$

then $\det(B)=(a-2018)(d-2018)-bc$ then we have: $a = 2018$ or $d=2018$ AND $b=0$ or $c=0$. And I took the possible candidate for $A^2$ when all of these happen at the same time so: $a=2018,d=2018,b=0,c=0\implies \operatorname{Tr}(A^{2018})=2\times2018^{1009}$ by induction.

Did I do everything right? I feel like my solution is not really that good.

Also can you give me some advice that might help me in these kind of situation with problems like this?

• How do you get $\operatorname{det}(B) = (a-2018)(d-2018) = bc$? Shouldn't it be $\operatorname{det}(B) = (a-2018)(d-2018) - bc = 0$? – Sambo Jul 8 '18 at 17:49
• @Sambo yeah it's $0$ but I added $bc$ both sides. – C. Cristi Jul 8 '18 at 17:50
• Why must one of $b$ or $c$ be $0$? If a sum is zero, it is not guaranteed that both terms are zero. – Michael Burr Jul 8 '18 at 17:52
• But then on one side you should have $\operatorname{det}(B) + bc$ – Sambo Jul 8 '18 at 17:53
• Here is one approach: Read up on what eigenvalues are. Put $A=\pmatrix{a &b\\c & d}$ and compute the eigenvalues by solving $\det(A-xI) = 0$. Then use the fact that if $A$ has eigenvalue $x$ then $A^2$ has eigenvalue $x^2$ so $\sqrt{2018}$ or $-\sqrt{2018}$ has to be an eigenvalue of $A$. Use this to show that $a+d = 0$ as otherwise $\sqrt{2018}$ would be rational. Once you have done this you will have the two eigenvalues. Finally $\text{Tr}[A^n]$ is the sum of the eigenvalues of $A^n$ which is just the sum of $x^n$ where $x$ are the eigenvalues of $A$. – Winther Jul 8 '18 at 18:51

Let $$A=\begin{bmatrix} \sqrt{2018}&0\\0&a\end{bmatrix}.$$ Then, $$\det(A^2-2018I_2)=0$$ and $$A^{2018}=\begin{bmatrix}2018^{1009}&0\\0&a^{2018}\end{bmatrix},$$ which does not have a trace independent of $a$. So, is there any additional information you have about $A$?

• So the problem is wrong? Also what is the intuition behind the matrix? – C. Cristi Jul 8 '18 at 17:59
• There is one more condition behind the problem $A$ must be in $M_2(\mathbb{Q})$. – C. Cristi Jul 8 '18 at 18:02
• That small addition makes the problem have a unique answer. Hint: look at the eigenvalues. – Michael Burr Jul 8 '18 at 18:08
• I haven't learned about eigenvalues yet... – C. Cristi Jul 8 '18 at 18:09

$\det(A^2-2018I_2)=0$ implies that $2018$ is an eigenvalue of $A^2$.

The spectral mapping theorem gives that $\sigma(A^2) = \sigma(A)^2$ so at least one of $\sqrt{2018}$ and $-\sqrt{2018}$ is in $\sigma(A).$

Now, because $A \in M_2(\mathbb{Q})$, its characteristic polynomial $p_A$ has rational coefficients and $\deg p_A = 2$.

We know that one of $\pm\sqrt{2018}$ is a root of $p_A$, and the minimal polynomial of $\pm\sqrt{2018}$ over $\mathbb{Q}$ is $x^2 - 2018$. Hence $x^2 - 2018$ divides $p_A$ and $p_A$ is monic so we conclude $p_A(x) = x^2 - 2018$.

Hence $\sigma(A) = \{-\sqrt{2018}, \sqrt{2018}\}$. The spectral mapping theorem again gives

$$\sigma(A^{2018}) = \sigma(A)^{2018} = \{(-\sqrt{2018})^{2018}, \sqrt{2018}^{2018}\} = \{2018^{1009}\}$$

The trace is the sum of eigenvalues so

$$\operatorname{Tr} A^{2018} = 2\cdot 2018^{1009}$$

• Why are we using the spectral mapping theorem here? Just conjugate $A$ to be upper-triangular, and note that doesn't change the determinant or trace. – Steve D Jul 8 '18 at 22:58