# Confused about finding non-brute force way to solve for matrix to the 2019th power

I am attempting to solve this problem, it has four parts. I solved part a (a trivial matrix problem), but the next three parts appear to be a bit confusing to me. I just would like some help getting started so I can see and observe this matrix and come up with a solution.

The Questions Note: The idea here is NOT to use brute force computation to get $$A^{2019}$$ matrix, instead use some observations which can significantly reduce computational work and will also give you an insight into such problems.

Obviously this is some huge numbered matrix, but I do not understand what observation will reduce this? My first thought was just to use a calculator and calculate $$A^{2019}$$ and then just multiply that by each vector. But that appears to be not the point of the question.

$$Let A = \left[\begin{array}{rrr} -4 & -6 & -12 \\ -2 & -1 & -4 \\ 2 & 3 & 6 \end{array}\right]$$

And Let $$u$$ = [6 5 -3], $$v$$ = [-2 0 1], and $$w$$ = [-2 -1 1].

b). Compute $$A^{2019}\mathbb v$$

• If $p(x)$ is the characteristic polynomial of $A$, then $p(A)=0$. If you divide $x^{2019}=p(x)q(x)+r(x)$, with $r$ of smaller degree than $p$, then $A^{2019}=p(A)q(A)+r(A)=r(A)$. – user647486 Mar 19 at 20:40
• Take into account that the quotient $q$ doesn't need to be computed. Only the remainder is necessary. The characteristic polynomial in your case is $-x^3 + x^2 + 2 x$. The remainder can be found by indeterminate coefficients. For example, evaluating the equation $x^{2019}=p(x)q(x)+r(x)$ at the roots of $p$. In your case $0,-1,2$. The remainder becomes, I think, $\frac{2^{2019}-4\cdot (-1)^{2019}}{6}x^2+\frac{2^{2019}+2\cdot(-1)^{2019}}{5}x$. – user647486 Mar 19 at 20:47
• So, you only need to compute $A$ and $A^2$ and plug them in there. – user647486 Mar 19 at 20:49
• Hint: Is there a simple relationship between $\mathbf v$ and $A\mathbf v$? – amd Mar 19 at 20:50
• What is $x$ ? Also you can decompose your matrix like this : $A=SJS^{-1}$ where $J$ is a diagonal matrix and then $A^{2019}=A J^{2019} S^{-1}$. – Alain Mar 19 at 20:51

I assume that you're asking about $$A^{2019}v$$.
To that end, observe that $$Av = 2v$$ (I'll leave it to you to verify that this is true).
It follows that $$A^2v = AAv = A(2v) = 2Av = 4v.$$
Similarly, $$A^3 v = 8v$$, and in general we have $$A^k v = 2^k v$$ for any positive integer $$k$$.