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If it is given that matrix $ A = \begin{bmatrix} \dfrac{5}{2} & \dfrac{3}{2}\\ -\dfrac{3}{2} & -\dfrac{1}{2} \end{bmatrix}$ then find the value of $A^{2022}$. Here is my try on it : $$\\$$ If we look at the pattern while multliplying A with itself it comes out to be something like this, $$A^2 = \begin{bmatrix} 4 & 3\\ -3 & -2 \end{bmatrix}$$ $$ A^3 = \begin{bmatrix} \dfrac{11}{2} & \dfrac{9}{2}\\ -\dfrac{9}{2} & -\dfrac{7}{2} \end{bmatrix}$$ $$A^4 = \begin{bmatrix} 7 & 6\\ -6 & -5 \end{bmatrix}$$ so $$ A^n = \begin{bmatrix} \dfrac{3n}{2}+1 & \dfrac{3n}{2}\\ -\dfrac{3n}{2} & -\dfrac{3n}{2}+1 \end{bmatrix} \tag{1}\label{eq1}$$ thus after substituting $n=2022$ we get $ A^{2022} = \begin{bmatrix} 3034 & 3033\\ -3033 & -3032 \end{bmatrix}$ $$\\$$ Is there is any other way to find the value of $A^{2022}$ without finding out the pattern and then general equation \eqref{eq1} like I did above?

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  • $\begingroup$ Various related questions: google.de/… $\endgroup$
    – Martin R
    Commented Aug 29, 2022 at 19:23
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    $\begingroup$ Remember that, by Cayley-Hamilton, the matrix makes its characteristic polynomial $p(x)$ vanish. Being a $2\times 2$ matrix, its characteristic polynomial has degree $2$. If you divide $x^{2022}=p(x)q(x)+r(x)$, the remainder $r(x)$ would have degree at most $1$. Note that $A^{2022}=r(A)$. So, all you need to do is to find the coefficients of $r(x)=ax+b$. $\endgroup$
    – plop
    Commented Aug 29, 2022 at 19:42

6 Answers 6

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Another option is to see that $A = I + B$, where
$B = \begin{pmatrix} \frac 3 2 & \frac 3 2 \\ -\frac 3 2 & -\frac 3 2 \end{pmatrix}$
and $B^2 = 0$.

So $A^n = I + nB$.

But that's a post-hoc explanation: it is not obvious to remark that when looking at the matrix - although such a simplification can be expected, in an exercise like this one, especially if it comes from a math competition where speed matters.

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    $\begingroup$ Note : It's a particular case of Dunford decomposition $\endgroup$
    – Lelouch
    Commented Aug 29, 2022 at 20:11
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The characteristic polynomial of $A$ is \begin{align} \det(A-\lambda I)=\left|\begin{array}{c} \frac{5}{2}-\lambda & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}-\lambda \end{array}\right|\\ = (5/2-\lambda)(-1/2-\lambda)+9/4 \\ = -5/4-2\lambda+\lambda^2+9/4 \\ = (\lambda-1)^2 \end{align} Cayley-Hamilton gives $(A-I)^2=0$. So, a binomial expansion works great because only two terms are non-zero: \begin{align} A^{2022} &= (I+(A-I))^{2022} \\ &=I+2022 (A-I) \\ &=2022 A - 2021 I \\ &=2022\left[\begin{array}{rr} \frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2} \end{array}\right] -\left[\begin{array}{rr}2021 & 0 \\ 0 & 2021\end{array}\right] \\ &=\left[\begin{array}{rr} 3034 & 3033 \\ -3033 & -3032 \end{array}\right]. \end{align}

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It's more or less linked to eigenvalue decomposition,but you can also use a good matrix polynomial $P$ such that $P(A)=0$.

For example, $P(X)=(X-1)^2$ (this is the characteristic polynomial here, it always works but it might not be the best choice in other cases). If you are not familiar with the theory, you can just verify by hand that $(A-I_2)^2$ is actually the null matrix ($I_2$ is identity matrix of size 2x2).

Then lets do the euclidean division of $X^n$ by $P(X)$ :

$ X^n = P(X)Q(X)+R(X)$ with $\deg(R) < \deg(P) = 2$ and thus $\deg(R)=1$. That means we know that $R(X)=aX+b$ where $a,b$ are two real numbers. The beauty here is that we absolutely dont need to compute $Q$ to achieve our goal.

Let's find what are $a$ and $b$ !

By evaluating the equality for $X=1$ we obtain $1^n = P(1)Q(1)+R(1)$, but $P(1)=0$ and $R(1)=a+b$ and thus $1=a+b$.

By evaluating the derivative polynomial for $X=1$ we obtain $n1^{n-1} = P'(1)Q(1) + P(1)'Q(1)+R'(1)$ . Here again $P'(1)=P(1)=0$ and thus $n=a$ and $b=1-a=1-n$.

(If you have not noticed, I used the fact that $1$ is a multiple eigenvalue to compute $a$ and $b$ ...)

That means $X^n = (X-1)^2 Q(X) + nX + 1-n$.

Now lets take $X=A$ (where $A$ is a matrix, not a real number), since $P(A)=0$ we have $$\boxed{A^n = nA - (n-1)I_2}$$

You can check that it corresponds to what you found.

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  • $\begingroup$ interesting approach $\endgroup$
    – G Cab
    Commented Aug 29, 2022 at 22:08
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An alternate and perhaps more straightforward way is to compute the eigenvalue decomposition of $A$, say $X\Lambda X^{-1}$. This is possible since your matrix has 2 eigenvalues (i.e. its geometric multiplicity is equal to its algebraic multiplicity).

Then, $$A^{2022} = (X\Lambda X^{-1})^{2022} = X\Lambda^{2022} X^{-1}.$$ Since $\Lambda$ is a diagonal matrix, $\Lambda^n$ is simply the diagonal elements raised to the $n$th power. Though for your specific $A$ it might be tedious calculating $\lambda_1^{2022}$ and $\lambda_2^{2022}$.

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    $\begingroup$ The characteristic polynomial is $\lambda^2-2\lambda+1$ and the geometric multiplicity to the only eigenvalue $1$ is $1$. $\endgroup$ Commented Aug 29, 2022 at 19:45
  • $\begingroup$ @MichaelHoppe Yes you are right, I had accidentally used $1/2$ instead of $-1/2$ when calculating the eigenvalues. My apologies. $\endgroup$
    – CBBAM
    Commented Aug 29, 2022 at 19:47
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COMMENT.-$A$ has a double eigenvalue $\lambda=1$, is not diagonalizable and its Jordan matrix decomposition is $$A=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$ so we have $$A^{2022}=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}^{2022}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$

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The matrix has determinant 1 and trace 2, so it is parabolic, so a conjugate of an upper triangular matrix (with diagonal elements equal to 1). Computing powers of parabolic matrices is fun and easy (though your solution is essentially that).

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