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$$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$.


I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?

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    $\begingroup$ Note that matrix $A$ is rank-$1$, i.e., $A = u v^\top$. $\endgroup$ Dec 20, 2020 at 16:39
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    $\begingroup$ Even if one eigenvalue is zero, this matric is diagonalizable, see wolframalpha.com/input/… $\endgroup$
    – user376343
    Dec 20, 2020 at 16:51
  • $\begingroup$ @RodrigodeAzevedo I think your comment is well worth writing up as an answer. $\endgroup$
    – saulspatz
    Dec 20, 2020 at 16:57

4 Answers 4

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Every matrix is annihilated by its characteristic polynomial. Since the characteristic polynomial is $$P(x)=x^2-2x,$$ we deduce $$A^2=2A.$$ Then $$A^{100}=(A^2)^{50}=(2A)^{50}=2^{50}(A^2)^{25}=\dots = 2^{99}\cdot A$$

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Since $\;A=x^ty\;$ where $\;y=(6,4)\;$ and $\;x=(1,-1)\;$ then $A^{100}=(x^ty)^{100}=x^t(yx^t)^{99}y\;$. Since $\;yx^t=2\;$ (it is an inner product) you have $x^t(yx^t)^{99}y=2^{99}x^ty=2^{99}A\;$

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Note that

$$A= 2B \text{ with } B=\begin{bmatrix} 3 & 2\\ -3 & -2\end{bmatrix} \text{ and } B^2 = B$$

Hence,

$$A^{100} = 2^{100}B = 2^{100}\begin{bmatrix} 3 & 2\\ -3 & -2\end{bmatrix}$$

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$$|\lambda I - A| = \lambda^2 - 2\lambda$$ A matrix satisfies its characteristic equation: $$A^2 -2A =0$$ $$A^2 = 2A$$ $$A^4 = 4A^2 = 8A$$ $$A^8 = 16A^4= 2^7 A$$ $$A^{16} = (A^4)^4 = 2^{12}A^4=2^{15}A$$ $$A^{32}= (A^16)^2 = 2^{30}A^2 = 2^{31}A$$ $$A^{64} = (A^16)^4 = 2^{60} A^4 = 2^{63}A$$ $$A^{100}= A^{64+32+4} = 2^{63+31+3}A^3 = 2^{97}A^3=2^{99}A$$ $$A^{100} \approx \pmatrix{3.8&2.5\\-3.8&-2.5}\cdot10^{30}$$

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