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$$ A = \begin{bmatrix} 1 & 1 & 13 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \\ \end{bmatrix} $$ Find $A^{14}+3A-2I$.

One way is to find $A^2$, then $A^4$, then $A^8$, then $A^{14}$.

Another way is using eigen values and diagonal matrix concept.

In our university exam, this question is given only for 5 marks. So I am wondering if there is any simple way to do it.

Appreciate any hint.

San
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    I don't know what "5 marks" means (since you gave no scale), but diagonalization seems like the right strategy. – Xander Henderson Dec 12 '17 at 14:22
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    I would say your approach to diagonalize would be best here. –  Dec 12 '17 at 14:23
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    I feel for a 5 mark question, solving by eigenvalues may be a very good technique. –  Dec 12 '17 at 14:26
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    What is the characteristic polynomial ? you should have surely start from there – Guy Fsone Dec 12 '17 at 14:41
  • The eigenvalues are not particularly simple for this matrix. Once you have the characteristic polynomial, you might try applying the Cayley-Hamilton theorem to reduce the degree of the polynomial in $A$ that you’re supposed to compute. The exponent of 14 makes me suspect that it’s a tedious computation either way. – amd Dec 12 '17 at 19:49

1 Answers1

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The characteristic polynomial of $A$ is $\lambda^3+20\lambda+10$, which doesn’t seem to have easily-computable (at least not by hand) roots. Diagonalization or using a decomposition of $A$ into a linear combination of eigenspace projections doesn’t seem like a fruitful way to go.

An alternative to consider is to use the Cayley-Hamilton theorem to reduce the degree of the polynomial that is to be computed: since $A^3+20A+10I=0$, then $A^{14}+3A-2I$ is equal to its remainder when divided by the characteristic polynomial. This will reduce the degree to at most two and seems like it’ll be less work overall than successively squaring $A$ and then performing a few more matrix multiplications to get to $A^{14}$.

The long division is tedious, but not difficult, and results in $$A^{14}+3A-2I = 56\,010\,000\,A^2+188\,000\,003\,A+79\,199\,998\,I.$$

amd
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