Let $$A=\begin {bmatrix} 1 & 3 & 1\\ 1 & 1 & -2\\ 1 & 2 & -1\\ \end{bmatrix} $$
I need to demonstrate this relation: $$\forall n \in \mathbb{N}, A^n \neq I_3$$ where $I_3$=identity matrix.
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it helps to know that $1$ is not an eigenvalue of $A.$ – abel May 09 '16 at 16:30
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If you have seen eigenvalues, you know that if $\lambda$ is an eigenvalue of $A$, $\lambda^n$ is an eigenvalue of $A^n$. But all eigenvalues of $A$ are such that $|\lambda| \neq 1$ ($\lambda \approx 1.839$ and $\lambda \approx -0.420 \pm 0.606 i$), thus it is impossible, whatever $n$, to have $|\lambda|^n=1$ (the eigenvalue of $I_3$).
Jean Marie
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You desire a "high school proof" but, at your level, beyond the definition of the product of matrices, what are the results you have seen in your lectures and that you can use ? – Jean Marie May 09 '16 at 17:43
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newton's binom, matrix inverse, transposed matrix but i don't know if this would help. Yes i checked – Emanuel May 09 '16 at 17:59
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Proof using mathematical induction:
The base case: for $n=1$ we have $$A^1 = \begin {bmatrix} 1 & 3 & 1\\ 1 & 1 & -2\\ 1 & 2 & -1\\ \end{bmatrix}$$ So $A^1 \ne I$ is true for the base case.
Induction hypothesis: Assume that $A^n \ne I$ is a correct mathematical proposition and let proof that $A^{n+1} \ne I$ is also correct. $$A^{n+1}=A^{n}\cdot A$$ According to the base case we have $A \ne I$ and according to the induction hypothesis we have $A^n \ne I$ , so $$A^n\cdot A \ne I \cdot I$$ $$\Leftrightarrow A^{n+1} \ne I$$
Hence $A^{n+1} \ne I$ is true. So $$\forall n \in \mathbb{N}: A^n\ne I$$
Ayoub Falah
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Ayoub Falah, you wrote nonsense, the implication $A\ne B& C\ne D \Rightarrow AB\ne CD $ is wrong of course. – bot May 10 '16 at 01:18
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Could you explain mathematically why should I delete it. Is there anything wrong? – Ayoub Falah May 10 '16 at 04:41
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@bot,@lastresot: but $A$ is clearly defined as $\begin {bmatrix} 1 & 3 & 1\ 1 & 1 & -2\ 1 & 2 & -1\ \end{bmatrix}$ – Ayoub Falah May 10 '16 at 10:53
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What I meant by a counterexample that you show that $\exists k \in \mathbb{N}:A^k \cdot A=I$ – Ayoub Falah May 10 '16 at 10:57
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@ Ayoub Falah, I mean what I say, but what do you mean asking an axample $\exists k\in \mathbb N: A^k\cdot A= I$? – bot May 10 '16 at 12:40
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