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Prove with induction if

$$A = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$$

then $\forall n \in \mathbb{N}$

$$A^n = \begin{pmatrix} n + 1 & n \\ -n & 1 - n \end{pmatrix}$$


For $n = 1$ we have

$$A^1 = \begin{pmatrix} 1 + 1 & 1 \\ -1 & 1 - 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}$$

so the base case is correct.

EDIT: we assume it holds true for $n$ (Thank you @gdcvdqpl !)

For $n+1$ we have

$$A^{(n+1)} = \begin{pmatrix} (n+1) + 1 & (n+1) \\ -(n+1) & 1 - (n+1) \end{pmatrix} = \begin{pmatrix} n + 2 & n + 1 \\ - n - 1 & - n \end{pmatrix}$$

And also

$$A^nA^1 = \begin{pmatrix} n + 1 & n \\ -n & 1 - n \end{pmatrix} \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 2(n+1) - n & n + 1 \\ -2n - (1 - n) & -n \end{pmatrix} = \begin{pmatrix} n + 2 & n + 1 \\ - n - 1 & - n \end{pmatrix} $$

So it also holds for $n+1$.

Is that correct ? We don't have solutions to this exercise. Feel free to point out any inconsistency. Thank you for helping me

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    $\begingroup$ you forgot to explicitly say that you assume the property to hold at rank $n$ in the induction step $\endgroup$
    – gdcvdqpl
    Sep 29, 2023 at 22:47
  • $\begingroup$ @gdcvdqpl uh yes in fact, I will add that immediately, thank you very much ! $\endgroup$
    – wengen
    Sep 29, 2023 at 22:48
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    $\begingroup$ Ooops... it was properly computed, I mis-typed it. Sorry. Again: MatrixPower[{{2, 1}, {-1, 0}}, n] // MatrixForm $$\pmatrix{n+1 & n \\ -n & 1-n}$$ $\endgroup$ Sep 30, 2023 at 1:51
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    $\begingroup$ @user21820 what ? why do you think it is wrong ? and it doesn't seem AI generated for me, why do you think that ? $\endgroup$
    – wengen
    Oct 4, 2023 at 9:55
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    $\begingroup$ @wengen: That is not my area of expertise. – With respect to this question: What is your question, if not a request for verifying your solution? $\endgroup$
    – Martin R
    Oct 4, 2023 at 13:03

1 Answer 1

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this question illustrates how finding the Jordan form, including the change of basis matrix, cleans things up.

$$\left( \begin{array}{rr} 1 & 0 \\ -1 & 1 \\ \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{rr} 1 & 0 \\ 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 1 \\ -1 & 0 \\ \end{array} \right) = A $$

and

$$\left( \begin{array}{rr} 1 & 0 \\ -1 & 1 \\ \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ 0 & 1 \\ \end{array} \right)^n \left( \begin{array}{rr} 1 & 0 \\ 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rr} 2 & 1 \\ -1 & 0 \\ \end{array} \right)^n = A^n $$

while

$$ \left( \begin{array}{rr} 1 & 1 \\ 0 & 1 \\ \end{array} \right)^n = \left( \begin{array}{rr} 1 & n \\ 0 & 1 \\ \end{array} \right) $$

this last identity could also be done by induction, but is an example of the binomial formula because $\left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ and $\left( \begin{array}{rr} 0 & 1 \\ 0 & 0 \\ \end{array} \right)$ commute, while the square (or any higher power) of the second one is the zero matrix.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

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    $\begingroup$ $\bigcirc^{24}$. ${}$ $\endgroup$
    – copper.hat
    Sep 30, 2023 at 2:06
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    $\begingroup$ @copper.hat Thank you. But I don't know what $\bigcirc^{24}$ means. The Urban Dictionary seemed not to have it. $\endgroup$
    – Will Jagy
    Sep 30, 2023 at 2:16
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    $\begingroup$ On my display the last two big circles are closer together that all the rest! $\endgroup$
    – copper.hat
    Sep 30, 2023 at 2:31
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    $\begingroup$ Oh, I get it now. I like putting a line to separate sections. If I change the size of the browser window, the line may spread over two lines. Or, I guess, squeeze in some places. $\endgroup$
    – Will Jagy
    Sep 30, 2023 at 2:36

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