# Prove with induction if $A= \begin{pmatrix}2&1\\-1&0 \end{pmatrix}$ $\forall n \in \mathbb{N}$ $A^n= \begin{pmatrix} n+1&n\\-n&1-n \end{pmatrix}$

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

• you forgot to explicitly say that you assume the property to hold at rank $n$ in the induction step Sep 29, 2023 at 22:47
• @gdcvdqpl uh yes in fact, I will add that immediately, thank you very much ! Sep 29, 2023 at 22:48
• Ooops... it was properly computed, I mis-typed it. Sorry. Again: MatrixPower[{{2, 1}, {-1, 0}}, n] // MatrixForm $$\pmatrix{n+1 & n \\ -n & 1-n}$$ Sep 30, 2023 at 1:51
• @user21820 what ? why do you think it is wrong ? and it doesn't seem AI generated for me, why do you think that ? Oct 4, 2023 at 9:55
• @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? Oct 4, 2023 at 13:03

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.

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