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

Question

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

Answer 1

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.

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