# Matrix demonstration $A^k$

Given a matrix $$A = \begin{bmatrix} 7 & 4\\ -9 & -5 \end{bmatrix}$$ $$\in \mathcal{M2\times2}\, (\mathbb{R})$$

Show that $$A^k = \begin{bmatrix} 1+6k & 4k\\ -9k & 1-6k \end{bmatrix}$$ for every $$k \in \mathbb{N}$$

• What have you tried so far? What approach do you think will work here? – user3482749 Jan 13 at 0:37
• Have you tried proof by induction? – user7530 Jan 13 at 0:37
• yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly. – PTSONIC Jan 13 at 0:40
• Can you start by computing $A^kA$ and simplifying the result? Shows us your work. – Git Gud Jan 13 at 0:46
• yes the result is this matrix I think $\begin{bmatrix} 7+6k & 4+4k\\ -9-9k & -5-6k \end{bmatrix}$ – PTSONIC Jan 13 at 1:01

Try a simple induction on $$k$$; it is clear that for $$k = 1$$,

$$A^1 = A = \begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 1 + 6 \cdot 1 & 4 \cdot 1 \\ -9 \cdot 1 & 1 - 6 \cdot 1 \end{bmatrix}; \tag 1$$

then assuming that for some $$k$$

$$A^k = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}, \tag 2$$

we find

$$A^{k + 1} = A^kA = \begin{bmatrix} 1 + 6 \cdot k & 4 \cdot k \\ -9 \cdot k & 1 - 6 \cdot k \end{bmatrix}\begin{bmatrix} 7 & 4 \\ -9 & -5 \end{bmatrix} = \begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \\ -63k - 9 + 54k & -36k - 5 + 30k \end{bmatrix}$$ $$= \begin{bmatrix} 7 + 6k & 4 + 4k \\ -9 -9k & - 5 -6k \end{bmatrix} = \begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \\ -9(k + 1) & 1 - 6(k + 1) \end{bmatrix}, \tag 3$$

which is simply (2) with $$k$$ replaced by $$k + 1$$; we thus infer the formula (2) holds for all $$k \ge 1$$, as desired. $$OE\Delta$$.

Let $$N = A-I$$ and observe that $$N^2=0$$. Since $$N$$ and $$I$$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+\binom k1N+\binom k2N^2+\dots = I+kN.$$

How did I hit upon this decomposition? I computed that $$1$$ was the only eigenvalue of $$A$$, but the only diagonalizable $$2\times2$$ matrices with repeated eigenvalues are multiples of the identity, so $$A$$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = P\begin{bmatrix}1&1\\0&1\end{bmatrix}P^{-1} = I + P\begin{bmatrix}0&1\\0&0\end{bmatrix}P^{-1}$$ for some invertible matrix $$P$$.

• Very nice indeed, endorsed, +1!!! – Robert Lewis Jan 13 at 1:47