1
$\begingroup$

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 $k \in \mathbb{N}$

$\endgroup$
  • $\begingroup$ What have you tried so far? What approach do you think will work here? $\endgroup$ – user3482749 Jan 13 at 0:37
  • 2
    $\begingroup$ Have you tried proof by induction? $\endgroup$ – user7530 Jan 13 at 0:37
  • $\begingroup$ 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. $\endgroup$ – PTSONIC Jan 13 at 0:40
  • 1
    $\begingroup$ Can you start by computing $A^kA$ and simplifying the result? Shows us your work. $\endgroup$ – Git Gud Jan 13 at 0:46
  • $\begingroup$ yes the result is this matrix I think $\begin{bmatrix} 7+6k & 4+4k\\ -9-9k & -5-6k \end{bmatrix} $ $\endgroup$ – PTSONIC Jan 13 at 1:01
2
$\begingroup$

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. $OE\Delta$.

$\endgroup$
2
$\begingroup$

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 $P$.

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.