Compute $ A^2, A^3, A^4.$ Find a formula for $A^k$

Question

Let

$$A=\pmatrix{1&1&1\\ 0&1&1 \\ 0&0&1} $$ Compute $A^2,A^3,A^4$

Find a formula for $A^k $ for every $k > 0$. You do not need to prove that the formula is correct.

Answer 1

\begin{eqnarray*} A^k= \begin{bmatrix} 1 &k & \frac{k(k+1)}{2} \\0 &1 & k \\ 0 & 0 &1 \\ \end{bmatrix} . \end{eqnarray*} Prove it by induction.

Answer 2

$$A^2=\pmatrix{1&2&3\\ 0&1&2 \\ 0&0&1} \implies A^3=\pmatrix{1&3&6\\ 0&1&3 \\ 0&0&1} $$

$$A^4=\pmatrix{1&4&10\\ 0&1&4 \\ 0&0&1} \implies A^k=\pmatrix{1&k&S_k\\ 0&1&k \\ 0&0&1} $$

Where $S_k=S_{k-1}+k$

$S_k=S_{k-1}+k=S_{k-2}+(k-1)+k=S_{1}+....+(k-2)+(k-1)+k$ $S_k=1+....+(k-2)+(k-1)+k$ it gives $S_k=\frac {k(k+1)}2$

$$A^k=\pmatrix{1&k& \frac {k(k+1)}2\\ 0&1&k \\ 0&0&1} $$

Answer 3

Or, $A = I + N,$ where $$ N = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) $$ and $$ N^2 = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) $$ and $N^3 = N^4 = N^5 = \cdots = 0.$

As you can see in the other answers, the fact that $I$ and $N$ commute means we can expand this, and the fact that $N^3 = 0$ means the (binomial) expansion stops very soon. $$ (I+N)^k = I + k N + \frac{k(k-1)}{2} N^2$$ because the binomial coefficient for $k$ elements taken $2$ at a time is $\frac{k(k-1)}{2} \; \; .$

$$ $$

The element in position $(1,3)$ becomes $$ k + \frac{k^2 - k}{2} = \frac{k^2 + k}{2} $$