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Prove that

\begin{align} \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix}^n &= \begin{pmatrix} \lambda^{n} & n\lambda ^{n-1}\\ 0 & \lambda^{n} \end{pmatrix} . \end{align}

I have tried the following using induction (Base case is ok).

Here the inductive step.

I suppose that there exists some $k \ge 1$ for the problem. And I want to know if it also holds for $k+1$,

\begin{align} \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix}^{k+1} &= \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix}^k \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix} \\ &= \begin{pmatrix} \lambda^{k} & k\lambda ^{k-1}\\ 0 & \lambda^{k} \end{pmatrix} \begin{pmatrix} \lambda & 1\\ 0 & \lambda \end{pmatrix} \\ &= \begin{pmatrix} \lambda^{k+1} & (k+1)\lambda^{k} \\ 0 & \lambda ^{k+1} \end{pmatrix}. \end{align}

Then is proven for $k+1$.

I'm not sure if the proof is ok. And I would like to know if there exists a proof without using induction.

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    $\begingroup$ The induction is correct. $\endgroup$
    – ljfirth
    Commented Jun 23 at 20:09
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    $\begingroup$ Fun fact this has a beautiful generalization for analytic functions of matrices $$f\left(\begin{pmatrix}\lambda & 1 \\ 0 & \lambda\end{pmatrix}\right) = \begin{pmatrix}f(\lambda) & f'(\lambda) \\ 0 & f(\lambda)\end{pmatrix}$$ $\endgroup$ Commented Jun 23 at 21:28

1 Answer 1

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Another approach can be along the following lines, let $$A=\begin{bmatrix}a &1\\0&a\end{bmatrix}=a\begin{bmatrix}1 &0\\0&1\end{bmatrix}+\begin{bmatrix}0 &1\\0&0\end{bmatrix}=aI+B$$ Then, $$A^n=(aI+B)^n=a^nI+na^{n-1}B+\binom{n}{2}a^{n-2}B^2+\dotsb + B^n$$ Observe that $B^2=\begin{bmatrix}0 &0\\0&0\end{bmatrix}$. So, $B^k=O$ for all $k \geq 2$. Thus, $$A^n=a^nI+na^{n-1}B=\begin{bmatrix}a^n &na^{n-1}\\0&a^n\end{bmatrix}$$

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