Matrix Power Formula

Question

Prove that for a fixed $a \in \mathbb{R}$ we have the matrix power formula for all $n \in \mathbb{Z}_+$: $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix}$$

How would we prove this? Right now we are doing work with proofs by induction... but how would I prove this?

Answer 1

For $n = 1$ we should have (according to the rule) $$ \begin{pmatrix}a & 1\\0 & a\end{pmatrix}^1 = \begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix} $$ but of course $$ \begin{pmatrix}a^1 & 1\cdot a^0\\0 & a^1\end{pmatrix} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} $$ so it's correct (this is called the base case and it's the starting point of the induction).

Now for the induction step. Assume that the rule is correct for all $n \leq k$ for some $k$ (this is called the induction hypothesis). We need to check, on the basis of this alone, that it also holds for $n = k+1$. We know that $$ \begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{k+1} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} $$ Now, $\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k$ we can rewrite (in red below), according to the induction hypothesis (remember, we assumed that the rule was true for all $n \leq k$). Therefore we have $$ \color{red}{\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^k}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} = \color{red}{\begin{pmatrix}a^k & ka^{k-1}\\0 & a^k\end{pmatrix}}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} $$ Now we just use matrix multiplication: $$ \begin{pmatrix}a^k & ka^{k-1}\\0 & a^k\end{pmatrix}\cdot \begin{pmatrix}a & 1\\0 & a\end{pmatrix} = \begin{pmatrix}a^k \cdot a + ka^{k-1}\cdot 0 & a^k\cdot 1 + ka^{k-1}\cdot a\\ 0\cdot a + a^k \cdot 0& 0 \cdot 1 + a^k \cdot a\end{pmatrix}\\ = \begin{pmatrix}a^{k+1} & (k+1)a^k\\0 & a^{k+1}\end{pmatrix} $$ which is exactly the rule for $n = k+1$, finishing the inductive step.

Answer 2

This is a classical example of proof by induction.

• Base Case: Check the formula works when $n=1$
• Inductive Step: Assume it works when $n=k$, show this implies it works for $n=k+1$
• Conclusion: If so then it works for all $n \in \mathbb{N}$

A hint for the Inductive Step is to write out the formula with $n=k$ and then multiply on the left by your matix to give an expression with the $(k+1)^{\mathrm{st}}$ power.

$$\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right)^{\!k} = \left(\begin{array}{cc} a^k & k\,a^{k-1} \\ 0 & a^k\end{array}\right) \implies \color{red}{\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right)}\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right)^{\!k} = \color{red}{\left(\begin{array}{cc} a & 1 \\ 0 & a \end{array}\right)}\left(\begin{array}{cc} a^k & k\,a^{k-1} \\ 0 & a^k\end{array}\right)$$

Answer 3

Verify that it is true for $n = 1$. This is the induction basis.

Then assume the result holds for $n$, so $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n = \begin{pmatrix}a^n & na^{n-1}\\ 0 & a^n\end{pmatrix}$$ This is the induction hypothesis.

Finally, show (with the help of the induction hypothesis) that the result holds for $n + 1$, so $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} = \begin{pmatrix}a^{n+1} & (n+1)a^n\\ 0 & a^{n+1}\end{pmatrix}$$

This can be easily proved: \begin{align} \begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} & = \begin{pmatrix}a & 1\\0 & a\end{pmatrix}\times\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^n\\ & = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} \begin{pmatrix}a^n & na^{n-1}\\ 0 & a^n\end{pmatrix}\\ & = \begin{pmatrix}a^{n+1} & na^n + a^n\\0 & a^{n + 1}\end{pmatrix}\\ & = \begin{pmatrix}a^{n+1} & (n+1)a^n\\0 & a^{n + 1}\end{pmatrix} \end{align} as required.

Answer 4

By induction:

for $n = 1 $ it's true.

For $n +1 $ $$\begin{pmatrix}a & 1\\0 & a\end{pmatrix}^{n+1} = \begin{pmatrix}a & 1\\0 & a\end{pmatrix} \begin{pmatrix}a^n & na^{n-1}\\0 & a^n\end{pmatrix} = \begin{pmatrix}a^{n+1} & a^n + na^n\\0 & a^{n+1} \end{pmatrix} = \begin{pmatrix}a^{n+1} & (n+1)a^n\\0 & a^{n+1} \end{pmatrix}$$

Answer 5

When you're proving it for $n$ (last step of induction), you can separate this matrix in $\left[ \begin {array}{cc} 0&1\\ 0&0\end {array} \right] $ and $\left[ \begin {array}{cc} a&0\\ 0&a\end {array} \right] $, and the sum of that two matrices gives the one on your exercise. That is: \begin{equation} \left[ \begin {array}{cc} 0&1\\ 0&0\end {array} \right]+\left[ \begin {array}{cc} a&0\\ 0&a\end {array} \right]=\left[ \begin {array}{cc} a&1\\ 0&a\end {array} \right] \end{equation}

Now, by newton's binomial, you have that for $n>1$, the matrix $\left[ \begin {array}{cc} 0&1\\ 0&0\end {array} \right] $ is the $0$ matrix, so working only with the first two terms of the newton's binomial, it will give you the desired answer. Shortly, you have to calculate

\begin{equation} \left(\left[ \begin {array}{cc} 0&1\\ 0&0\end {array} \right]+\left[ \begin {array}{cc} a&0\\ 0&a\end {array} \right]\right)^n \end{equation} by newton's binomial.