Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm a little rusty and I've never done a mathematical induction problem with matrices so I'm needing a little help in setting this problem up.

Show that $$\begin{bmatrix}1&1\\1&1\end{bmatrix}^{n} = \begin{bmatrix}2^{(n-1)}&2^{(n-1)}\\2^{(n-1)}&2^{(n-1)}\end{bmatrix}$$ for every $n\ge 1$.

share|improve this question
    
yes, just realized I left something off –  cele Jun 25 at 15:25
1  
First show that it's true for $n=1$ (obvious). Then assume that it's true for $n$, and compute the value at $n+1$ by multiplying out the matrices. –  katrielalex Jun 25 at 15:28
    
@gnometorule, after looking at this problem with a professor I know, they suggested induction. As I've mentioned, I'm a little rusty, just getting back into higher math so I went with their suggestion –  cele Jun 25 at 15:28
    
nice little problem! 1up –  John Smith Jun 25 at 15:30
    
My comment was written when you were still missing the n-exponent. –  gnometorule Jun 25 at 15:36

2 Answers 2

up vote 4 down vote accepted

The case $n=1$ is clear since $2^0 = 1$. So suppose that $$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^n = \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix} \quad \quad *$$ for some $n \geq 1$ and let us prove that $$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$ We have $$\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n+1} = \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix}^{n} \begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix} \overset{*}{=} \begin{pmatrix} 2^{n-1}&2^{n-1} \\ 2^{n-1} & 2^{n-1}\end{pmatrix}\begin{pmatrix} 1&1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2\cdot 2^{n-1}&2\cdot 2^{n-1} \\ 2\cdot 2^{n-1} & 2\cdot 2^{n-1}\end{pmatrix} =\begin{pmatrix} 2^{n}&2^{n} \\ 2^{n} & 2^{n}\end{pmatrix}. $$

And thus the relation is true for every $n \in \mathbb{N}$

share|improve this answer
    
Thank you, your work has made it so much more clearer for me to follow. –  cele Jun 25 at 15:38

It is clearly true for $n=1$. Assume it's true for $n$. Then $$\begin{pmatrix}1&1\\1&1\end{pmatrix}^{n+1}=\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}2^{(n-1)}&2^{(n-1)}\\2^{(n-1)}&2^{(n-1)}\end{pmatrix}=\begin{pmatrix}2^{(n-1)}+2^{(n-1)}&2^{(n-1)}+2^{(n-1)}\\2^{(n-1)}+2^{(n-1)}&2^{(n-1)}+2^{(n-1)}\end{pmatrix}=\begin{pmatrix}2^n&2^n\\2^n&2^n\end{pmatrix}$$

So it's true for $n+1$. By induction it is true for all $n$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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