Mathematical Induction on two Matrices

[Siann122]

Homework Statement

(1 1)^n = (1 n)
(0 1) (0 1)

Prove this through mathematical induction.

Homework Equations

The Attempt at a Solution

I've replaced n with 1, so I've done that far.
Then I said k = n.
Then replaced all n with (k+1).
I'm really stuck on actually proving this though as it's two matrices. I suspect it may have something to do with matrix multiplication?
Thanks for any help guys.

 

[Fredrik]

If you know that $A^n=B(n)$, then what can you say about $A^{n+1}$?

FAQ post about how to type math at Physics Forums.

 

[Siann122]

If you know that $A^n=B(n)$, then what can you say about $A^{n+1}$?

FAQ post about how to type math at Physics Forums.

So Aⁿ⁺¹ = B(n+1). (Or at least to do induction I would assume this to be true).

Still have kind of lost me though, how do I apply this to the question?

 

[Fredrik]

So Aⁿ⁺¹ = B(n+1).

No. In your problem that would be what you're supposed to prove. In other problems it wouldn't even be true.

Don't you see any way to use the assumption $A^n=B(n)$ to say something about $A^{n+1}$?

 

[Siann122]

Don't you see any way to use the assumption $A^n=B(n)$ to say something about $A^{n+1}$?

That's my primary issue.

Normally if there was Aⁿ⁺¹ I'd assume that A * Aⁿ = B(n). I don't really know where else to go with that, I simplified both matrices down.

 

[Fredrik]

That's my primary issue.

Normally if there was Aⁿ⁺¹ I'd assume that A * Aⁿ = B(n).

The equality $AA^n=B(n)$ doesn't follow from the assumption $A^n=B(n)$. In fact it contradicts it (for most choices of A and B, including the one in your problem).

 

[Siann122]

The equality $AA^n=B(n)$ doesn't follow from the assumption $A^n=B(n)$. In fact it contradicts it (for most choices of A and B, including the one in your problem).

Then I'm honestly stuck given that I'm not exactly sure what you're asking, as I've never done induction with matrices before and my book doesn't explain how to do it with matrices specifically. I know that's a piss poor excuse but wrapping my head around it as a matrix as opposed to say a set or a statement seems to be significantly more difficult.

 

[Fredrik]

I don't know what obstacles you have put up for yourself, so I'm not sure what I can tell you to get you past them. The problem I asked you to solve is extremely trivial. In particular you don't have to know anything about induction to solve it. Maybe knowing that will help?

What does the equality $A^n=B(n)$ tell you about $A^{n+1}$?

This is no harder than the following problem: Suppose that x and y are real numbers such that $x^5=y$. What does this assumption tell us about $x^6$?

 

[CAF123]

If (1 1) is a matrix, then how can (1 1)² even make sense?

 

[Siann122]

Any replies I'm making from here on in are me blundering in the dark. This logic thing I'm really struggling with because I seem to not have any :P.

As for CAF123's response, because the matrix is now (1x1 1x1) = (1 1). I don't get how that relates at all because the only way that that works is because it's using 1's in the matrix, any other number and it wouldn't make sense?

 

[CAF123]

As for CAF123's response, because the matrix is now *(1x1 1x1) = (1 1)* . I don't get how that relates at all because the only way that that works is because it's using 1's in the matrix, any other number and it wouldn't make sense?

How did you get this *? That is not how matrix multiplication works. I don't see how we can define the matrix product (1 1)(1 1). Do you agree or am I missing something?

 

[Fredrik]

CAF123, the problem is to prove that for all positive positive integers n, we have
$$\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}^n=\begin{pmatrix}1 & n\\ 0 & 1\end{pmatrix}.$$
Siann122, can you at least try to answer the question about the real numbers x and y?

 

[CAF123]

CAF123, the problem is to prove that for all positive positive integers n, we have
$$\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}^n=\begin{pmatrix}1 & n\\ 0 & 1\end{pmatrix}.$$

I see, I was wondering what those trailing (0 1) matrices were, thanks.

 

[HallsofIvy]

Sian122, If
\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}^n= \begin{pmatrix}1 & n \\ 0 & 1 \end{pmatrix}
then
\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}^{n+1}= \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}^n= \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}1 & n \\ 0 & 1\end{pmatrix}
What is that last product equal to?