Inductive proof for matrix multiplication problem

[jaejoon89]

Homework Statement

Compute

[1 1]^n
[ 1]

This is what my book has. I am assuming the blank means 0 (please tell me if this is not the case).

Homework Equations

None really (matrix multiplication)

The Attempt at a Solution

I did n = 1, 2, 3, 4 and found a patten. It looks like:

[1 1]^n
[ 1]
=
[1 n]
[ 1]

But how do I show that? Obviously showing it for a couple cases doesn't mean it's actually true. Inductively, I can show the n = 2 case is true. but I don't really know how to show the n+1 case is true other than write

[1 1]^n+1
[ 1]
=
[1 n+1]
[ 1]

but that seems a little unsatisfactory or like I'm skipping steps or something. How do you do it?

 

[Mark44]

Assume that the statement is true for an exponent of n. Then use this to show that the (n + 1)st power of your matrix produces the right value.

The blank means 0, I'm pretty sure.

 

[jaejoon89]

Right. But how do you do that?

Letting n be the natural numbers and assuming the following statement is true

[1 1]^n
[ 1]
=
[1 n]
[ 1]

Thus,

[1 1]^n+1
[ 1]
=
[1 n+1]
[ 1]I don't really know how to show it beyond that - is there some abstract way to write out the intermediate steps? As it is, it seems somewhat unsatisfactory.

 

[Mark44]

I'm putting the 0 in the lower left corner since it looks silly to leave it out.

[1 1]^(n+1)
[0 1]

=
[1 1]^n [1 1]
[0 1] [0 1]

= ??
This is where you use what you have assumed.