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# Matrix Question using Proof by Induction watch

1. This is my working for 7(i).

so it appears a formula for could be

To prove this by induction let

Now is obviously true by definition.

Assume is true.

Can I conclude that the general matrix is proved by induction from this? What is the inductive assumption?
2. If I could use latex, I would explain. But I can't so sorry.
3. (Original post by TheEd)
To prove this by induction let
A pedantic point. p(n) is a sentence, not your actual matrix itself. So p(n) is the statement "A^n = blah".

(Original post by TheEd)
[p(n):]

[...]

Assume is true.

Firstly, I assume you mean A^(k+1) here (see above note). Your inductive hypothesis is that p(k) is true, i.e. that . You have to prove from here that A^(k+1) is what you said it is, by actually multiplying A^k by A.
4. to find A^(k+1), multiply A^k by A
5. ..don't remember any of that in the film before! Badum dum!
6. (Original post by generalebriety)
A pedantic point. p(n) is a sentence, not your actual matrix itself. So p(n) is the statement "A^n = blah".

Firstly, I assume you mean A^(k+1) here (see above note). Your inductive hypothesis is that p(k) is true, i.e. that . You have to prove from here that A^(k+1) is what you said it is, by actually multiplying A^k by A.
Thanks
7. The method is as follows:

Show that P(1) is true.
Assume that P(k) is true where k is an element of the set of positive integers.
Show that P(k+1) is true by multiplying A^k by A.
P(1) is true, P(k) is true and implies that P(k+1) is true, so P(n) is true for n∈Z^+.

Sorry for my LaTeX deficiency.
8. (Original post by CocoPop)
The method is as follows:

Show that P(1) is true.
Assume that P(k) is true where k is an element of the set of positive integers.
Show that P(k+1) is true by multiplying A^k by A.
P(1) is true, P(k) is true and implies that P(k+1) is true, so P(n) is true for n∈Z^+.

Sorry for my LaTeX deficiency.
For the second matrix the general matrix is:

.

Proving it by induction you get:

.

But I don't see what this proves?
9. Addition formulae for sine and cosine.
10. (Original post by TheEd)
Proving it by induction

You don't quite seem to understand the concept. You haven't proved anything by multiplying A^k by A; you now need to show that A^(k+1) is equal to what you predicted it would be before (i.e. that p(k+1) is true). Do this by using trig addition formulas.
11. (Original post by generalebriety)

You don't quite seem to understand the concept. You haven't proved anything by multiplying A^k by A; you now need to show that A^(k+1) is equal to what you predicted it would be before (i.e. that p(k+1) is true). Do this by using trig addition formulas.
Ah I get it now Thanks for the clarification!
12. Sorry for not making that clear before. A good way of showing your method is by referring to the "left hand side" (LHS) and "right hand side" (RHS) of the equation.

So you could say that:

LHS = A^(k+1), here you can show what your predicted outcome is according to the statement for A^n.

Then on the Right hand side you could say that A^(k+1)=A^k * A

So:

RHS = A^k * A

You can then show that LHS = RHS... Don't know if this is the method that other people use, but I find it helpful.
13. (Original post by CocoPop)
Sorry for not making that clear before. A good way of showing your method is by referring to the "left hand side" (LHS) and "right hand side" (RHS) of the equation.

So you could say that:

LHS = A^(k+1), here you can show what your predicted outcome is according to the statement for A^n.

Then on the Right hand side you could say that A^(k+1)=A^k * A

So:

RHS = A^k * A

You can then show that LHS = RHS... Don't know if this is the method that other people use, but I find it helpful.
Notationally it's a bit bad. It's quite obvious that A^(k+1) = A^k * A regardless of how you think the answer should come out looking. But whatever works for you, I guess...

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