Linear Algebra: For A=[[7 4][-9 -5]], show A^k=[[1+6k 4k][-9k 1-6k]]

diogomgf

Junior Member
Joined
Oct 19, 2018
Messages
127
Hello every one,

I got stuck in another linear algebra problem:

Let A2x2 = [ 7 4]
[-9 -5]

Show that Ak = [1+6k 4k]
[-9k 1-6k]

for any k in N.

Thank you
 
Hello every one,

I got stuck in another linear algebra problem:

Let A2x2 = [ 7 4]
[-9 -5]

Show that Ak = [1+6k 4k]
[-9k 1-6k]

for any k in N.

Thank you
This problem will be solved most efficiently through "induction". Please share your work so that we know where you are stuck.
 
This problem will be solved most efficiently through "induction". Please share your work so that we know where you are stuck.

I am stuck trying to figure out how to approach the general proof.

Substituting k for 2, will prove it is correct. Also for 3, etc... (by induction as you said).

But how to prove it is for any N?

I think my problem is mostly a formal problem...
 
I am stuck trying to figure out how to approach the general proof.

Substituting k for 2, will prove it is correct. Also for 3, etc... (by induction as you said).

But how to prove it is for any N?

I think my problem is mostly a formal problem...
You assume that it is true for N-1 AND THEN you show it is true for N (possibly using your assumption)
 
I am stuck trying to figure out how to approach the general proof.

Substituting k for 2, will prove it is correct. Also for 3, etc... (by induction as you said).
No. Trying to prove something is true "for all n" by trying actually to do "all n" is not only impossible (since there are infinitely-many such "n") but also is not induction!

To learn, in general, how induction works, please try here. Then:

But how to prove it is for any N?
You have shown that the rule works for a specific value; namely, for n = 2 (or k = 2, since they're calling it "k" rather than the standard "n").

Now make the assumption that, for k = n, the formula works. That is, that:

. . . . .\(\displaystyle A^n\, =\, \left[\begin{array}{rr}1+6n&4n\\-9n&1-6n\end{array}\right]\)

Then try to prove the k = n+1 case, perhaps by multiplying A^n by another copy of A. See where that leads.... ;)
 
No. Trying to prove something is true "for all n" by trying actually to do "all n" is not only impossible (since there are infinitely-many such "n") but also is not induction!

To learn, in general, how induction works, please try here. Then:


You have shown that the rule works for a specific value; namely, for n = 2 (or k = 2, since they're calling it "k" rather than the standard "n").

Now make the assumption that, for k = n, the formula works. That is, that:

. . . . .\(\displaystyle A^n\, =\, \left[\begin{array}{rr}1+6n&4n\\-9n&1-6n\end{array}\right]\)

Then try to prove the k = n+1 case, perhaps by multiplying A^n by another copy of A. See where that leads.... ;)

Thanks alot for the answer, it really helped me alot!
 
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