Linear Algebra - Proof by induction - Matrices

[Florian]

I tried a matrix and I got this pattern:



So, I know that after the multiplication of A^k the 1-entries of A shift one position to the right. And I know, that if k = n, than A^k = A^n = (0).

My problem is, what do I have to prove exactly? Do I have to prove, that the entries are shifting or that A^n = 0?

Thanks,
Flo

 

[mmm4444bot]

The meaning of the linear-algebra aspect of this exercise is probably over my head, but my first question to you would be: "Do you understand the underlying concept of proofs by induction?"

(1) You show that it's true when n = 1

(2) You assume that it's true for any n = k (the induction hypothesis)

(3) You now use the induction hypothesis to show that it's also true for k + 1 (when it's true for k)

This proves that it's true for any n because you've shown that it's true for 1 and also for 1 + 1, so it's true for n = 2, as well as n = 1.

But k could be 2. So, you've shown that it's true for 2 + 1. Therefore, it's true for n = 3.

But k could be 3. So, you've shown that it's true for n = 4.

I could go on to make similar statements for all natural numbers n, but I don't have to.

That's induction.

 

[Florian]

The concept of proofs by induction is clear to me. That's not the problem. My problem is, to show that this matrix is nilpotent without using determinants because it did not occur in my linear algebra lecture.
So, I am quarreling with a proof where I don't have any lemmata, etc. But thanks for your help.

 

[daon]

If the object of the exercise is to show A is nilpotent then you must show there exists an n such that A^n=[0]. That is not what you said in your original post. From that post it seems the objective is to come up with a general formula for A^k.