Proof by Induction Question on Determinants and Eigenvalues

[Abuda]

Thanks, although I still haven't managed to factorise the expression although I did type it up in LaTeX!

Homework Statement

Prove by induction that the following statement is true for all positive integers n.

If $\lambda$ is an Eigenvalue of the square matrix $A$, then $\lambda^n$ is an eigenvalue of the matrix $A^n$

Homework Equations

I think the relevant equations are:

$$\det(A-\lambda I)=0$$

(where I is an identity matrix)

$$\det(AB)=\det(A)\det(B)$$

The Attempt at a Solution

So, in mathematical terms I converted the question into the proposition p(n). I got,

$$p(n):\det(A^n - \lambda^nI) = 0$$

Now I assumed that p(n) is true for some value k, so I got,

$$p(k):\det(A^k - \lambda^kI) = 0$$

I know that p(1) is true so all I need to do is prove that p(k)=>p(k+1)

So I need to show that $\det(A^{k+1} - \lambda^{k+1}I) = 0$ by using the assumption.

I tried a couple of things here that got me no where. I multiplied both sides of the assumption by det(A) and got:
$$\det(A^{k+1} - A\lambda^k I) = 0$$

But I saw no continuation. I also tried $A^n=PD^nP^{-1}$ and got something incredibly messy but it didnt work for me.

Thanks for all help in advance.

 

[tiny-tim]

Welcome to PF!

Hi Abuda! Welcome to PF!

(try using the X² icon just above the Reply box )

Hint: factor Aⁿ - Lⁿ

(though I don't see what that has to do with proof by induction )

 

[Abuda]

I edited the question to make it readable as you suggested tiny-tim! And thanks for the welcome. But I'm still confused due to my being not able to factorise it :S

 

[Dick]

I edited the question to make it readable as you suggested tiny-tim! And thanks for the welcome. But I'm still confused due to my being not able to factorise it :S

You are probably better off doing this without the determinants. If x is an eigenvector of A with eigenvalue lambda, then Ax=(lambda)x. Try doing induction with that definition.

 

[Abuda]

You are probably better off doing this without the determinants. If x is an eigenvector of A with eigenvalue lambda, then Ax=(lambda)x. Try doing induction with that definition.

Thanks for the simple answer Dick! I think I got the answer. I got:

$$P(n): A^nx=\lambda^n x$$

P(1) is true

Assume P(k) is true.

$$P(k): A^kx=\lambda^k x$$

Multiplying both sides of p(k) by A yields:

$$A^{k+1}x=\lambda^{k}Ax$$

Now we sub in $Ax=\lambda x$ from the definition of eigenvalues.

$$A^{k+1}x=\lambda^{k+1}x$$

Therefore, p(k) implies p(k+1) and hence p(n) is true for all n. It seems that I spent a lot of time on the method using the determinant approach but I realize now that the determinant approach couldn't have worked easily as the substitution of Ax=lambda x couldn't have been carried out...Thanks heaps Physics Forums. Its a pleasure to have joined your website!
Alex

 

[tiny-tim]

just for the record, my idea was Aⁿ - Lⁿ = (A^n-1 + A^n-2L + … + L^n-1)(A - L)

 

[Abuda]

just for the record, my idea was Aⁿ - Lⁿ = (A^n-1 + A^n-2L + … + L^n-1)(A - L)

Oh Yeah! Because we know that Det(AB)=Det(A)Det(B) and Det(A-L)=0 we can prove it directly without induction using the determinant approach. Great idea. I tried to factorize it briefly before I posted the question but I didnt even know the rule you posted so I had no success!..Thanks again
Alex

 

[tiny-tim]

Hi Alex!

… I didnt even know the rule you posted …

oooh, in that case for future possible use you also need to learn …

Aⁿ + Lⁿ = (A^n-1 - A^n-2L + … - AL^n-1 + L^n-1)(A - L)

(with alternating + and - signs, and it only works for odd n )

of course, these work for ordinary numbers also

 

[Abuda]

Hi Alex!


oooh, in that case for future possible use you also need to learn …

Aⁿ + Lⁿ = (A^n-1 - A^n-2L + … - AL^n-1 + L^n-1)(A - L)

(with alternating + and - signs, and it only works for odd n )

of course, these work for ordinary numbers also

Thanks Tim, Excellent, I put them down in book! :)