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! :)