- #1
Abuda
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Thanks, although I still haven't managed to factorise the expression although I did type it up in LaTeX!
Prove by induction that the following statement is true for all positive integers n.
If [itex]\lambda[/itex] is an Eigenvalue of the square matrix [itex]A[/itex], then [itex]\lambda^n[/itex] is an eigenvalue of the matrix [itex]A^n[/itex]
I think the relevant equations are:
[tex]\det(A-\lambda I)=0[/tex]
(where I is an identity matrix)
[tex]\det(AB)=\det(A)\det(B)[/tex]
So, in mathematical terms I converted the question into the proposition p(n). I got,
[tex]p(n):\det(A^n - \lambda^nI) = 0[/tex]
Now I assumed that p(n) is true for some value k, so I got,
[tex]p(k):\det(A^k - \lambda^kI) = 0[/tex]
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 [itex]\det(A^{k+1} - \lambda^{k+1}I) = 0[/itex] 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:
[tex]\det(A^{k+1} - A\lambda^k I) = 0[/tex]
But I saw no continuation. I also tried [itex]A^n=PD^nP^{-1}[/itex] and got something incredibly messy but it didnt work for me.
Thanks for all help in advance.
Homework Statement
Prove by induction that the following statement is true for all positive integers n.
If [itex]\lambda[/itex] is an Eigenvalue of the square matrix [itex]A[/itex], then [itex]\lambda^n[/itex] is an eigenvalue of the matrix [itex]A^n[/itex]
Homework Equations
I think the relevant equations are:
[tex]\det(A-\lambda I)=0[/tex]
(where I is an identity matrix)
[tex]\det(AB)=\det(A)\det(B)[/tex]
The Attempt at a Solution
So, in mathematical terms I converted the question into the proposition p(n). I got,
[tex]p(n):\det(A^n - \lambda^nI) = 0[/tex]
Now I assumed that p(n) is true for some value k, so I got,
[tex]p(k):\det(A^k - \lambda^kI) = 0[/tex]
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 [itex]\det(A^{k+1} - \lambda^{k+1}I) = 0[/itex] 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:
[tex]\det(A^{k+1} - A\lambda^k I) = 0[/tex]
But I saw no continuation. I also tried [itex]A^n=PD^nP^{-1}[/itex] and got something incredibly messy but it didnt work for me.
Thanks for all help in advance.
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