Recent content by Abuda

  1. A

    Which Limit Law should I refer to in my solution?

    Thank you very much for helping me and fixing up my latex skills. Your explanation about canceling sounds good to me and the other notes about the algebra of limits. Alex
  2. A

    Which Limit Law should I refer to in my solution?

    Homework Statement Evaluate the limit below indicating the appropriate Limit Law(s) implemented. [tex]\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}[/itex] 2. The attempt at a solution [tex]\lim_{x\rightarrow 0.5}\frac{2x^2+5x-3}{6x^2-7x+2}=\lim_{x\rightarrow...
  3. A

    Proof by Induction Question on Determinants and Eigenvalues

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

    Proof by Induction Question on Determinants and Eigenvalues

    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
  5. A

    Proof by Induction Question on Determinants and Eigenvalues

    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...
  6. A

    Proof by Induction Question on Determinants and Eigenvalues

    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
  7. A

    Proof by Induction Question on Determinants and Eigenvalues

    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...
Back
Top