Difference between revisions of "HWK4MA527Fall2013" - Rhea

Revision as of 10:04, 14 September 2013

Homework 4 collaboration area

MA527 Fall 2013

Question from Ryan Russon

Ok, what in the world are we supposed to do with p. 345 #1? From what I gather, The problem asks us to find the eigenvectors of this matrix P, A (the given matrix), and then some new 'A' where A = P^-1AP and show that x = Py, where y is an eigenvector of P and x is an eigenvector of both A and this new A. The eigenvalues of P are really crummy... are we allowed to use a calculator or MATLAB or something to help us with this one? Maybe I am just over-complicating this problem... HELP!

I could be mistaken, but I think question #1 should read "if y is an eigenvector of 'A' , show that x = Py are eigenvectors of A". Theorem 3 on p. 340 says that "if x is an eigenvector of A, then y = P^-1x is an eigenvector of 'A' corresponding to the same eigenvalue". If that is the case, x = Py, and the only calculation you need to do on P is to find P^-1 for  'A' = P^-1AP. --Andrew Sauseda 23:22, 9 September 2013 (UTC)

from James Ayling: I agree with you Andrew

from Ryan Russon:

That makes A LOT more sense. Thanks guys!

from Ryan Leemhuis:

Glad I wasn't the only one to come here with this question. I agree...error in the book. This was realized after calculating the eigenvalues of P and finding they are not very "user friendly"

from James Ayling: Any suggestions on how to go about answering Page 345 Questions 24 and 25?

Answer from Eun Young :

Q. 24. By thm 4, $ A= X D X^{-1} $ where $ X^{-1} = X^T $.

If we set $ X^T x = y $, then we have $ Q = y^T D y $. We just transformed Q to the canonical form. See P.343.

So, $ Q = x^T A x = y^T D y = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2 $ where $ X^T x = y $.

Hence, the values of Q are controlled by the sings of the eigenvalues.

We know that there is a one-to-one correspondence between all nonzero x and all nonzero y since X is invertible.

Hence, if Q > 0 for all nonzero x, Q > 0 for all nonzero y.

Consider y = ( 1 0 0 ... ). Then, the first eigenvalue should be positive.

Consider y = ( 0 1 0 ...). Then, the second eigenvalue should be positive.

In this manner, all eigenvalues are positive.

Conversely, if all eigenvalues are positive, Q >0 for all nonzero y. So, Q >0 for all nonzero x.

You can show the others similarly.

Q.25. Prob22 => $ 4x_1^2 + 12 x_1 x_2 + 13 x_2^2 = 16 $ <=> $ Q = x^T A x $ where $ A = \begin{bmatrix} 4 \ \ 6 \\ 6 \ \ 13 \end{bmatrix} $.

To show that this is positive definite, you need to check if $ a_{11} > 0 $ and $ det(A) > 0 $.

From James Ayling: Thanks Eun for the clarification.

Question From bminchhoff: for problem 25 I understand that all principle minors must be positive for positive definiteness but how do we know if principle minors not positive are negative definite or indefinite?

Question - 25: To show whether its negative definite or indefinite (In case it is not positive definite)- I think Calculating eigen values will help. As one of the eigen value in this case is positive and other one is negative, that satisfies the condition for indefinite.(Kunal Bansal)

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