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from [[User:Jayling|James Ayling]]: Any suggestions on how to go about answering Page 345 Questions 24 and 25? | from [[User:Jayling|James Ayling]]: Any suggestions on how to go about answering Page 345 Questions 24 and 25? | ||
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| + | Answer from [[User:Park296|Eun Young]] : | ||
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| + | Q. 24. By thm 4, <math>A= X D X^{-1}</math> where <math>X^{-1} = X^T</math>. | ||
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| + | If we set <math> X^T x = y </math>, then we have <math> Q = y^T D y </math>. We just transformed Q to the canonical form. See P.343. | ||
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| + | So, <math> Q = x^T A x = y^T D y = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2 </math> where <math>X^T x = y</math>. | ||
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| + | Hence, the values of Q are controlled by the sings of the eigenvalues. | ||
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| + | We know that there is a one-to-one correspondence between all nonzero x and all nonzero y since X is invertible. | ||
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| + | Hence, if Q > 0 for all nonzero x, Q > 0 for all nonzero y. | ||
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| + | Consider y = ( 1 0 0 ... ). Then, the first eigenvalue should be positive. | ||
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| + | Consider y = ( 0 1 0 ...). Then, the second eigenvalue should be positive. | ||
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| + | In this manner, all eigenvalues are positive. | ||
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| + | You can show the others similarly. | ||
Revision as of 06:36, 12 September 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 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.
You can show the others similarly.
