| Line 2: | Line 2: | ||
[[2013 Fall MA 527 Bell|MA527 Fall 2013]] | [[2013 Fall MA 527 Bell|MA527 Fall 2013]] | ||
| + | |||
---- | ---- | ||
| + | |||
'''Question''' from [[User:Rrusson|Ryan Russon]] | '''Question''' from [[User:Rrusson|Ryan Russon]] | ||
| Line 9: | Line 11: | ||
<br> | <br> | ||
| − | 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<sup>-1</sup>x 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<sup>-1</sup> for 'A' = P<sup>-1</sup>AP. --[[User:Asauseda|Andrew Sauseda]] 23:22, 9 September 2013 (UTC)<br> | + | 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<sup>-1</sup>x 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<sup>-1</sup> for 'A' = P<sup>-1</sup>AP. --[[User:Asauseda|Andrew Sauseda]] 23:22, 9 September 2013 (UTC)<br> |
| − | <br> | + | <br> |
| − | from [[User:Jayling|James Ayling]]: | + | from [[User:Jayling|James Ayling]]: I agree with you Andrew |
| − | from [[User:Rrusson|Ryan Russon]]: | + | from [[User:Rrusson|Ryan Russon]]: |
| − | That makes A LOT more sense. Thanks guys! | + | That makes A LOT more sense. Thanks guys! |
| − | from [[User: | + | from [[User:Rleemhui|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" | + | 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 [[User:Jayling|James Ayling]]: | + | from [[User:Jayling|James Ayling]]: Any suggestions on how to go about answering Page 345 Questions 24 and 25? |
| + | <br> Answer from [[User:Park296|Eun Young]] : | ||
| − | + | Q. 24. By thm 4, <span class="texhtml">''A'' = ''X''''D''''X''<sup> − 1</sup></span> where <span class="texhtml">''X''<sup> − 1</sup> = ''X''<sup>''T''</sup></span>. | |
| − | + | If we set <span class="texhtml">''X''<sup>''T''</sup>''x'' = ''y''</span>, then we have <span class="texhtml">''Q'' = ''y''<sup>''T''</sup>''D''''y''</span>. We just transformed Q to the canonical form. See P.343. | |
| − | + | So, <math> Q = x^T A x = y^T D y = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2 </math> where <span class="texhtml">''X''<sup>''T''</sup>''x'' = ''y''</span>. | |
| − | + | ||
| − | So, <math> Q = x^T A x = y^T D y = \lambda_1 y_1^2 + \cdots + \lambda_n y_n^2 </math> where < | + | |
Hence, the values of Q are controlled by the sings of the eigenvalues. | 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. | + | 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 | + | 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 = ( 1 0 0 ... ). Then, the first eigenvalue should be positive. | ||
| Line 44: | Line 45: | ||
Consider y = ( 0 1 0 ...). Then, the second eigenvalue should be positive. | Consider y = ( 0 1 0 ...). Then, the second eigenvalue should be positive. | ||
| − | In this manner, | + | In this manner, all eigenvalues are positive. |
| − | Conversely, if all eigenvalues are positive, Q | + | 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. | + | You can show the others similarly. |
| − | Q.25. | + | Q.25. Prob22 => <math>4x_1^2 + 12 x_1 x_2 + 13 x_2^2 = 16</math> <=> <span class="texhtml">''Q'' = ''x''<sup>''T''</sup>''A''''x''</span> where <math>A = \begin{bmatrix} 4 \ \ 6 \\ 6 \ \ 13 \end{bmatrix}</math>. |
| − | To show that this is positive definite, you need to check if < | + | To show that this is positive definite, you need to check if <span class="texhtml">''a''<sub>11</sub> > 0</span> and <span class="texhtml">''d''''e''''t''(''A'') > 0</span>. |
| − | From [[User:Jayling|James Ayling]]: | + | From [[User:Jayling|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''' 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? | ||
| + | |||
---- | ---- | ||
| − | + | ||
| + | From - Kunal Bansal: Sec: 8.4, Ques - 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. | ||
[[2013 Fall MA 527 Bell|Back to MA527, Fall 2013]] | [[2013 Fall MA 527 Bell|Back to MA527, Fall 2013]] | ||
[[Category:MA527Fall2013Bell]] [[Category:MA527]] [[Category:Math]] [[Category:Homework]] [[Category:Linear_algebra]] | [[Category:MA527Fall2013Bell]] [[Category:MA527]] [[Category:Math]] [[Category:Homework]] [[Category:Linear_algebra]] | ||
Revision as of 10:10, 14 September 2013
Homework 4 collaboration area
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 = XT.
If we set XTx = y, then we have Q = yTD'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 XTx = 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 = xTA'x where $ A = \begin{bmatrix} 4 \ \ 6 \\ 6 \ \ 13 \end{bmatrix} $.
To show that this is positive definite, you need to check if a11 > 0 and d'e't(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?
From - Kunal Bansal: Sec: 8.4, Ques - 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.
