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== Homework 4 collaboration area == | == Homework 4 collaboration area == | ||
| − | [[2013 Fall MA 527 Bell|MA527 Fall 2013]] | + | [[2013 Fall MA 527 Bell|MA527 Fall 2013]] |
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is leading principal minor number two. That's all of them that need checking. | is leading principal minor number two. That's all of them that need checking. | ||
| − | However, in the case of a 2x2, it is so easy to find the eigenvalues and diagonalize that this test might not be needed. | + | However, in the case of a 2x2, it is so easy to find the eigenvalues and diagonalize that this test might not be needed. |
Note that a matrix A is positive definite if and only if -A is negative definite. | Note that a matrix A is positive definite if and only if -A is negative definite. | ||
| − | <br> '''RESPONSE from Mickey Rhoades '''[[User:Mrhoade|'''Mrhoade'']]'''''''''''' | + | <br> '''RESPONSE from Mickey Rhoades '''[[User:Mrhoade|'''Mrhoade''']]''''''''''''''''' |
| − | I've researched this a bit and here is what I have found: Let A be a symmetric n × n matrix. D<sub>k</sub> is a leading principle minor of order k and ∆<sub>k</sub> is a principle minor of order k. Leading principle minors are a subset of the principle minors. Then we have:<br>1) A is positive definite ⇔ D<sub>k</sub> > 0 for all leading principal minors<br>2) A is negative definite ⇔ (−1) | + | I've researched this a bit and here is what I have found: Let A be a symmetric n × n matrix. D<sub>k</sub> is a leading principle minor of order k and ∆<sub>k</sub> is a principle minor of order k. Leading principle minors are a subset of the principle minors. Then we have:<br>1) A is positive definite ⇔ D<sub>k</sub> > 0 for all leading principal minors<br>2) A is negative definite ⇔ (−1)k D<sub>k</sub> > 0 for all leading principal minors |
| − | 3) A is positive semidefinite ⇔ ∆<sub>k</sub> ≥ 0 for all principal minors<br>4) A is negative semidefinite ⇔ (−1) | + | 3) A is positive semidefinite ⇔ ∆<sub>k</sub> ≥ 0 for all principal minors<br>4) A is negative semidefinite ⇔ (−1)k ∆<sub>k</sub> ≥ 0 for all principal minors |
| − | For the matrix in problem 22, the leading principle minors are D<sub>1</sub> = +4, and D<sub>2</sub> = +16 which are all all positive. Their is also a non-leading principle minor of ∆<sub>1 </sub>= +13. For the matrix in problem 23, the leading principle minors are D<sub>1</sub> = -11 and D<sub>2</sub> = -2028 and a non-leading pronciple minor of ∆<sub>1</sub> = +24. This is clearly not positive definite under condition 1 or negative definite under condition 2 because (-1) | + | For the matrix in problem 22, the leading principle minors are D<sub>1</sub> = +4, and D<sub>2</sub> = +16 which are all all positive. Their is also a non-leading principle minor of ∆<sub>1 </sub>= +13. For the matrix in problem 23, the leading principle minors are D<sub>1</sub> = -11 and D<sub>2</sub> = -2028 and a non-leading pronciple minor of ∆<sub>1</sub> = +24. This is clearly not positive definite under condition 1 or negative definite under condition 2 because (-1)2 x -2028 is not > 0. Condition 3 is also not met nor is condition 4. We should then conclude that it is indefinite. Also, indefiniteness is defined as: indefinite if xTAx > 0 for some x∈Rn, and < 0 for some other x∈Rn. <br> This is the best I can come up with as its still not crystal clear to me. - Mick ---- <br> |
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| − | <br> '''Question''' from [[User:Rrusson|Ryan Russon]] | + | <br> '''Question''' from [[User:Rrusson|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 | + | 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! |
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| − | 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 | + | 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. --[[User:Asauseda|Andrew Sauseda]] 23:22, 9 September 2013 (UTC)<br> |
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<br> Answer from [[User:Park296|Eun Young]] : | <br> Answer from [[User:Park296|Eun Young]] : | ||
| − | Q. 24. By thm 4, <span class="texhtml">''A'' = ''X''''D''''X'' | + | Q. 24. By thm 4, <span class="texhtml">''A'' = ''X''''D''''X'' − 1</span> where <span class="texhtml">''X'' − 1 = ''XT''</span>. |
| − | If we set <span class="texhtml">'' | + | If we set <span class="texhtml">''XTx'' = ''y''</span>, then we have <span class="texhtml">''Q'' = ''yTD''''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">'' | + | 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">''XTx'' = ''y''</span>. |
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. | ||
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You can show the others similarly. | You can show the others similarly. | ||
| − | Q.25. Prob22 => <math>4x_1^2 + 12 x_1 x_2 + 13 x_2^2 = 16</math> <=> <span class="texhtml">''Q'' = '' | + | Q.25. Prob22 => <math>4x_1^2 + 12 x_1 x_2 + 13 x_2^2 = 16</math> <=> <span class="texhtml">''Q'' = ''xTA''''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 <span class="texhtml">''a''<sub>11</sub> > 0</span> and <span class="texhtml">''d''''e''''t''(''A'') > 0</span>. | 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>. | ||
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y2'=+(16/200)*y1-(4/200)*y2 | y2'=+(16/200)*y1-(4/200)*y2 | ||
| − | The eigenvalues and eigenvectors I get from this are 0, -0.1 and [0.25 1]<span class="texhtml" | + | The eigenvalues and eigenvectors I get from this are 0, -0.1 and [0.25 1]<span class="texhtml">''T''</span>, [-1 1]<span class="texhtml">''T''</span>. |
When I solve for the initial conditions I get C1=-40 and C2=240. | When I solve for the initial conditions I get C1=-40 and C2=240. | ||
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| − | [[2013 Fall MA 527 Bell|Subrina: Regarding question 18 Section 4.3]] | + | [[2013 Fall MA 527 Bell|Subrina: Regarding question 18 Section 4.3]] |
Yes we have to consider the mixed water rate for both tank. For tank T1, it does not matter though. But for tank T2, it makes a significant difference. | Yes we have to consider the mixed water rate for both tank. For tank T1, it does not matter though. But for tank T2, it makes a significant difference. | ||
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Response from [[User:Jayling|Jayling]]: Thanks Subrina that now makes sense to me. If you look at it in terms of fertiliser only then the injection of pure water does not matter, however I did not take into account the exit of the mixed water from T2. I get the same results as you now Subrina. Keith you may have come unstuck with the algebra, I came unstuck with the flow rate equations... Thanks for everyone's contributions so far. | Response from [[User:Jayling|Jayling]]: Thanks Subrina that now makes sense to me. If you look at it in terms of fertiliser only then the injection of pure water does not matter, however I did not take into account the exit of the mixed water from T2. I get the same results as you now Subrina. Keith you may have come unstuck with the algebra, I came unstuck with the flow rate equations... Thanks for everyone's contributions so far. | ||
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| + | Response from Mickey Rhoades [[User:Mrhoade|Mrhoade]] | ||
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| + | I got the same answer. The inflow of pure water just ensures that the denominator of 200 gallons in the tank remains constant. This problem would be much more complicated if this was not the case. I got the same. y<sub>1 </sub>= 100e<sup>-.04t</sup> and y<sub>2</sub> = 200e<sup>-04t</sup> | ||
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| − | I got the same equations but goofed my algebra, you're right. Thanks all for inputs. Keith | + | I got the same equations but goofed my algebra, you're right. Thanks all for inputs. Keith |
| − | [[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 15:34, 17 September 2013
Homework 4 collaboration area
Question from student in regard to page 345 #25: I found a general formula for principal minors of a 3X3 matrix. Is there a general formula for the principal minors of a 2X2 matrix?
Answer: a11 is leading principal minor number one and
a11a22 − a12a21
is leading principal minor number two. That's all of them that need checking.
However, in the case of a 2x2, it is so easy to find the eigenvalues and diagonalize that this test might not be needed.
Note that a matrix A is positive definite if and only if -A is negative definite.
'RESPONSE from Mickey Rhoades Mrhoade''''''''''''
I've researched this a bit and here is what I have found: Let A be a symmetric n × n matrix. Dk is a leading principle minor of order k and ∆k is a principle minor of order k. Leading principle minors are a subset of the principle minors. Then we have:
1) A is positive definite ⇔ Dk > 0 for all leading principal minors
2) A is negative definite ⇔ (−1)k Dk > 0 for all leading principal minors
3) A is positive semidefinite ⇔ ∆k ≥ 0 for all principal minors
4) A is negative semidefinite ⇔ (−1)k ∆k ≥ 0 for all principal minors
For the matrix in problem 22, the leading principle minors are D1 = +4, and D2 = +16 which are all all positive. Their is also a non-leading principle minor of ∆1 = +13. For the matrix in problem 23, the leading principle minors are D1 = -11 and D2 = -2028 and a non-leading pronciple minor of ∆1 = +24. This is clearly not positive definite under condition 1 or negative definite under condition 2 because (-1)2 x -2028 is not > 0. Condition 3 is also not met nor is condition 4. We should then conclude that it is indefinite. Also, indefiniteness is defined as: indefinite if xTAx > 0 for some x∈Rn, and < 0 for some other x∈Rn.
This is the best I can come up with as its still not crystal clear to me. - Mick ----
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'xwhere $ 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 (for prob - 23)- 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.
Hi all. I am facing a problem with Question 18, Section 4.3 (mixing problem). The way I understand it, the pure water entry of 12 gal/min into T1 should not reflect into the fertilizer rate change equations? Ofcourse it keeps the volume constant in both tanks. If I continue in this way, I find that the eigen vectors to the two Eigen values are identical. When I go to solve for the initial value conditions I am not able to proceed in getting values for the constants (getting c1+c2=-100, 2*c1+2*c2=200). I am getting stuck, please advise. Thanks! &&&
Response from Jayling: Yeah I was going a bit cross eyed too with this problem. However I did not get identical eigenvalues. Can you tell me what your fertiliser rate equations for T1 and T2 are? For me when you just consider the "fertiliser" the rate equations I get are:
y1'=-(16/200)*y1+(4/200)*y2
y2'=+(16/200)*y1-(4/200)*y2
The eigenvalues and eigenvectors I get from this are 0, -0.1 and [0.25 1]T, [-1 1]T.
When I solve for the initial conditions I get C1=-40 and C2=240.
The issue that I have with this is similar to the above person's issue: The fact that the amount of pure water rate going into the system is identical to the amount of mixed water rate leaving the system, does this imply that we can treat the fertiliser system as closed? Or should we consider a pure water rate such as y3' entering the system, and then instead you are solving a 3x3 system rather than a 2x2 system? Or am I as usual overthinking this?
Don't we need to consider the mixed water rate leaving T2? In this case there will be a -(12/200)*y2 term in equation for y2'. - Keith Rodrigues (same person as original question asker!)
Subrina: Regarding question 18 Section 4.3
Yes we have to consider the mixed water rate for both tank. For tank T1, it does not matter though. But for tank T2, it makes a significant difference.
I got bellow results
y1'=-(16/200)*y1+(4/200)*y2+12*0
y2'=+(16/200)*y1-(4+12/200)*y2
The eigenvalues and eigenvectors I get from this are -0.12, -0.04 and [1 -2]T, [1 2]T.
When I solve for the initial conditions I get C1=0 and C2=100.
Response from Jayling: Thanks Subrina that now makes sense to me. If you look at it in terms of fertiliser only then the injection of pure water does not matter, however I did not take into account the exit of the mixed water from T2. I get the same results as you now Subrina. Keith you may have come unstuck with the algebra, I came unstuck with the flow rate equations... Thanks for everyone's contributions so far.
Response from Mickey Rhoades Mrhoade
I got the same answer. The inflow of pure water just ensures that the denominator of 200 gallons in the tank remains constant. This problem would be much more complicated if this was not the case. I got the same. y1 = 100e-.04t and y2 = 200e-04t
I got the same equations but goofed my algebra, you're right. Thanks all for inputs. Keith
