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Here are some of the ways I looked at the problem to make sense of it: | Here are some of the ways I looked at the problem to make sense of it: | ||
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Remark from [[User:Bell|Steve Bell]]: | Remark from [[User:Bell|Steve Bell]]: | ||
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tank one at time t and y2 as the amount in tank two. Then | tank one at time t and y2 as the amount in tank two. Then | ||
Latest revision as of 13:05, 12 December 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
Response from AP Here are some of the ways I looked at the problem to make sense of it:
- I looked at this in terms of accumulation of mass. Mass of fertilizer entering - mass of fertilizer leaving is the amount of accumulation. The pure water has no fertilizer and therefore does not affect the mass of fertilizer entering or leaving. Therefore there is no affect on the accumulation or “de-accumulation” (if that is a word) of mass of fertilizer
- The units: The ratio of gal/min flow over gallons reduces to 1/min. The question asks for “the contents” of the tanks. The only way the units will work out is if the Y terms are in lbs to get lbs/min. Since the final equation is a function of time the units of minutes will cancel and leave us with lbs, a.k.a “the contents”
- When setting up the equations, y1’ is for one tank and y2’ is for the other. The y2 and y1 for each equation are the mass in and mass out but not respectively. If you call y1 “in” for y1’, you must call y1 out for y2’. The reason is the inflow of one tank is the outflow of the other.
- Since we swapped y1 and y2, we have to be careful to put them into the correct positions in the matrix.
- I am open for challenges or corrections of this thinking. Please.
Remark from Steve Bell:
AP, I think you need to think of y1 as the amount of fertalizer in tank one at time t and y2 as the amount in tank two. Then
y1' = (rate in) - (rate out)
and you can figure out those rates from the diagram. The rate coming in from tank 2 will depend on y2, and the pure water flowing in won't contribute. The rate out will depend on y1. Let me know if you want me to go over this problem in the review next week.
