# Which statements are true given a solution to the system x'=Ax?

• JessicaHelena
In summary, the solution [x, y] = e^{-3t} [-2, -1] to the system $x' = Ax$ has an eigenvalue of -3 and [4, 2] as an eigenvector of A. The trajectory of this solution in the phase plane with axes x, y is part of a straight line, and the phase portrait of the system is an attracting node. The eigenvalues and eigenvectors of A cannot be determined without additional information, and therefore, it is not possible to confirm if e^{-3t} [-1, -2] is another solution or if the system is stable.
JessicaHelena
Summary:: Suppose that [x, y] = e^{-3t} [-2, -1] is a solution to the system $x' = Ax$, where A is a matrix with constant entries. Which of the following must be true?

a. -3 is an eigenvalue of A.
b. [4, 2] is an eigenvector of A.
c. The trajectory of this solution in the phase plane with axes x, y is part of a straight line.
d. e^{-3t} [-1, -2] is another solution to the same system.
e. the phase portrait of the system is an attracting node.
f. the system is stable.

Please refer to the attached screenshot for clearer question; however, the checked answers there are not my final answers.

This is a star node, and hence ##\lambda = -3##. (Hence ##a## is right.) Since the eigenvalue is negative, it is an attracting node (which makes ##e## correct).
I graphed the parametric function, and the result was a straight line extending only in the 3rd quadrant. Hence ##c## is right.
I wasn't sure of a good mathematical way to get the matrix A, but by trial and error, got that A = [0, -6; -3, 3]. Checking for ##Av = \lambda v##, I got the right answer, which makes me believe I have the right A. From this, trA = 3, and det A = -18. Checking against the rule of stability, we get that this system is not stable. (hence ##f## is not an answer).

I am not sure how exactly to do b and d, but using the rule that ##Av = \lambda v##, I don't believe they are right.

Hence I would check a, c, and e. Is this way of thinking right?

[Moderator's note: Moved from a technical forum and thus no template.]

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JessicaHelena said:
Since the eigenvalue is negative, it is an attracting node (which makes eee correct).
What about the other eigenvalues of A?

JessicaHelena said:
I wasn't sure of a good mathematical way to get the matrix A, but by trial and error, got that A = [0, -6; -3, 3].
That would be because there is no mathematical way. You do not have enough information to determine A.

JessicaHelena said:
I am not sure how exactly to do b and d, but using the rule that Av=λvAv=λvAv = \lambda v, I don't believe they are right.
Is [-2,-1] an eigenvector of A?

@Orodruin

Orodruin said:
What about the other eigenvalues of A?
I had thought that this was a repeated eigenvalue, since the second constant c2 = 0.
Orodruin said:
Is [-2,-1] an eigenvector of A?
I had thought that the way solutions are formatted is
[x, y] = c_1 e^{##\lambda_1 t##} *eigenvector_1 + c_2 e^{##\lambda_2 t##}*eigenvector_2.
From this, I thought that c_2 = 0 (which led me to the conclusion that this was a star node) and that [-2, -1] was an eigenvector of A.

JessicaHelena said:
I had thought that this was a repeated eigenvalue, since the second constant c2 = 0.
It just says that this is a solution. A priori, this does not tell you anything about the other eigenvalues.

JessicaHelena said:
I had thought that the way solutions are formatted is
[x, y] = c_1 e^{##\lambda_1 t##} *eigenvector_1 + c_2 e^{##\lambda_2 t##}*eigenvector_2.
From this, I thought that c_2 = 0 (which led me to the conclusion that this was a star node) and that [-2, -1] was an eigenvector of A.

Yes, ##c_2## is zero, that is why you know nothing about the second eigenvalue. However, it is correct that [-2 -1] is an eigenvector. Are [4 2] and [-2 -1] linearly independent?

@Orodruin

Orodruin said:
is zero, that is why you know nothing about the second eigenvalue. However, it is correct that [-2 -1] is an eigenvector. Are [4 2] and [-2 -1] linearly independent?
Well, yes, which actually makes b true as well.

Hmm, so. So far I know that

a. -3 is an eigenvalue of A.
b. [4, 2] is an eigenvector of A.
are right.

c. The trajectory of this solution in the phase plane with axes x, y is part of a straight line.
should be right because I graphed the parametric function.

But since I don't know the matrix A, and there isn't a surefire way to figure out one (from what I gather),

d. e^{-3t} [-1, -2] is another solution to the same system.
e. the phase portrait of the system is an attracting node.
f. the system is stable.

will not necessarily be right because
d: we don't know A, so I cannot really use ##Av = \lambda v## to check
e: While negative eigenvalues do make attracting node, if the system has another eigenvalue that is non-negative, the phase portrait wouldn't be an entirely attracting node
f. I need that trA < 0 and detA > 0 for the system to be stable, but I don't know A.

Is this way of thinking right?

Looks ok to me.

Yes—that turned out to be right. Thank you!

## 1. What is the purpose of the system x'=Ax?

The system x'=Ax is a differential equation used to model and understand the behavior of a system over time. It describes how the state of a system, represented by the vector x, changes over time when subjected to external influences, represented by the matrix A.

## 2. What does the solution to the system x'=Ax represent?

The solution to the system x'=Ax represents the value of the state vector x at any given time, t. It provides a complete description of the behavior and evolution of the system over time.

## 3. Can a solution to the system x'=Ax be unique?

Yes, a solution to the system x'=Ax can be unique if the matrix A is invertible. In this case, there is only one possible solution that satisfies the initial conditions for the state vector x.

## 4. How can the solution to the system x'=Ax be calculated?

The solution to the system x'=Ax can be calculated by finding the eigenvalues and eigenvectors of the matrix A, which can then be used to construct the general solution. The initial conditions can then be used to find the specific solution for a given system.

## 5. Are there any real-world applications of the system x'=Ax?

Yes, the system x'=Ax has many real-world applications in fields such as physics, engineering, and economics. It is commonly used to model physical systems like chemical reactions, population growth, and electrical circuits, among others. It is also used in control systems and optimization problems in engineering and economics.

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