Eigenvalues with an added part

  • Thread starter TheTopGun
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In summary, if you know the eigenvalues of matrix 'A', you can solve the DE ##\dot{x} = Ax + b## using a matrix exponential.
  • #1
TheTopGun
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Homework Statement


Hi, I have an electrical circuit, from which I have derived 4 equations to work out the current I of the circuit. To solve I need to put the equations into a matrix and find the eigenvalues & vectors, great I can do that. However there is an additional matrix on the end for voltage values. Previously it was assumed the circuit was DC so all voltages would = 0, as you can see in the picture below. Now its AC and I don't know what to do with the 'extra' matrix as its now a function and has to be considered.

This probably makes little sense as I admit a lot is going over my head.

Homework Equations



This picture best describes the situation.

https://www.dropbox.com/s/5y2oe359a84f7za/IMG_20141005_202343113.jpg?dl=0

The Attempt at a Solution



I can find the eigenvalues for matrix 'A' but I don't know where the extra bit comes into play...

Thankyou for any guidance.
 
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  • #2
Treat the problem as a 1st order differential equation, there are two solutions which must be added together, the homogeneous part then the inhomogeneous portion. Look in your differential equations text or the CRC Math Handbook. Looking in my Boyce and Diprima, third ed (1977), Section 7.10, the solution of your type of problem is given in detail.
 
  • #3
TheTopGun said:

Homework Statement


Hi, I have an electrical circuit, from which I have derived 4 equations to work out the current I of the circuit. To solve I need to put the equations into a matrix and find the eigenvalues & vectors, great I can do that. However there is an additional matrix on the end for voltage values. Previously it was assumed the circuit was DC so all voltages would = 0, as you can see in the picture below. Now its AC and I don't know what to do with the 'extra' matrix as its now a function and has to be considered.

This probably makes little sense as I admit a lot is going over my head.

Homework Equations



This picture best describes the situation.

https://www.dropbox.com/s/5y2oe359a84f7za/IMG_20141005_202343113.jpg?dl=0

The Attempt at a Solution



I can find the eigenvalues for matrix 'A' but I don't know where the extra bit comes into play...

Thankyou for any guidance.

If you know the eigenvalues of ##A##, you can solve the DE ##\dot{x} = Ax + b## using a matrix exponential:
[tex] x(t) = e^{At} x(0) + \int_0^t e^{A(t-s)} b(s) \, ds [/tex]
The matrix exponential can be computed from the eigenvalues and their multiplicities:
(1) If all eigenvalues ##r_1, r_2, \ldots, r_n## of ##A## are distinct, there exist matrices ##E_1, E_2, \ldots, E_n## such that
[tex] e^{At} = \sum_{i=1}^n E_i e^{r_i t} [/tex]
More generally, for any analytic function ##f## we have
[tex] f(A) = \sum_{i=1}^n E_i f(r_i)[/tex]
where the ##E_i## are the same for any function. One way to find the ##E_i## is to apply this to various ##f## for which ##f(A)## is readily computable:
[tex] f(x) = 1 = x^0 \Rightarrow f(A) = I = \sum_{i=1}^n r^0 E_i = \sum_{i=1}^n E_i\\
f(x) = x \Rightarrow f(A) = A = \sum_{i=1}^n r_i^1 E_i = \sum_{i=1}^n r_i E_i \\
f(x) = x^2 \Rightarrow f(A) = A^2 = \sum_{i=1|}^n r_i^2 E_i,\\
\vdots
[/tex]
Here ##I## is the identity matrix.
For example, for a 4x4 matrix with four distinct eigenvalues ##r_1,r_2, r_3,r_4## we have
[tex] I = E_1 + E_2 + E_3 + E_4\\
A = r_1 E_1 + r_2 E_2 + r_3 E_3 + r_4 E_4\\
A^2 = r_1^2 E_1 + r_2^2 E_2 + r_3^2 E_3 + r_4^2 E_4\\
A^3 = r_1^3 E_1 + r_2^3 E_2 + r_3^3 E_3 + r_4^3 E_4
[/tex]
After solving for the ##E_i## we have
[tex] e^{Aw}= E_1 e^{r_1 w} + E_2 e^{r_2 w} + E_3 e^{r_3 w} + E_4 e^{r_4 w}[/tex]

(2) If not all eigenvalues of ##A## are distinct, it is a bit more complicated, but the first order of business would be to check if, indeed, your eigenvalues are all distinct. After that, complications can be dealt with if needed.
 

1. What are eigenvalues with an added part?

Eigenvalues with an added part, also known as generalized eigenvalues, are a set of numbers that describe the behavior of a linear transformation on a vector space. They are calculated by finding the roots of the characteristic polynomial of the transformation matrix.

2. How are eigenvalues with an added part used in science?

Eigenvalues with an added part are used in various scientific fields, such as physics, chemistry, and engineering. They are particularly useful in studying the behavior of systems that involve multiple variables and interactions, as well as in analyzing the stability and control of complex systems.

3. Can eigenvalues with an added part be negative?

Yes, eigenvalues with an added part can be negative, positive, or zero. The sign of the eigenvalue is important in determining the type of transformation that the matrix represents, such as rotation, scaling, or shearing.

4. How do eigenvalues with an added part relate to eigenvectors?

Eigenvalues with an added part are closely related to eigenvectors, which are the corresponding vectors that are unaffected by the transformation. The eigenvalue represents the scaling factor by which the eigenvector is multiplied when transformed by the matrix.

5. Are eigenvalues with an added part unique?

Yes, eigenvalues with an added part are unique for a given transformation matrix. However, a matrix may have multiple eigenvectors with the same eigenvalue. In this case, the eigenvectors form a basis for the corresponding eigenspace.

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