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Systems Of Linear D.E's, Complex Eigenvalues

  • #1
1. Find the General Solution of the given system

[ -1 -1 2 ] X = X'
[ -1 1 0 ]
[ -1 0 1 ]

det(A-lambda*Identity matrix) = 0, solve for eigenvalues/values of lambda
(A-lambda*Identity matrix|0)



The eigenvalues we got are 1 and 1 +/- i. The matrix generated for (A-(1-i)I|0) is

[ i -1 2 | 0 ]
[ -1 i 0 | 0 ]
[ -1 0 i | 0 ]

Where do we go from here??
 
Last edited:

Answers and Replies

  • #2
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4,858
I think you have made a mistake on your eigenvalue calculations. The eigenvalues I get are [itex]\lambda = 0[/itex] (of multiplicity 2) and [itex]\lambda = 1[/itex].

After you have found the eigenvectors, you need to do this for each one:
Substitute the [itex]\lambda [/itex] in the matrix A - [itex]\lambda[/itex]I.
Row reduce this matrix. You are guaranteed that the row-reduced matrix will have at least one row of zeroes.
Of the nonzero rows that remain, solve for x1 in terms of the other variable(s). Solve for xx in terms of the other variables.

As an example, if you ended up with this:
[1 2 0]
[0 1 -1]
[0 0 0]

This says that
x1 = -2x2
x2 = x3
x3 = x3

The three variables on the left side are given in terms of x2 and x3 on the right side. If you choose x2 = 1 and x3 = 0, that gives you one solution. If you choose x2 = 0 and x3 = 1, that gives you another solution. Every possible solution is a linear combination of values for you choose x2 and x3.
 
  • #3
[ 1 -1 2 ] X = X' .........that first one in the upper left is not negative :( sorry!
[ -1 1 0 ]
[ -1 0 1 ]
 

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