The following system is given: x ˙ = y + z y ˙ = x + z z ˙ = x + yThe first thing I did was to find out the eigenvalues. I found out, that -1 is a doubled and 2 a single eigenvalue, so λ 1 , 2 = − 1 , λ 3 = 2In the excercises ago, the ideas were to determine y = e λ x u _ . so I tried to do the following: ( 0 − λ 1 1 1 0 − λ 1 1 1 0 − λ ) Is this step right? I tried to find a scheme as in the excercises ago and in the line x˙=y+z I don't have an x but one y and one z. x ˙ = y + zWhen inserting λ 1 = − 1 I have A − λ E = 0 _ → ( 1 1 1 1 1 1 1 1 1 ) = 0 _ which means that x i + y i + z i = 0 f o r i = 1 , 2 , 3 . Here is the point on which I don't know how to go on. One solution is the trivial one, so x=y=z=0. Can I use this solution?I think that I have to use something like y = C 1 ∗ ( u 1 ∗ e λ 1 x u 2 ∗ e λ 1 x u 3 ∗ e λ 1 x ) + C 2 ( . . . ) + C 3 ( . . . ) in the case λ 1 = 1, but how to I get my u here exactly?

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The following system is given:      x    ˙    =  y  +  z        y    ˙    =  x  +  z        z    ˙    =  x  +  yThe first thing I did was to find out the eigenvalues. I found out, that -1 is a doubled and 2 a single eigenvalue, so      λ          1      ,      2        =  −  1  ,            λ    3    =  2In the excercises ago, the ideas were to determine   y  =      e          λ      x            u    _  . so I tried to do the following:      (                            0          −          λ                          1                          1                                      1                          0          −          λ                          1                                      1                          1                          0          −          λ                      )  Is this step right? I tried to find a scheme as in the excercises ago and in the line x˙=y+z I don&#039;t have an x but one y and one z.      x    ˙    =  y  +  zWhen inserting       λ    1    =  −  1 I have  A  −  λ  E  =      0    _    →      (                            1                          1                          1                                      1                          1                          1                                      1                          1                          1                      )    =      0    _  which means that       x    i    +      y    i    +      z    i    =  0     f  o  r     i  =      1    ,    2    ,    3  . Here is the point on which I don&#039;t know how to go on. One solution is the trivial one, so x=y=z=0. Can I use this solution?I think that I have to use something like  y  =      C    1    ∗      (                                        u            1                    ∗                      e                                          λ                1                            x                                                                        u            2                    ∗                      e                                          λ                1                            x                                                                        u            3                    ∗                      e                                          λ                1                            x                                            )    +      C    2        (                            .          .          .                      )    +      C    3        (                            .          .          .                      )  in the case       λ    1    =  1, but how to I get my u here exactly?

Madisyn Avery · Accepted Answer

The equations can be written as       p    ˙    =  A  p, with   p  ∈        R    3   and   A  =      [                            0                          1                          1                                      1                          0                          1                                      1                          1                          0                      ]  .Note that   A  =  v      v    T    −  I, where   v  =  (  1  ,  1  ,  1      )    T  , so it has one eigenvalue at 2 corresponding to the eigenvector   v, and two at -1 corresponding to the eigenspace   {  v      }    ⊥  . (Note that   A is symmetric hence has an orthonormal basis of eigenvectors.)Hence if you write   p  =  α  v  +  w  ∈        R    3  , where   w  ⊥  v, you will have   A  p  =  α  2  v  −  w, and, in general,       A    k    p  =  α      2    k    v  +  (  −  1      )    k    w, from which we see that       e          A      t        p  =  α      e          2      t        v  +      e          −      t        w.The projection of a point   p  ∈        R    3   onto   v is straightforward to compute.The point here is that you don&#039;t need to explicitly find eigenvectors for the eigenspace  {  v      }    ⊥  .

The following system is given: <mover> x ˙ </mover> = y

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