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We have the differential equation $ \dot x = Ax$,

$$ A =\begin{pmatrix} 8 & 12 & -2 \\ -3 & -4 & 1 \\ -1 & -2 & 2 \\ \end{pmatrix} $$

  • I calculated the determinant $det(A-\lambda E) = - (\lambda - 2)^3$, to find the eigenvalues $\lambda $ for the eigenvectors $v$ of our given matrix $A$ from the characteristic polynomial mentioned above.

  • We have the only eigenvalue: $\lambda _{1,2,3} = 2$ and the eigenvector: $$ v_1 =\begin{pmatrix} -1 \\ 2 \\ 0 \\ \end{pmatrix} $$

  • A solution of the differential equations generally looks like $x(t) = Ce^{\lambda _1}v_1 + Be^{\lambda _2}v_2 $ where $C,B$ are some constants, but I don't know how to get it in my case, because I already know the solution.

  • The solution is according to my textbook:

$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \end{pmatrix} = e^{2t} \left \{ A \begin{pmatrix} 2 \\ -1 \\ 0 \\ \end{pmatrix} + B \left [ \begin{pmatrix} 2 \\ -1 \\ -1 \\ \end{pmatrix} + t\begin{pmatrix} 2 \\ -1 \\ 0 \\ \end{pmatrix} \right ] + C \left [ \begin{pmatrix} -1 \\ 1 \\ 2 \\ \end{pmatrix} + \begin{pmatrix} 2 \\ -1 \\ -1 \\ \end{pmatrix}t + \begin{pmatrix} 2 \\ -1 \\ 0 \\ \end{pmatrix}t^2/2 \right ] \right \} $$

-- Where $A,B,C$ are the constants

  • I don't understand the part with that constant $C$ where is the parameter $t$, or globally: what to do with just one eigenvalue and eigenvector of the matrix $A$.

Can anyone explain me this problem please?

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    $\begingroup$ You have a deficient matrix and cannot find three linearly independent eigenvectors and in this case, need two generalized eigenvectors. See Example $3$ for the form of the solution in these notes: mathcs.holycross.edu/~spl/old_courses/304_fall_2008/handouts/…. There are also many examples in MSE. $\endgroup$ – Moo Apr 17 '17 at 0:17
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As Moo has explained in his comment to your question, the matrix is deficient and can’t be diagonalized, so you have to go to the Jordan decomposition instead. If your textbook is giving you such a problem to solve, I’m sure that this has been covered somewhere in the preceding material. The link Moo provided has a good explanation of the process (see Example 3 in particular, as he said), and a search on this site turns up this explanation and many others, so I’m not going to go through that here.

The Jordan basis that the textbook used is evidently $(2,-1,0)^T$, $(2,-1,-1)^T$ and $(-1,1,2)^T$, so we have the decomposition $$A=PJP^{-1}=\left[\begin{array}{r}2&2&-1\\-1&-1&1\\0&-1&2\end{array}\right]\begin{bmatrix}2&1&0\\0&2&1\\0&0&2\end{bmatrix}\left[\begin{array}{r}2&2&-1\\-1&-1&1\\0&-1&2\end{array}\right]^{-1}.$$ Just as with a diagonalizable matrix, the $P$s cancel in powers of this expression, so $e^{tA}=Pe^{tJ}P^{-1}$. We can write $J$ as the sum of the diagonal matrix $2I$ and the nilpotent matrix $N=\tiny{\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}}$, so that $e^{tJ}=e^{t(2I+N)}=e^{2tI}e^{tN}$. The last equality doesn’t hold for matrices in general, but does when they commute, as is the case here. The first exponential is simply $e^{2t}I$, and we can compute the second by using the series expansion of the exponential $$e^{tN}=I+tN+{t^2\over2!}N^2+{t^3\over3!}N^3+\cdots$$ This series is truncated after three terms because, as you can verify for yourself, $N^3=0$. Putting this all together, the solution to the differential equation is $$e^{tA}\mathbf C=e^{2t}P\left(I+tN+\frac12t^2N^2\right)P^{-1}\mathbf C=e^{2t}P\begin{bmatrix}1&t&\frac12t^2\\0&1&t\\0&0&1\end{bmatrix}P^{-1}\mathbf C,$$ where $\mathbf C$ is a vector of constants to be determined by the boundary conditions. Since these constants are arbitrary, we can absorb $P^{-1}$ into them, so this becomes $$e^{2t}\left[\begin{array}{r}2&2&-1\\-1&-1&1\\0&-1&2\end{array}\right]\begin{bmatrix}1&t&\frac12t^2\\0&1&t\\0&0&1\end{bmatrix}\begin{bmatrix}A\\B\\C\end{bmatrix}.$$ Expand this product using the fact that the columns of a matrix product are linear combinations of the left-hand factor’s columns, and you end up with the textbook solution.

Although going through a full Jordan decomposition computation builds character, it’s not really necessary to do so in order to compute the exponential of $A$. This matrix can be decomposed directly into the sum of a scalar multiple of the identity and a nilpotent matrix as follows: The Cayley-Hamilton theorem tells us that $N=A-2I$ is nilpotent of order 3, so as above, $$e^{tN}=I+tN+\frac12t^2N^2.$$ Writing $A=2I+N$, the solution to the differential equation is therefore $$e^{2t}\begin{bmatrix}1+6t+t^2&12t+2t^2&-2t\\-3t-\frac12t^2&1-6t-t^2&t\\-t&-2t&1\end{bmatrix}\begin{bmatrix}C_1\\C_2\\C_3\end{bmatrix}.$$ This might not look the same as the book solution, but remember that the constants are arbitrary, so with a bit of fiddling and renaming, the two solutions can be made to look the same.

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  • $\begingroup$ Thank you very much. This example is actually from the textbook I found at library and we did not talk about that in school properly. I am curious but unfortunately I didn´t have a linear algebra II. $\endgroup$ – Leif Apr 17 '17 at 20:37
  • $\begingroup$ @Leif The pdf Moo references is a very good and concise presentation of the Jordan form and how to work with it. $\endgroup$ – amd Apr 17 '17 at 20:46

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