# The differential equation with a matrix

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?

• 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. – Moo Apr 17 '17 at 0:17

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.