$$\begin{aligned} y_1' &= -y_1 - 4y_2 + 2y_3 \\ y_2' &= 2y_1 + 5y_2 - y_3 \\ y_3' &= 2y_1 + 2y_2 + 2y_3 \end{aligned}$$

I get the eigenvalues $\lambda=1,2,3$ and for the first two I get the eigenvector $\mathbf{0}$. For the last, I get two redundant rows that I'm not sure how to handle.

A similar thing happens when I'm solving single equations and I have to use the approach that $y=u(x)e^{\lambda t}$ so I guess something similar happens here, but I'm just not sure how that looks on paper.

I've tried another approach where I recombine the system into the DE $y'''=3y+2y'+2y''$ but then my solution heads in a different direction in terms of sin and cos.

There's fundamental understanding missing here (or sloppy work). How do I approach this?

EDIT: one of my signs was wrong in the last equation. Fixed now.

  • $\begingroup$ Please check again the equations and your matrices that all coefficients and signs thereof are the same in every instance. Trying some variants, while 3 as eigenvalue seems rather certain, I do not get $1,2$, or even that both of the other eigenvalues are positive. $\endgroup$ Sep 15, 2021 at 19:00
  • $\begingroup$ $0$ is, by definition, not an eigenvector. Are you lacking dimensions somewhere? Perhaps the geometric multiplicity of an eigenvalue is less than its algebraic multiplicity, giving you a degenerate system? $\endgroup$ Sep 15, 2021 at 19:13
  • $\begingroup$ The Eigensystem command in Mathematica yields different eigenvalues. Maybe you should double-check your algebra? $\endgroup$ Sep 15, 2021 at 19:15
  • $\begingroup$ Ok, will do. It's one of those fuzzy-headed days so I'm sure it's user error. Just needed to ask out loud as a sanity check. $\endgroup$ Sep 15, 2021 at 20:02
  • $\begingroup$ Ok, for the current sign pattern you get eigenvalues $0,3,3$, not $1,2,3$. Check your calculations again. $\endgroup$ Sep 17, 2021 at 5:34

2 Answers 2


Define the vector

$x = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} $


$ x' = A x \hspace{24pt} (1)$

Now you have to find the eigenvalues and eigenvectors of $A$. Then you can write

$A = P D P^{-1} \hspace{24pt} (2) $

where $D$ is a diagonal matrix whose diagonal elements are the eigenvalues, and the columns of $P$ are the eigenvectors in respective order.

Plug in equation $(2)$ into equation $(1)$:

$x' = P D P^{-1 } x \hspace{24pt} (3) $

so that,

$ P^{-1} x' = D P^{-1} x \hspace{24pt} (4) $

Define the vector

$y = P^{-1} x \hspace{24pt} (5) $

And this in equation $(4)$ to obtain:

$y' = D y \hspace{24pt} (6)$

The solution of $(6)$ is

$y(t) = e^{Dt} y(0) \hspace{24pt} (7) $

The matrix $e^{Dt}$ is easily computable because $D$ is a diagonal matrix.

Finally, plug in $(5)$ into $(7)$:

$ x(t) = P e^{Dt} P^{-1} x(0) \hspace{24pt} (8) $


The eigenvalues you get by solving: \begin{align*} \det(A-\lambda I)&=0\\ \left|\begin{matrix} -1-\lambda &-4 &2\\ 2 &5-\lambda &-1\\ 2 &2 &-2-\lambda \end{matrix}\right|&=0\\ (-1-\lambda)[(5-\lambda)(-2-\lambda)+2]+4[2(-2-\lambda)+2]+2[4-2(5-\lambda)]&=0\\ -\lambda^3+2\lambda^2+7\lambda-12&=0\\ 3,\frac{-1\pm\sqrt{17}}{2}&=\lambda. \end{align*} As the eigenvalues are all distinct, the matrix is diagonalizable, and you should be able to proceed from here: get your diagonal matrix set up, and you can write the solution in terms of that. That is, once you can write $A=PDP^{-1},$ where $D$ is diagonal (and just equal to the eigenvalues on the main diagonal, whereas $P$ equals your eigenvectors lined up), then \begin{align*} \dot{\mathbf{y}}(t)&=A\mathbf{y}(t)\\ \mathbf{y}(t)&=e^{At}\,\mathbf{y}(0)\\ \mathbf{y}(t)&=e^{PDP^{-1}t}\,\mathbf{y}(0)\\ \mathbf{y}(t)&=Pe^{Dt}P^{-1}\,\mathbf{y}(0). \end{align*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .