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I am attempting to solve a system of equations using a method from a textbook. The problem comes in when I find multiple copies of the same eigenvalue, and I am struggling to find the 3 eigenvectors.

Here is the question:

Solve

$x'(t) = Bx(t)$

where $B$ is \begin{bmatrix} -1 & 2 & -1 \\ 0 & -1 & 2 \\ 0 & 0 & -1\end{bmatrix}

The answer is stated to be:

$x = (a + (2b - c)t + 2ct^2)e^{-t}$

$y = (b + 2ct)e^{-t}$

$z = ce^{-t}$

$\textbf{Attempt at solution:}$

Since $B$ is an upper triangular matrix it's clear that the determinant is $(-1 - \lambda)^3 $ which gives us $\lambda_{1,2,3} = -1$ as eigenvalues.

After plugging in $\lambda = -1$ for $(B - \lambda I)$ The resulting matrix is

\begin{bmatrix} 0 & 2 & -1 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{bmatrix}

Therefore, after multiplying this matrix by- \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}

and setting this equal to 0, I get that:

$2x_2 - x_3 = 0$ and $2x_3 = 0$

I then get the eigenvector $v_1 = (1,0,0)$ (I believe this is correct)

I am not sure how to proceed from this point, in terms of finding 2 more eigenvectors to reach that solution. If someone could provide a walkthrough from this point and show how the answer is reached, that would be incredibly helpful! This is for self-study.

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  • $\begingroup$ You’ve found that the eigenspace of $-1$ is only one-dimensional, so the matrix isn’t diagonalizable. You’re not going to find any more linearly independent eigenvectors. To proceed in the way you started, you’ll need to compute the Jordan decomposition of the matrix, or use the method of decomposing it into the sum of diagonal and nilpotent matrices. $\endgroup$ – amd Jul 28 '17 at 23:07
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Method 1: Notice that the third equation is decoupled from the other two.

We have $$z' = -z \implies z(t) = c e^{-t}$$

Substituting $z(t)$ into the second equation, we have

$$y' = -y + 2 z = -y + 2 c e^{-t} \implies y(t) = (b + 2ct)e^{-t}$$

Substituting $y(t)$ and $z(t)$ into the first equation, we have

$$x' = -x + 2 y - z = -x + 2(b + 2ct)e^{-t} - c e^{t}$$

Solving we get

$$x(t) = (a + (2b - c)t + 2ct^2)e^{-t}$$

Method 2: Eigenvalues / Eigenvectors

This is a deficient matrix (as you discovered). Are you familiar with generalized eigenvectors and the Jordan form?

The eigenvalue (triple) is $\lambda = -1$ and the RREF of $[A + I]v _1 = 0$ gives

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}v_1 = 0 \implies v_1 = (1, 0, 0)$$

Unfortunately, we cannot get any more linearly independent eigenvectors, so we need to find generalized ones. Following these Jordan Matrix Notes, we have the RREF of $[A + I]v_2 = v_1$ of (shown as augmented matrix)

$$\begin{bmatrix} 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

We can choose

$$v_2 = \left(0, \dfrac{1}{2}, 0 \right)$$

Repeating for the RREF of $[A + I]v_3 = v_2$, we get

$$\begin{bmatrix} 0 & 1 & 0 & \frac{1}{8} \\ 0 & 0 & 1 & \frac{1}{4} \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

We can choose

$$v_3 = \left(0, \dfrac{1}{4}, \dfrac{1}{8} \right)$$

Referring to the notes, we can now write

$$X(t) = e^{-t}\left[c_1 v_1 + c_2 (v_2 + t v_1) + c_3 \left(v_3 + t v_2 + \dfrac{t^2}{2!} v_1\right)\right]$$

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  • $\begingroup$ Your first method is what I actually tried first. I got the substitution for z(t) correct, and I understand the solution for x(t) as well. In the step "substituting z(t) into the second equation..." where did the "t" in "2ct" come from in the evaluation of y(t)? Thanks! $\endgroup$ – rubyquartz Jul 29 '17 at 0:19
  • $\begingroup$ The equation for $y'(t)$ is linear and you can solve that using an Integrating Factor. The result of that is what I show. Another approach $y' + y = 2ce^{-t}$. We see that the complementary solution shares an $e^{-t}$ with the particular, so we choose $y_p = k t e^{-t}$ using Undetermined Coefficients to compensate. Review (in detail): tutorial.math.lamar.edu/Classes/DE/… $\endgroup$ – Moo Jul 29 '17 at 0:20
  • $\begingroup$ Ok, thank you- I am familiar with eigenvalues and the Jordan form. If it's not too much to ask would you mind adding that to your answer? $\endgroup$ – rubyquartz Jul 29 '17 at 0:26

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