Calculating the Modal Matrix for A

[boo]

TL;DR Summary

Modal matrix

Mentor note: The Tex shown below had to be modified a fair amount to conform to the MathJax on this site.

Trying to calculate the modal matrix for the following

$A =\begin{pmatrix} 1 && 1/2 && 1/2 \\ 0 && 1/2 && -1/2 \\ 0 && 1/2 && 1 .5 \end{pmatrix}$

there are two eigenvectors for this matrix

$A =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

and

$A =\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$
using the traditional formula for the generalized eigenvector does not work

$Av2 = \lambda v2 + v1$

and, yet, MATLAB has no problem calculating them as\\

$A =\begin{pmatrix} 1/2 \\ -1/2 \\ 1/2 \end{pmatrix}$

$A =\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$
$A =\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}$

 

[Mark44]

You don't list the eigenvalues, or the multiplicities of the eigenvectors that you show. I've worked some on this problem and have found that that the only eigenvalue is $\lambda = 1$ of multiplicity 3.
This website (https://ximera.osu.edu/linearalgebra/textbook/eigenvalueProperties/generalizedEigenvectors , third example) shows a 3x3 matrix with a similar situation as in your problem.

Comment. You should distinguish between the matrix A and the eigenvectors, both of which you label as A. I would call the eigenvectors $v_1, v_2$, and $v_3$, and not keep using A for everything.

Also, I had to edit your post to remove all of the \begin{equation*} \end{equation*} stuff. AFAIK, the subset of LaTeX we use, MathJax, doesn't support these commands.

 

[boo]

Sorry, struggled with the Latek code which I don't use. The eigenvalues are 1 with algebraic multiplicity 3 and geometric multiplicity 2. The null space of (A - \lambda I) has 2 eigenvectors v1 = (1 0 0) and v2 = (0 -1 1) ( or any linear combination of these). Looking for the third (generalized) eigenvector (v3) by setting either
(A - \lambda I)*v3 = v2 or (A - \lambda I)*v2 = v3 gives no solution (inconsistent equations).

[Post edited by a Mentor to fix a small typo]

 

[boo]

Don't see any way to edit these posts which is frustrating when I see a typo. Anyway that last equation should be (A - \lambda I)*v3 = v2

 

[Mark44]

Sorry, struggled with the Latek code which I don't use. The eigenvalues are 1 with algebraic multiplicity 3 and geometric multiplicity 2. The null space of (A - \lambda I) has 2 eigenvectors v1 = (1 0 0) and v2 = (0 -1 1) ( or any linear combination of these). Looking for the third (generalized) eigenvector (v3) by setting either
(A - \lambda I)*v3 = v1 or (A - \lambda I)*v2 = v3 gives no solution (inconsistent equations).

Again, take a look at the link I posted, especially the third example. I think it will be helpful.

Don't see any way to edit these posts which is frustrating when I see a typo.

There's a limited time window for which you can edit a post. After that time elapses, you can't revise your post.

 

[pasmith]

All you know is that $(A-I)v_3 \in \ker(A - I)$. Since $\ker(A-I)$ is two-dimensional, the most you can say is $$(A - I)v_3 = \alpha v_1 + \beta v_2.$$ We also need $\{v_1, v_2, v_3\}$ to be a basis of $\mathbb{R}^3$. An obvious way to extend $v_1 = (1,0,0)^T$ and $(0,-1,1)^T$ to a basis of $\mathbb{R}^3$ is to add $v_3 = (0,0,1)^T$. Can you find $\alpha$ and $\beta$ such that $(A - I)v_3 = \alpha v_1 + \beta v_2$?

 

[Mark44]

I've finally had a chance to dig into this problem in some detail, and realize that the suggestions I made don't do any good. The suggestion from @pasmith seems like it might bear fruit.

 

[berkeman]

Don't see any way to edit these posts which is frustrating when I see a typo. Anyway that last equation should be (A - \lambda I)*v3 = v2

I fixed the v1 --> v2 typo for you. In the future if you need a typo fixed and you are no longer able to edit your post, just click the "Report" link in your post and ask the Mentors to make the edit for you.

 

[boo]

Yes, I checked out that link and none of the problems addressed this issue. They were all easily solved using the conventional technique of choosing one or another of the eigenvectors obtained in "step one" in order to find the generalized eigenvectors in step 2. This works "most" of the time but occasionally (as in the example I gave) it results in an inconsistent equation(s) when you try to row reduce and solve for the generalized eigenvector. I agree that pasmith has the correct approach. I think that the final solution is that you have to find a linear combination of eigenvectors that, as part of a Jordan chain, do not result in inconsistent equations when solving for the generalized eigenvectors. Instead of going back to the original matrix here's another one with exactly the same problem that avoids dealing with all of these fractions:$$A = \begin{pmatrix}0 & -2 & 0 \\ 2 & 4 & 0 \\ 2 & 2 &2 \end{pmatrix}$$
This has eigenvalues of 2 with algebraic multiplicity 3. The dimension of the null space of $\left(\mathbf{A}~-~\lambda~I\right)$ is 2. Solving for these two eigenvectors we have $v_1 = \begin{pmatrix} 0\\0\\1\end{pmatrix}~ and ~v_2 = \begin{pmatrix} -1\\1\\0\end{pmatrix}$. Setting up the augmented matrix with $v_1$ gives us $$ \begin{pmatrix}-2 & -2 & 0 & 0\\ 2 & 2 & 0 & 0\\ 2 & 2 &0 &1\end{pmatrix}$$ which has no solution. For $v_2$ we have $$ \begin{pmatrix}-2 & -2 & 0 & -1\\ 2 & 2 & 0 & 1\\ 2 & 2 &0 &0\end{pmatrix}$$ which is also inconsistent. So I tried a linear combination of $v_1$ and $v_2$ like this:$$ \begin{pmatrix}-2 & -2 & 0 & a\\ 2 & 2 & 0 & b\\ 2 & 2 &0 & c\end{pmatrix}$$ Doing row reduction on this (augmented) matrix gives us $$ \begin{pmatrix}1 & 1 & 0 & -a/2\\ 0 & 0 & 0 & b+a\\ 0 & 0 &0 & c+ a\end{pmatrix}$$ So clearly, in order to have a solution we must have $b = c = 0$. So $a = 1~ b = -1 ~and~ c = -1$. This vector is $$v_3 = \begin{pmatrix} 1\\-1\\-1\end{pmatrix}$$ which is a linear combination of our original 2 vectors $v_1~ and~ v_2$. This gives us 3 eigenvectors $v_4 = \begin{pmatrix} -0.5\\0\\0\end{pmatrix} v_5 = \begin{pmatrix} -1\\1\\0\end{pmatrix}~ and~ v_6 = \begin{pmatrix} 0\\0\\1\end{pmatrix}$. So choosing as our basis set $v_3~and~v_4$ and our original $v_1$. ($v_5$ and $v_6$ just reproduce $v_2$ and $v_1$). Where $v_3 ~and ~v_4$ are a Jordan chain. Our modal matrix is then $$S =\begin{pmatrix}0 & 1 & -0.5 \\ 0 & -1 & 0 \\ 1 & -1 &0 \end{pmatrix}.$$ Where $S^{-1}*A*S$ is our Jordan form:$$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 &2 \end{pmatrix}$$ Also, as a side note, observe that although any scalar multiple of an eigenvector is also an eigenvector this is not the case for generalized eigenvectors. So one can not substitute a scalar multiple of $v_4$. Sorry for the long post and hopefully there are no typos here.

 

[boo]

So going back to my original problem and using this technique I find two eigenvectors in the null space of $\mathbf A~-~\lambda I$ to be $\begin{pmatrix}1\\0\\0\end{pmatrix}$and$\begin{pmatrix}0\\-1\\1\end{pmatrix}$ but that the first eigenvector in the Jordan chain must have the form $C*\begin{pmatrix}1\\-1\\1\end{pmatrix}$. Choosing $C = 1$ we see that this eigenvector is the sum of the original two eigenvectors. Call it $v_1=\begin{pmatrix}1\\-1\\1\end{pmatrix}$.So solving for the generalized eigenvector we get one additional (generalized) eigenvector $v_2=\begin{pmatrix}0\\2\\0\end{pmatrix}$. Adding the original eigenvector $v_3=\begin{pmatrix}1\\0\\0\end{pmatrix}$ we have the modal matrix $$S = \begin{pmatrix}1 &0&1\\-1&2&0\\1&0&0\end{pmatrix}$$Which results in the Jordan form for the matrix $\mathbf A$
$$\mathbf J = \begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}$$