0
$\begingroup$

Investigate the two-dimensional linear system $ \begin{bmatrix} x' \\ y' \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix}$ where $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ for the case when $D=det(A)=0$ and determine the normal forms that arise.

The characteristic polynomial is $P(\lambda) = det(A-\lambda I) = \lambda^2 - S\lambda +D$ with $S= trace(A)= a_{11}+a_{22}$. Then for $D=det(A)=0$, the characteristic polynomial becomes $P(\lambda) = det(A-\lambda I) = \lambda^2 - S\lambda = \lambda(\lambda - S)$ so we get eigenvalues $\lambda = 0$ and $\lambda = S$.

I am not sure how to proceed to find the Normal Forms.

$\endgroup$
0
$\begingroup$

There are two possibilities to consider: $S=0$ and $S\ne0$.

Taking the second possibility first, you’ve got a $2\times2$ matrix with 2 distinct eigenvalues. The possibilities for the Jordan normal form of such a matrix are pretty limited. It’s going to have two Jordan blocks, obviously, so what does this mean for the sizes of those blocks?

If $S=0$ there again two possibilities to consider: either $A$ is diagonalizable, in which case $A=0$ (can you see why?) or it is not. In the latter case, how many Jordan blocks does this matrix need and what are their sizes? The answers will whittle down the possibilities for the normal form quite dramatically.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Would the blocks also be 2 by 2 ? $\endgroup$ – Sam Whelchel Apr 4 '17 at 3:01
  • $\begingroup$ @SamWhelchel no, since that would make the matrix at least $4\times4$. $\endgroup$ – amd Apr 4 '17 at 5:42
  • $\begingroup$ I see, then each block would only have one element because the matrix is 2 X 2. $\endgroup$ – Sam Whelchel Apr 4 '17 at 11:24
  • $\begingroup$ @SamWhelchel Right. Can you see what the normal form of the matrix has to be? $\endgroup$ – amd Apr 4 '17 at 17:46
  • $\begingroup$ I think it would be the form $ \begin{bmatrix} \lambda & 0 \\ 0 & \mu \end{bmatrix} $ where $\lambda$ and $\mu$ are the eigenvalues, but I am still messing with the matrix. $\endgroup$ – Sam Whelchel Apr 5 '17 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.