0
$\begingroup$

I want all possible complex matrices that have characteristic polynomial $(x-1)^5(x+3)$ and minimal polynomial of $(x-1)^2(x+3)$. These matrices must not be similar to each other.

Now, the characteristic polynomial shows us that this is probably a 6x6 matrix with 5 $1$s in the diagonal and one $-3$ at the bottom right corner. The minimal polynomial probably tells us that the block corresponding to $\lambda=1$ is maximum size 2 meaning $\begin{bmatrix} 1 & 1\\0 & 1\end{bmatrix}$ and block corresponding to $\lambda = -3$ is simply $[-3]$ (1 x 1 block).

There are a few things I am confused about from here. First is how many of those 2 x 2 blocks are there. Can we even tell if there are how many blocks from the given information? Are answers to this question simply a different combinations of how many of these blocks there are and where these blocks are around the diagonal?

The second question deals with the fact that the question asked for complex matrices. Do some of my matrices have to have entries with $i$ in it? If so, how should I even proceed?

If you could enlighten me in some sort of way, I would greatly appreciate it. Thank you.

$\endgroup$
0
$\begingroup$

From the characteristic polynomial, the sum of sizes of the Jordan blocks for eigenvalue $1$ is $5$, the sum of sizes of Jordan blocks for eigenvalue $-3$ is $1$, and those are all the eigenvalues. Thus the matrix is $6 \times 6$, and its Jordan form has a single $-3$ on the diagonal (whether you put this at top left or bottom right, or even between two blocks for eigenvalue $1$, is up to you). From the minimal polynomial, the largest Jordan block for eigenvalue $1$ has size $2$. Thus you could have either two such blocks of size $2$ and one of size $1$, or one of size $2$ and three of size $1$. Again, the arrangement of the blocks does not matter. Matrices with different arrangements are similar.

As for your second question, "complex" does not mean "not real". It just means that the entries of the matrix are all complex numbers. Your matrices in Jordan form are already "complex". You don't need to put any $i$'s in them.

$\endgroup$

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.