Timeline for Classifying Matrices
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13 at 16:42 | vote | accept | Important_man74 | ||
| Jan 13 at 8:15 | comment | added | Asigan | In my answer, an alternative approach is one by one check (for example in case two) whether $(\lambda,\mu)=(-1,-1)$, $(-1,0)$, $(0,-1)$, $(0,0)$ are solutions, and the same for other two cases. But I think that is even more complicated than directly solve them from equations. | |
| Jan 13 at 8:13 | comment | added | Asigan | @Important_man74 Ok, I have written a complete answer to show how the Jordan form is applied to solve such kind of problem. As for your other questions, (1) Since $A$ is of size $3\times 3$, it must have characteristic polynomial of degree $3$, and therefore can never be $x^2(x+1)^2$. (2) I confess in my previous comment, my cases are redundant. It can be reduced to three cases. (3) Yes, $A$ can not admit eigenvalues other than $0$ and (I think rather than $1$ you mean) $-1$. In fact in my answer I suppose the eigenvalues are unknown and then show they are $-1$ or $0$. | |
| Jan 13 at 8:05 | answer | added | Asigan | timeline score: 1 | |
| Jan 13 at 2:30 | comment | added | Important_man74 | @Asigan sorry for the late response but i am still a bit confused. Is it true that all of the matrices here share the characteristic polynomial $x^2(x+1)^2$? I'm not really sure where each of the 6 cases you mention arise from. Also, I am not quite sure about my comment above about the possible valid minimal polynomials. Finally, isn't it only possible that a matrix which satisfies these relations can only have 1 or 0 as an eigenvalue? | |
| Jan 5 at 11:50 | comment | added | Asigan | In \begin{equation}\begin{pmatrix} \lambda & 1 & \\ & \lambda & \\ & & \mu \end{pmatrix},\end{equation} $\lambda$ has geometric multiplicity $1$, algebraic multiplicity $2$, while $\mu$ has both multipllicity $1$. As for $\mathtt{diag}\{\lambda,\lambda,\mu\}$, $\lambda $ has both multiplicity $2$, and $\mu$ has both multiplicity $1$. ($\mathtt{diag}\{\lambda,\lambda,\mu\}$ has three Jordan blocks $\{\lambda\},\{\lambda\},\{\mu\}$.) | |
| Jan 5 at 11:50 | comment | added | Asigan | @Important_man74 "the algebraic multiplicity of the eigenvalue tells us how many times it appears on the diagonal and the geometric tells us how many blocks they are in" is correct. | |
| Jan 5 at 9:55 | comment | added | Important_man74 | So the algebraic multiplicity of the eigenvalue tells us how many times it appears on the diagonal and the geometric tells us how many blocks they are in? By your example $\lambda$ has algebraic multiplicity 2 and geometric multiplicity 1 and $\mu$ has both multiplicities equal to 1? | |
| Jan 5 at 8:55 | comment | added | Asigan | The rest is (lengthy but easy) case work: If $A$ has three eigenvalues, then $J$ must be $\mathtt{diag}\{\lambda,\mu, \omega\}$. If $A$ has two, then \begin{equation*} J=\begin{pmatrix} \lambda & 1 & \\ & \lambda & \\ & & \mu \end{pmatrix} \text{ or } \mathtt{diag}\{\lambda,\lambda,\mu\}. \end{equation*} If $A$ has one, then there are three cases: "$1+1+1$, $2+1$, $3$" (corresponding to the size of each Jordan block). Please work through these six situations :). | |
| Jan 5 at 8:54 | comment | added | Asigan | Are you familiar with Jordan's canonical form? If you are, then suppose $J$ is the Jordan's canonical form of $A$, with $J=PAP^{-1}$. Then we have \begin{align*} &A^4 + 2A^3 + A^2 = 0 &\\ \iff & P(A^4 + 2A^3 + A^2 )P^{-1}= 0 & (\text{since $P0P^{-1}=0$})\\ \iff & J^4 + 2J^3 + J^2 = 0 & (\text{since $PA^nP^{-1}=PAP^{-1}PAP^{-1}\cdots PAP^{-1}=J^n$}) \end{align*} The same argument gives $J^2+J\ne 0$. | |
| Jan 5 at 8:27 | comment | added | Milten | You can find the characteristic polynomial from the Jordan normal form. You don't need it to list the matrices. | |
| Jan 5 at 8:18 | comment | added | Important_man74 | Ah I see, so the only valid minimal polynomials would be $x^2$, $x^2(x+1)$,$x(x+1)^2$, or $x^2(x+1)^2$. How does the characteristic polynomial come into play here? | |
| Jan 5 at 7:59 | comment | added | Milten | And yes, from that point I would probably list the possible cases of Jordan normal forms. | |
| Jan 5 at 7:55 | comment | added | Milten | I don't have the solution, but since $A(A+1)\ne0$, the minimal polynomial cannot be a divisor of $x(x+1)$. There are now very few possible options for the minimal polynomial. | |
| Jan 5 at 7:46 | history | asked | Important_man74 | CC BY-SA 4.0 |