if $f(x)$ is an irreducible polynomial of degree $4$ over a field $K$, then I thought that means it gives way to a $4$ dimensional $K$-vectorspace over its splitting field $L$. But this seems not to be the case
Given a field $F$ and a polynomial $P \in F[x]$ such that $P$ is irreducible over $F$. Let $L_P$ be the splitting field of $P$ and $F$. Does $\operatorname{dim}_F({L_F}) = \deg(P)$ hold?
If looking for a the minimal polynomial of $\alpha$ in a field $F$ is it sufficient to find a polynomial $P$ ...
Let $n$ be a natural number $\geq 2$ and $A$ a matrix $\epsilon M_{n}(K)$. We suppose the matrices $A$ and $I_{n}+A$ are invertible.
Show that $I_{n}+A^{-1}$ is invertible and also $A(I_{n}+A)^{-1}$
Matrix $I_{n}$ is an indentity matrix (right?), but what is let's say $I_{3}$?
My effort so far...
If I'm required for a matrix $A$ to find a minimal polynomial and matrices $B$ and $P$ invertible s.t. $P^{-1} A P$ is in Jordan Canonical Form, how would I go about it?
Oh I see. We've only used the eigenvalues and their multiplicity so far to determine the different possibilities for the Jordan Canonical Form, @Martin?
Ok. Since these were the only two possibilities, the Jordan form must be $J_2$.
(Because similar matrices have the same minimal polynomials.)
So the information about $\chi_A$ and $m_A$ was sufficient (in this example) to determine the Jordan form.
I should stress that this is not always the case. It can happen that we know both $\chi_A$ and $m_A$ but Jordan normal form is not uniquely determined by this information. (Although I think that this can happen only for $4\times4$ or larger matrices.)
Do you agree that we at least found $J$ for this matrix @Khallil?
But in the case that the information about the characteristic and minimal polynomials is insufficient i.e. all of them are the same, then we'd need to substitute in $J_1$ and $J_2$ into the polynomials and see if they're zero, @Martin?
(Yep, I agree that we've found $J$. Aren't we also tasked with finding $P$?)
So we have two problems: a) Sometimes we will not find J just from min. poly and char. poly. b) We also want to find P.
But before discussing how to do that let us have a look on a different thing.
We will get back to this matrix again.
But let us assume that we have this Jordan matrix: $J=\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$
One $3\times3$ block.
Or even better, let us make it more general.
Let us say the only eigenvalu is $\lambda$ and we want to say something about $J=\begin{pmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{pmatrix}$.
It is clear that $\chi_J(x)=(x-\lambda)^3$.
We want to say something about $m_J(x)$.
For this purpose, let us have a look at $J^2, J^3, \dots$
Yes. And this gives us some connection between Jordan form and minimal polynomial.
We can have several eigenvalues. But let us concentrate on one of them.
For example if $\lambda$ has multiplicity 3.
The there are three possibilities for the part of Jordan matrix corresponding to this matrix.
a) If we have only one $3\times3$ block, then there will be factor $(x-\lambda)^3$ in the minimal polynomial, since this is the first power when this block becomes zero.
b) If we have $2\times 2$ block and $1\times1$ block, then there will be factor $(x-\lambda)^2$ in the minimal polynomial.
c) If all block are $1\times1$ then we will get only $(x-\lambda)$.
So if we notice this could you answer the following question:
What is the characteristic and minimal polynomial of $J_1= \begin{pmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}$?
What is the characteristic and minimal polynomial of $J_2= \begin{pmatrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{pmatrix}$?
Can you say how these polynomials look like for the above matrices, @Khallil?
And in both cases the minimal polynomial will be $(x-\lambda)^2$.
Here's why:
$J-\lambda I$ is in the first case $\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
In the second case $\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$
If we calculate $(J-\lambda I)^2$ we can ignore the two $2\times 2$ blocks which consist fo zero.
Basically we will have there - in the second case - twice the matrix $\begin{pmatrix}0&1\\0&0\end{pmatrix}^2$.
Which is equal to zero, because the one "moved away from the diagonal" and got out of this block.
What I am getting at is that for $4\times4$ matrices we can have different Jordan forms, even though minimal and characteristic polynomial are the same.
Why and how is the Jordan Canonical form of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n >3$?
Thanks.
Comparison says that the columns have to be eigenvectors. So if some matrix has this Jordan form, the eigenspace for the eigenvalue $2$ must be 2-dimensional.
The more interesting case is $J_2=\begin{pmatrix} -2 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}$.
This was what we were able to say about $J$ using only charpoly.
We knew that it is either $J=J_1$ or $J=J_2$.
(We were able to decide using min. polynomial, but we do not want to use it now.)
So we see that in both cases $A\vec v_1=-2\vec v_1$ and $\vec v_1$ must be an eigenvector corresponding to the eigenvalue $-2$.
That's the easy part.
We also have $A\vec v_2=2\vec v_2$ and the second column $\vec v_2$ must be an eigenvector corresponding to the eigenvalue $2$.
So far it should be similar as for diagonal matrix. The third column is different.
And if you actually try to calculate the eigenvectors for the eigenvalue 2, i.e., if you solve the system $(A-2I)\vec v=\vec 0$, you will see that in this case the eigenspace is one-dimensional.
This is what tells us that it is $J_2$ and not $J_1$. If it were $J_1$, we would need two linearly independent eigenvectors for 2.
Is at least this clear? That we can find $\vec v_1$ and $\vec v_2$ in the same way as we did for diagonalization.
Let me know if we can move to $\vec v_3$ @Khalil
Ok, I am not sure whether you are still here.
In any case, we want to $A\vec v_3=\vec v_2+2\vec v_3$.
Things can get more complicated than this. But since here we only have one-dimensional eigenspace, the choice of $\vec v_2$ does not matter. For each eigenvector $\vec v_2$ we can find a solution $\vec v_3$.
For $4\times4$ matrices it can be more complicated, but maybe it is better not to complicate things and only concentrate on things needed for this particular matrix.
I'd say it is. Assuming that you will be working or some more complicated exercises (like $4\times4$ matrices) where you do not know J directily from $\chi$ and $m$.
And it is relevant even of this example - if you decide to compute directly P and J and to not try to find $m_A$ first.
(If we did not calculate $m_A$ first, we would not know J beforehand at the moment when we start looking for P.)
ok, it is already after midnight here, so I should get some sleep
Why and how is the Jordan Canonical form of a matrix in $M_3(\mathbb C)$ fully determined by its characteristic and minimal polynomials? And why does it fail for $n >3$?
Thanks.
Linear & Abstract algebra
Linear & Abstract algebra