Finding Jordan normal form J and the matrix P

Khallil

Oct 17, 2015 19:55

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?

Martin Sleziak

Oct 17, 2015 20:05

I will ask the same thing as the last time: Maybe some of the questions tagged jordan-normal-form might help?

But maybe you can write here what is $A$ and what you already know about that matrix.

Khallil

Oct 17, 2015 20:30

$A = \begin{pmatrix} 0 & 1 & 1 \\ 2 & 1 & -1 \\ -6 & -5 & -3 \end{pmatrix}$

Thanks, @Martin. I've looked at some questions but I think the fudamental building up of the theory is lacking in my text.

Martin Sleziak

Ok.

First do you know how to find characteristic polynomial?

You probably should be able to find it - since have been were using it to find eigenvalues already.

You have defined it either as $\chi_A(x)=\det(xI-A)$ or as $\chi_A(x)=\det(A-xI)$.

Both definitions occur in various text. The only difference between them is sign.

Do you know how to calculate char. poly @Khallil?

Khallil

Yep, that's exactly how I've defined it, @MartinSleziak.

Martin Sleziak

For this particular matrix WA says that it will be $(x-2)(x+2)^2$.

This means that the eigenvalues are $\pm2$ and $-2$ has multiplicity 2.

We also know that minimal polynomial divides characteristic polynomial. And that all eigenvalues are roots of minpoly.

So in this case we have only two possibilities: $m_A(x)=(x-2)(x+2)$ or $m_A(x)=(x-2)(x+2)^2$.

So we could simply try whether we get zero if we plug our matrix into these polynomials.

Khallil

By plug in you mean replace $x$ with $A$, @MartinSleziak?

Martin Sleziak

To be more precise, we want to know whether $(A-2I)(A+2I)$ is zero.

The only difference is that absolute coefficients are multiplied by $A^0=I$.

Khallil

Oct 17, 2015 20:40

I haven't defined absolute coefficients yet, @MartinSleziak.

Martin Sleziak

Because if we simply replace $x$ by $A$ in $x-2$ we would get $A-2$, which does not make sense. (It is difference between matrix and nubmer.)

I mean that if you want to plug $A$ into $x^3-x+3$ then you will calculate $A^3-A+3$.

By "absolute coefficient" I meant coefficient of $x^0$. (What is correct Englesh term for the last coefficient of a polynomial?)

Khallil

I think it's just referred to as the constant term, @Martin.

Martin Sleziak

Right. So if $c$ is constant term, you have $cI$ in that place, if you want to plug a matrix into a polynomial.

Which kinf of makes sense - you replaced $x^0$ by $A^0$.

So one way to find minimal polynomial would be this way.

We have finitely many possibilities.

And we could try which of these polynomials returns zero for the matrix $A$.

I mean this: If $(A-2I)(A+2I)=0$, then $(x-2)(x+2)$ is minimal polynomial.

$$(A-2I)(A+2I)=
\begin{pmatrix}
-2 & 1 & 1 \\
2 &-1 & -1 \\
-6 & -5 & -5
\end{pmatrix}
\begin{pmatrix}
2 & 1 & 1 \\
2 & 3 & -1 \\
-6 & -5 & -1
\end{pmatrix}$$

The first two rows are zero.

But not the third one.

I got this:

$\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
8 & 4 & 4
\end{pmatrix}$

If we multiply this by $(A+2I)$ again, we get zero.

So we see that $(x-2)(x+2)$ is not minimal polynomial. We get $m_A(x)=(x-2)(x+2)^2$.

Khallil

I agree!

Martin Sleziak

Ok, so we have minimal polynomial

We can get some information about $J$ from minimal polynomial. (But we cannot always determine it.)

It works the other way round. If we find $J$, then we know minimal polynomial.

Khallil

Oct 17, 2015 20:50

May I ask what you mean by $J$, @Martin?

Martin Sleziak

Jordan form.

We want $J=P^{-1}AP$.

But this case is simple.

Good think to know is that similar matrices have the same minimal polynomial.

I.e., $m_A(x)=m_J(x)$.

And that $J$ is composed of Jordan blocks.

Khallil

So $P^{-1} A P$ being equal to $J$ means that $J$ and $A$ are similar?

Martin Sleziak

Do you know what Jordan blocks are? And how Jordan matrix can look in general?

@Khallil That's correct.

Khallil

Yep, I'm familiar with how they look. Each block contains an eigenvalue and the super diagonal only contains 1 entries, @MartinSleziak.

I think the number of blocks depends on the number of eigenvalues?

Martin Sleziak

I mean, if we know that eigenvalues are -2,2,2 then there are only two possibilities for J.

Since -2 has multiplicity one, it only can have block of the size $1\times1$.

But for the eigenvalue 2 there are two possibilities. We can have two $1\times 1$ blocks or one $2\times2$ blocks.

So if we know only characteristic polyn. and nothing else we can already say that there are only two possibilities for $J$.

$J=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ or $J=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$

I will rewrite it so that the blocks are better visible.

In the first possibility we have three blocks:

$J=\begin{pmatrix}
\boxed{-2} & 0 & 0 \\
0 & \boxed{2} & 0 \\
0 & 0 & \boxed{2}
\end{pmatrix}$

I the second possibility the whole lower right part is one block:

$\left(\begin{array}{c|cc}
-2 & 0 & 0 \\\hline
0 & 2 & 1 \\
0 & 0 & 2
\end{array}\right)$

@Khalil Is it clear what I mean by blocks and why this are the only possibilities?

Khallil

Oct 17, 2015 21:00

I can't really tell what you mean by the squared entries, @Martin.

Martin Sleziak

Jordan matrix always consists of blocks which you place there diagonally.

Khallil

Are the squared blocks supposed to correspond to the eigenvalues?

Martin Sleziak

The first matrix $J_1=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$
has three $1\times1$ blocks on the diagonal.

The second matrix $J_2=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$ has two blocks.

The first one is $1\times1$ in the upper left corner. The block for the eigenvalue 2 has size $2\times2$.

Khallil

Oh, I see what you're doing now!

Martin Sleziak

Which is why there is 1 above diagonal.

Let us consider another example: What can we say if characteristic polynomial is $(x--2)^3$. (For some different matrix.)

Then the possibilities for the Jordan form would be:

a) diagonal matrix $\begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$

b) two blocks, one of size 1 and the other one of size two; i.e. $\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$

c) or one $3\times3$ block $\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$

So this is what we can say about $J$ directly from characteristic polynomial.

We can enumerate all possibilities for the Jordan form.

In particular, if $n\times n$ matrix has $n$ distinct eigenvalues, the Jordan form will be diagonal matrix.

Khallil

Oct 17, 2015 21:06

So how can we choose the correct Jordan Form or is it not unique, @Martin?

Martin Sleziak

It is unique.

But so far the only information we used was the char. polynomail.

So is it more-or -less clear how I can get $J_1$ and $J_2$ if I know characteristic polynomial? @Khallil

Khallil

Yes, it's clear so far.

But isn't it from the minimal polynomial?

Martin Sleziak

No, the only thing we used was char. poly.

If we know min. poly, we can reduce the number of possibilities further.

And in this case we have only two possibilities, so we will be able to say which one is correct.

I will write again that we want to decide between $J_1=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ and $J_2=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$.

Khallil

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?

Martin Sleziak

@Khallil Exactly.

But if we know minimal polynomial, it can help us further.

Could you say what is minimal polynomial of $J_1$ and what is minimal polynomial of $J_2$?

We know that char. poly is $(x+2)(x-2)^2$ in both cases.

So min. poly can be either that or $(x-2)(x+2)$.

And it is not very difficult to see that $(J_1-2I)(J_1+2I)$ is zero.

$(J_1-2I)(J_1+2I)=
\begin{pmatrix}
-4 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{pmatrix}$

So minimal polynomial of $J_1$ is $(x-2)(x+2)$?

Is this one clear @Khallil

Khallil

Oct 17, 2015 21:19

Sorry for the late reply, @Martin.

Yes, that's clear.

How about the minimal polynomial of $J_2$, @Martin?

Martin Sleziak

Yes, we should look at that too.

Although we already know that $J_1$ and $A$ are not similar, since they have different minimal polynomials. So the Jordan form must be $J_2$.

But in more complicated exercises you might have more than two possibilities, So it might be useful to see what happens there.

$(J_2-2I)(J_2+2I)=
\begin{pmatrix}
-4 & 0 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 \\
0 & 4 & 1 \\
0 & 0 & 4
\end{pmatrix}$

So this product is not zero.

This cannot be minimal polynomial.

But since $(J_2-2I)^2=
\begin{pmatrix}
-4 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$, we get zero if we calculate $(J_2-2I)^2(J_2+2I)$.

Now it is probably correct. I seem to be mixing up the signs.

Khallil

I'm slightly confused now, @Martin.

Martin Sleziak

Ok, do you see that $(J_2+2I)(J_2-2I)$ is a non-zero matrix?

Khallil

Yep, I see that it's non-zero.

Martin Sleziak

And you are right that I have made a mistake. I did not square the $-4$ there.

So I should have written $(J_2-2I)^2=
\begin{pmatrix}
16 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$.

Wrong sign again... edited.

We do not have to do this - we know from Cayley-Hamilton theorem that $\chi_A(A)=0$ for any matrix.

So we know that $(J_2-2I)^2(J_2+2I)$ must be zero.

But we might check that anyway.

That was what I was trying to do.

And it is zero.

$(J_2-2I)^2(J_2+2I)=
\begin{pmatrix}
16 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
0 & 0 & 0 \\
0 & 4 & 1 \\
0 & 0 & 4
\end{pmatrix}=0$

The main thing is: $\chi_{J_1}(x)=(x-2)(x+2)$ and $\chi_{J_2}(x)=(x-2)^2(x+2)$.

@Khallil Do you agree at least that the characteristic polynomials look like this?

Sorry I meant minimal polynomials.

Khallil

Oct 17, 2015 21:33

Yep, I agree that those are the minimal polynomials.

Martin Sleziak

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?

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 $P$?)

Martin Sleziak

@Khallil Yes, and we want to get to that.

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 $J^2, J^3, \dots$

Directly by multiplications you should get:

$J^2=\begin{pmatrix}
\lambda^2 & 2\lambda & 1 \\
0 & \lambda^2 & 2\lambda \\
0 & 0 & \lambda^2
\end{pmatrix}$

Sorry, my mistake. This might be interesting too, but to get $m_J(x)$ I need something different.

So let us have a look at powers of $J-\lambda i$.

$J-\lambda I=\begin{pmatrix}
0& 1 & 0 \\
0 & 0 & 1 \\
0 & 0 &0
\end{pmatrix}$

Can you calculate $(J-\lambda I)^2$ and $(J-\lambda I)^3$?

Sorry for the confusion - that I started calculating something different first.

You should get that:

$(J-\lambda I)^2=
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$

$(J-\lambda I)^3=
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$

Did you get such results @Khallil

And can you see there a pattern - how the ones are "moving away from the diagonal?

Khallil

They are moving up a diagonal each time.

Is that a general pattern I should be noticing?

Martin Sleziak

It would work in the same way for higher dimensions.

$J-\lambda I=
\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}$

$(J-\lambda I)^2=
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}$

So for $4\times 4$ we will get $(J-\lambda I)^3\ne 0$ and $(J-\lambda I)^4=0$.

Khallil

Oct 17, 2015 21:48

Oh, I see. Yes, that seems very reasonable when you generalise it.

Martin Sleziak

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?

Khallil

$\chi_{A} (x) = (x-\lambda)^4$ for both $J_1$ and $J_2$ I believe, @Martin.

Martin Sleziak

Yes.

And what about minimal polynomial?

Khallil

It can be any of the $(x-\lambda)^{i}$ for $i=1,2,3,4$.

Martin Sleziak

And I am sorry.

I meant $J_2=
\begin{pmatrix}
\lambda & 1 & 0 & 0 \\
0 & \lambda & 0 & 0 \\
0 & 0 & \lambda & 1 \\
0 & 0 & 0 & \lambda
\end{pmatrix}$

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.

I will add this link:

5

Q: Jordan Canonical Form determined by characteristic and minimal polynomials in dimension $3$, but not beyond

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.

If you prefer, you can read up on this later and we can get to the computation of $P$.

Khallil

Oct 17, 2015 22:05

For sure!

Martin Sleziak

ok

So let us get back to our original problem.

I should say that we do not have to calculate $m_A$ to get $J$.

Khallil

We don't?

Martin Sleziak

We will get to it in a moment.

Khallil

Oh because $P$ will facilitate the lack of $m_A$?

Martin Sleziak

Yes. Of course, once we know $J$, we also know $m_A$.

You probably remember how we discussed finding invertible matrix $P$ such that $AP=PD$.

Diagonalization of a matrix

Oct 10 at 13:46, 1 hour 10 minutes total – 193 messages, 3 users, 0 stars

Bookmarked 18 hours ago by Martin Sleziak

This is a similar problem and let us try similar approach.

Khallil

Oct 17, 2015 22:08

Of course. That was really insightful! I still remember the proof.

Martin Sleziak

We want $P^{-1}AP=J$, which is the same as $AP=PJ$. (And we want $P$ to be invertible.)

And again let us denote columns of $P$ as $\vec v_1, \vec v_2, \vec v_3$.

I.e., $P=\begin{pmatrix}\vec v_1&\vec v_2&\vec v_3\end{pmatrix}$

In the same way as the last time we have $AP=\begin{pmatrix}A\vec v_1&\vec Av_2&A\vec v_3\end{pmatrix}$.

PJ is a bit more complicated. The last time we had diagonal matrix, so it was simpler.

Khallil

Now we have a super diagonal of 1s.

Martin Sleziak

Let us have a look at both possibilities.

For $J_1=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}$ we would get

$PJ_1=\begin{pmatrix}\vec v_1&\vec v_2&\vec v_3\end{pmatrix}\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{pmatrix}=\begin{pmatrix}-2\vec v_1&2\vec v_2&2\vec v_3\end{pmatrix}$.

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 $J_2=\begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$.

Can we calculate $PJ_2=\begin{pmatrix}\vec v_1&\vec v_2&\vec v_3\end{pmatrix} \begin{pmatrix}
-2 & 0 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{pmatrix}$?

Sorry, I had 1 in the wrong place. I have edited it.

What I am saying is that we get $PJ_2=\begin{pmatrix}-2\vec v_1&2\vec v_2&\vec v_2+2\vec v_3\end{pmatrix}$.

I should probably leave you some time to think about this @Khallil.

But the reasoning is based simply on how we multiply matrices.

Ping me if we can move on. (Or if there is need to explain this in detail let me know.)

Khallil

Can I ask why $J_1$ doesn't have any 1s on the super diagonal, @Martin?

Martin Sleziak

If you recall these were the two possibilities for the Jordan form.

Khallil

Oct 17, 2015 22:20

Oh, the 3 blocks because of the three eigenvalues (one of which has multiplicity 2)?

Martin Sleziak

Yes.

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 $A\vec v_3=\vec v_2+2\vec v_3$.

Khallil

Oct 17, 2015 22:36

Sorry, @Martin I'm still here!

Was reading through and am still making sense of it.

Martin Sleziak

This is equivalent $(A-2I)\vec v_3=\vec v_2$.

Which means that we take some eigenvenvector $\vec v_2$ and then solve the above system to find $\vec v_3$.

This is called generalized eigenvector.

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.

Khallil

Did the $A v_1 = -2 v_1$ come from the $AP = PJ$ and the $J$ we worked out before, @Martin?

Martin Sleziak

@Khallil Yes.

However, you get $Av_1=-2v_2$ for both possibilities we have for J.

Khallil

Is that significant?

Martin Sleziak

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

So I will have to leave @Khallil

Khallil

Oct 17, 2015 22:53

Ah, that's a shame

Thank you for the help!

Martin Sleziak

I hope you will somehow be able to get started. (Perhaps with help of some posts on main or some other materials.)

Good luck!

Khallil

I'm finally beginning to understand this stuff. I'll try and follow a similar methodology and get there!

See you soon, @Martin

:-}

Martin Sleziak

Oct 18, 2015 04:08

I will add some random selection of examples from the main site.

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Find Jordan normal form J and also the matrix P

A few problems exactly of the type which we tried above:

2

Q: How do I write this matrix in Jordan-Normal Form

I have the matrix $A=\begin{pmatrix}2&2&1\\-1&0&1\\4&1&-1\end{pmatrix}$, I want to write it in Jordan-Normal Form. I have $x_1=3,x_2=x_3=-1$ and calculated eigenvectors $v_1=\begin{pmatrix}1\\0\\1\end{pmatrix},v_2=\begin{pmatrix}1\\-4\\5\end{pmatrix},v_3=\begin{pmatrix}0\\0\\0\end{pmatrix}$. But...

matrices eigenvalues-eigenvectors jordan-normal-form

5

Q: What is wrong with this Jordan normal form computation?

The question I am working on is to compute the Jordan normal form of $$A := \begin{pmatrix} 2 & 1 & 5 \\ 0 & 1 & 3\\ 1 & 0 & 1\end{pmatrix}.$$ The characteristic polynomial and minimal polynomial of $A$ is $x^{2}(x - 4)$. Then the Jordan normal form of $A$ is given by $$J := \begin{pmatrix} 0 & 1 &...

linear-algebra matrices solution-verification jordan-normal-form

0

Q: Finding Jordan canonical form of a $4\times 4$ matrix

Let $$ A = \begin{pmatrix} 1&-2&1&0 \\ 1&-2&1&0 \\ 1&-2&1&0 \\ 1&-1&0&0 \end{pmatrix}$$ I need to find the Jordan canonical form and the minimal polynomial. Now I sense that their is a shortcut (due to the triple repeated rows) but I cannot quite work out what it is? I know that it has an e...

linear-algebra matrices jordan-normal-form

2

Q: Finding Jordan form

Find Jordan form of the following matrix: $$\left(\begin{matrix}4&-5&2 \\ 5&-7&3\\ 6&-9&4 \end{matrix}\right)$$ So I got stuck pretty much trying to find the eigenvalues. Related question: Is the characteristic polynomial of the characteristic matrix, equals to the characteristic polynomial of...

linear-algebra matrices eigenvalues-eigenvectors jordan-normal-form

0

Q: Finding Jordan Basis of a matrix (3X3)

Having trouble finding the jordan base for this matrix \begin{pmatrix} 1 & 1 &0 \\ 0 &1 &1 \\ 0& 0 &2 \end{pmatrix} I know that the Characteristic polynomial is : (t-1)^2(t-2) I started with eigenvalues λ=1 I found that the minimal k is 2 and: dim(ker(I-A))=...=Span({\begin{pmatrix} 1\\ 0 \\ 0 \e...

jordan-normal-form

2

Q: Jordan similar matrix

I have matrix $B = \begin{bmatrix}1 & 1 & -2 & 0\\2 & 1 & 0 & 2 \\ 1 & 0 & 1 & 1 \\ 0 & -1 & 2 & 1\end{bmatrix}$. I found the characteristic polynomial $(1-x)^4$ and was able to get my Jordan Matrix $J = \begin{bmatrix}1 & 1 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\end{bmatrix}$. ...

linear-algebra matrices jordan-normal-form

1

Q: How to determine the Jordan form and give a Jordan base for a matrix?

given is $\begin{pmatrix} 3&0&-1&0&0 \\ 1&3&0&1&0 \\ 0&0&3&0&0 \\ 0&0&0&3&0 \\ 0&0&0&0&-3 \end{pmatrix}$ I have to determine the jordan form and also give a jordanbase. I got this so far: Eigenvalues are $l_1 = 3, l_2 = -3$ whereas $P_A(x) = (x-3)^4 * (x+3)$. Eigenvectors are for $l_1 = 3$: ...

linear-algebra matrices jordan-normal-form

^ Some comments in the solution given here seem a bit imprecise to me. But the result as such seems to be correct. (And of course I might be wrong in my doubts.)

0

Q: Find matrix that implements Jordan normal form

I have a matrix $$B=\begin{pmatrix}1&2&3\\0&4&5\\0&0&6\end{pmatrix}$$ I have calculated the eigenvectors: $$\{\begin{pmatrix}-1\\0\\0\end{pmatrix},\begin{pmatrix}-\frac{2}{3}\\-1\\0\end{pmatrix},\begin{pmatrix}-\frac{8}{5}\\-\frac{5}{2}\\-1\end{pmatrix}\}$$ However, I am trying to find the matr...

linear-algebra eigenvalues-eigenvectors

2

Q: How to put a matrix in Jordan canonical form, when it has a multiple eigenvalue?

I have a question that reads: Put the matrix \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix} in Jordan Canonical Form. Moreover, in each case, find the appropriate transition matrix to the basis in which the original matrix assumes its Jordan form. I'm having a lot of tro...

linear-algebra matrices jordan-normal-form

3

Q: Find Jordan form of a $3\times 3$ matrix

$$\left( \begin{array}{ccc} 0 & 1 & 2 \\ -5 &-3 & -7 \\ 1 & 0 & 0 \end{array} \right) $$ I figured out the eigenvalues are all -1 from the characteristic polynomial, but I'm not sure how to find the 1's on the subdiagonal. I know they're 1, but I'm not sure how that's determined. I also know th...

linear-algebra eigenvalues-eigenvectors jordan-normal-form

4

Q: Help in finding the Jordan canonical form of a matrix

Determine the Jordan Canonical Form of the following matrix: $$A=\begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 4\\ \end{bmatrix}$$ I am trying to determine the Jordan Basis first. For that purpose I am trying to find out the generalized Eigenvectors of this matrix. Corresponding to $1$, ...

linear-algebra matrices jordan-normal-form

What can be said about J from the minimal polynomial?

We talked about the fact that if we know minimal polynomial and characteristic polynomial, we can say some things about J. (Sometimes even completely determine it.)

Of course, only from this information nothing can be said about P.

6

Q: Why does the largest Jordan block determine the degree for that factor in the minimal polynomial?

Let $A$ be a square matrix, so $A$ has some Jordan Normal form. Then $A$ has a minimal polynomial, say $m(X)=\prod_{i=1}^k (t-\lambda_i)^{m_i}$. Wikipedia says The factors of the minimal polynomial $m$ are the elementary divisors of the largest degree corresponding to distinct eigenvalues....

linear-algebra matrices jordan-normal-form

8

Q: Finding the Jordan canonical form of this upper triangule $3\times3$ matrix

I am supposed to find the Jordan canonical form of a couple of matrices, but I was absent for a few lectures. \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{bmatrix} Since this is an upper triangular matrix, its eigenvalues are the diagonal entries. Hence $\lambda_{1,2}=1$ and $\lam...

linear-algebra matrices jordan-normal-form

5

Q: Possible Jordan Canonical Forms Given Minimal Polynomial

I was supposed to find all possible Jordan canonical forms of a $5\times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester. I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block w...

linear-algebra abstract-algebra matrices jordan-normal-form

2

Q: If we know the eigenvalues of a matrix $A$, and the minimal polynom $m_t(a)$, how do we find the Jordan form of $A$?

We have just learned the Jordan Form of a matrix, and I have to admit that I did not understand the algorithm. Given $A = \begin{pmatrix} 1 & 1 & 1 & -1 \\ 0 & 2 & 1 & -1 \\ 1 & -1 & 2 & -1 \\ 1 & -1 & 0 & 1 \end{pmatrix} $, find the Jordan Form $J(A)$ of the matrix. So what I did so far: ...

linear-algebra matrices jordan-normal-form

1

Q: Find the jordan form the matrix

let characteristic polynomial $P_A(x)=(x+2)^4(x-3)^2$ and minimal polynomial $m_A=(x+2)^3(x-3)$ find the jordan form that possible. we know $q_6=\frac{f_6}{f_5}$ ($f_i$ is gcd{det of i x i submatrices which isnt equal to 0}) $q_6=\frac{f_6}{f_5}=\frac{(x+2)^4(x-3)^2}{f_5}=(x+2)^3(x-3)^2$ so ...

linear-algebra jordan-normal-form

4

Q: Help in finding the Jordan canonical form of a matrix

Determine the Jordan Canonical Form of the following matrix: $$A=\begin{bmatrix} 1 & 2 & 3\\ 0 & 4 & 5\\ 0 & 0 & 4\\ \end{bmatrix}$$ I am trying to determine the Jordan Basis first. For that purpose I am trying to find out the generalized Eigenvectors of this matrix. Corresponding to $1$, ...

linear-algebra matrices jordan-normal-form

2

Q: How to turn this matrix to Jordan normal form?

Matrix $A$ is $ \left( \begin{array}{ccc} 3 & 0 & 8 \\ 3 & -1 & 6 \\ -2 & 0 & -5 \end{array} \right)$ and I need to find a matrix P such that $P^{-1} A P = J$ where $J$ is a Jordan matrix. My trial is: Calculate the $det(λI - A) = (λ+1)^{3}$ and its elementary divisor is $(λ+1)^{3}$ as well; ...

linear-algebra matrices jordan-normal-form

2

Q: find the Jordan form and $P$ such that $P^{-1}AP = J$.

Consider the matrix $$A = \left(\begin{array}{cccc} -11&0&-9\\32&1&24\\16&0&13 \end{array}\right)$$ I want to find the Jordan form of $A$, with $1$-s at the bottom and the jordan basis, which is $P$ columns such that $P^{-1}AP = J$. I evaluated the charechteristic polynomial which is $f_A(x) = ...

linear-algebra diagonalization jordan-normal-form

Some other interesting posts

3

Q: 3x3 matrices completely determined by their characteristic and minimal polynomials

How do you show that two 3x3 matrices with the same characteristic and minimal polynomials both conjugate to the same Jordan normal form, assuming no knowledge of the eigenspaces? I know that it is possible to determine completely the Jordan normal form of a matrix only with its minimal and cha...

linear-algebra matrices jordan-normal-form minimal-polynomials

5

Q: Jordan Canonical Form determined by characteristic and minimal polynomials in dimension $3$, but not beyond

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-algebra matrices

@Khallil Maybe reading of some posts there might clarify some things. (After all we only talked about one example.)

There are things we did not talk about at all. There are other methods.

I suppose in those (and some other answers) you can find plenty of useful stuff.

Martin Sleziak

Oct 18, 2015 05:57

BTW as I mentioned I did not want to make things too complicated, so I tried to avoid things not needed in your example.

But I should at least be honest and say which are some important things which I did not even touch:

1) How to look for generalized eigenvectors if we have more than one eigenvector.

2) If we want only J and not P, another approach is to user ranks of $(A-\lambda I), (A-\lambda I)^2, \dots$.

3) Powers of the matrix $A-\lambda I$ can be used to find generalized eigenvectors. So we may start by solving the system with matrix $(A-\lambda I)^k$ for some k, if we prefer this approach.

4) There are probably tons of other stuff which I do not even recall right now.

Diagonalization of a matrix

$Mathematics$ Linear & Abstract algebra

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