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Completing the basis of a matrix, Jordan form related

 Quote by fluidistic I see. If you don't mind me asking, why is the kernel of ##(A+2I)^2## any interest?
Good question! The kernel of $(A+2I)^2$ is exactly the subspace of all generalized eigenvectors of A with eigenvalue -2.
So if v is in the kernel of $(A+2I)^2$, then $(A+2I)(A+2I)v=0$. So $(A+2I)v$ is an eigenvector of A or it is zero. In the last case, if $(A+2I)v=0$, then v is an eigenvector.

In general, given an eigenvector $\lambda$ with algebraic multiplicity n, then the kernel $(A-\lambda I)^n$ is exactly the space of all generalized eigenvectors. Note however, that you don't always need to calculate this kernel. For example: sometimes a smaller n also suffices. This is the case with our special matrix A. We see there that -2 has algebraic multiplicity 3, so we see that the kernel $(A+2I)^3$ is exactly the space of generalized eigenvectors. But the kernel of $(A+2I)^2$ is that space too (and certainly it is easier to calculate $(A+2I)^2$ than $(A+2I)^3$!). The reason that n=2 works here is exactly because there are two linear independent eigenvectors.

In general, we have that

$$Ker(A-\lambda I)\subseteq Ker(A-\lambda I)^2\subseteq Ker(A-\lambda I)^3 \subseteq ...$$

There exists a certain m (with m smaller or equal than the algebraic multiplicity) such that this sequence stabilizes. So for all k, we have $Ker(A-\lambda I)^m=Ker (A-\lambda I)^{m+k}$.

In our situation of our special matrix, we had

$$Ker(A+2I)\subseteq Ker(A+2I)^2 = Ker(A+2I)^3 = Ker(A+2I)^4 = ...$$

so the sequence stabilizes at 2. How do we know this without calculating?

We see that the $Ker(A-\lambda I)^k$ is an "increasing" sequence. That is: at every step k, we add more vectors to the space. We add vectors until the sequence stabilizes at m.

It is a theorem that the dimension of $Ker(A-\lambda I)^m$ (with m the plac where it stabilizes) is exactly the algebraic multiplicity of $\lambda$.

How can we use this information? Let us say that we are in our previous situation in the case that $\lambda=-2$. We see that $Ker(A+2I)$ has dimension 2 (because this is exactly the space of eigenvectors). We know that at the place where the chain stabilizes, we have $Ker(A+2I)^m$ has dimension 3 (since the algebraic multiplicity of -2 was 3). So we immediately get that the chain stabilizes at m=2.

As another example, let's say that A is a 5x5 matrix and such that there is one eigenvalue $\lambda$ that has multiplicity 5.
Let's say that we have only 1 linear independent eigenvector.
So $Ker(A-\lambda I)$ has dimension 1. Looking at $Ker(A-\lambda I)^2$, this must then have dimension 2. Then $Ker(A-\lambda I)^3$ has dimension 3. So we see that the chain stabilizes at m=5.
What does this imply practically?? This implies that we will find a vector v such that
$$\{v,(A-\lambda I)v,(A-\lambda I)^2v,(A-\lambda I)^3v,(A-\lambda I)^4v\}$$
gives rise to the Jordan normal form.

What if we found 2 linear independent eigenvectors? Then $Ker(A-\lambda I)=2$. Now there are two possibilities which can arise:
1) $Ker(A-\lambda I)^2 =3$. In this case, the chain stabilizes at m=4. So we see that in this case, we find a vector v and an eigenvector w such that
$$\{v,(A-\lambda I)v,(A-\lambda I)^2 v, (A-\lambda I)^3 v, w\}$$
gives rise to the Jordan normal form.

2) $Ker(A-\lambda I)^2=4$. In this case. The chain stabilizes at m=3. So the Jordan basis is given by
$$\{v,(A-\lambda I)v,(A-\lambda I)^2 v, w, (A-\lambda I)w\}$$
for suitable v and w.

OK, I made a loooooooooong digression, sorry. But I hope it is a bit clearer now
 Recognitions: Gold Member It's over 1 am for me so I will analyze your last post tomorrow. You made me remind of the way to get the Jordan form I've been taught. 1)First get all the eigenvalues of A. 2)Take an eigenvalue that you have not taken yet and do ##(A-\lambda I)^m## for m=1, 2, etc. and calculate each time the dimension of the kernel of ##(A-\lambda I)^m## until that number remains constant for some ##m_0 \leq n##. So you get ##dim (\ker (A-\lambda I)^m)=dim (\ker (A-\lambda I)^{m_0})## for all ##m \geq m_0##. For m=1, 2, ..., ##m_0##, calculate ##d_m=2 dim (ker [(A-\lambda I)^m])- dim (ker [(A-\lambda I)^{m-1}])- dim (ker [(A-\lambda I)^{m+1}])##. So that ##d_m >0## is equal to the number of times that the Jordan block ##J_m (\lambda )## of dimension mxm with lambda on its diagonal appear in the Jordan form. The geometric multiplicity of lambda is equal to the sum of the ##d_m##'s. The algebraic multiplicity is equal to the sum of the products ##md_m##. 3)Go to 2) until you've covered all lambda's.

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