# 4. Jordan Theorem

### Jordan form

Not every matrix can be diagonalized. The Jordan form of a matrix is a sensible generalization of a diagonal matrix, such that (nearly) every matrix can be transformed to it.

Instead of single numbers on the diagonals, in the Jordan form, we allow there matrices called Jordan blocks. A Jordan block is a matrix which has a fixed number on the diagonal, and ones immediately above the diagonal, e.g.

So this is an example of a matrix in the Jordan form:

### Jordan theorem

Jordan theorem states that for every matrix such that its characteristic polynomial can be decomposed into expressions of degree one, there exists a similar matrix which is in Jordan form. In particular, it is true over an algebraically closed field, e.g over .

Translating it to the language of linear mappings , for every mapping, the characteristic polynomial of which can be decomposed into expressions of degree one, there exists a basis (called a Jordan basis), such that the matrix of this mapping with respect to this basis is in Jordan form.

In particular, it means, that if are the numbers of columns in which the Jordan blocks start, then vectors are eigenvectors for eigenvalues , and the rest of the vectors (for , but less than or in the last block) are such that (it is implied by the matrix, the one immediately over the diagonal is responsible for adding the previous vector). Vectors are the first in their blocks, and are the last in their blocks.

### Jordan basis

Let us think how to find a Jordan basis (and prove Jordan Theorem). We are going to do it recursively.

1. Find at least one eigenvalue of mapping .
2. Consider a mapping . Notice that:
• a Jordan basis for is also a Jordan basis for . Indeed, if , then , but if , then , thus instead of looking for a basis for we may look for a basis for .
• zero is an eigenvalue of , because if is an eigenvector of for eigenvalue , then . Thus, the dimension of the kernel of is greater than zero.
• Therefore, the dimension of the image is less than , where is the dimension of the whole space.
3. Find (recursively) a Jordan basis of defined on a space with less dimensions than the original one (obviously, if the space has dimension , then any vector is an eigenvector so it is a Jordan basis). Let this basis be . This requires finding the image of first.
4. Now, in subsequent steps we are going to add new vectors to this basis. Some of the vectors are eigenvectors for wigenvalue zero. Let us denote the indices of those vectors by . Those vectors are important, because:
• they are in the kernel, which is the eigenspace for eigenvalue zero, so they are a basis of the kernel of , i.e. , in particular the dimension of equals ,
• are the starting vectors of some Jordan blocks for . Let the ending vectors of those blocks be .
5. The dimension of the kernel is obviously . Complete the basis of of to a basis of the whole kernel . Since those are vectors from the kernel, they are eigenvectors for , we can take them to our Jordan basis (e.g. at its end).
6. Thus, we already have vectors. We need more vectors. The idea is that we will extend blocks starting with , and ending with by one vector. Actually, we have to insert after each vectors , for such that . It indeed extends each of those blocks, since , and is the previous vector, and is the eigenvalue of those blocks. This is how we get the Jordan basis which we are looking for.

7. Indeed it is a basis, because it turns out to be linearly independent. If

then

But , but this implies that , because

is a linear combination of , but

is a linear combination of .
Thus

which implies that , because it follows that is in the intersection of the kernel and the image and thus has to be the zero vector by definition of . But then because are linearly independent, which means that the system we are considering is actually linearly independent.

E.g. let us find the Jordan basis for

Let be a mapping defined by this matrix. We check its eigenvalues. , so the only eigenvalue is . Thus let

Therefore,

Consider . We get that

But and , and since , we get that and is a Jordan basis for , because those vectors are eigenvectors for eigenvalue zero. Both are starting and ending vectors of their blocks. We have to find and such that and . We can take e.g. and . Then, is a Jordan basis both for , and for , and

and

### Powers of matrices in Jordan form

It is easy to notice that it is relatively easy to calculate powers of a matrix a Jordan form. The zeroes under the diagonal and outside of Jordan blocks are still there, and a power of a matrix in a Jordan form has simply powers of its Jordan blocks on the diagonal.

You may notice that

In particular, it becomes easy if , because that the -th power of a Jordan block has only zeros except for the places places over the diagonal, where there are ones (if ).

Generally, the rank of a -th power of such a block equals for and zero otherwise. But the rank of the power of any other Jordan block is always equal to its size.

### Finding Jordan form using powers

Notice that it means that the rank of Jordan blocks for eigenvalue zero decreases with taking higher and higher powers by one, until it reaches zero. This sensitivity (and the fact that the rank of a matrix does not depend on a basis) can be used to determine a similar matrix in the Jordan form without calculating a Jordan basis.

Indeed,

equals to the number of Jordan blocks for eigenvalie of size (because exactly those blocks decrease their rank by one if the power changes from to ).

### Jordan form and similar matrices

Every matrix over an algebraically closed field has a similar matrix in Jordan form. Such a matrix is unique up to the order of blocks. This can be used to determine whether two matrices are similar. They are if and only if they have the same matrix in Jordan form (up to order of blocks).

### Calculating powers of matrices

Similarly as in the case of diagonalizable matrices, we can use Jordan form to calculate a power of a matrix, if the characteristic polynomial of which can be decomposed.

E.g. let us calculate

We have

So the matrix in Jordan form is

because it is the only case for eigenvalue different from the identity.
It is easy to notice that

It is clear that the eigenvector for is , and . Let us find , such that . It is e.g. . Thus, is a Jordan basis.
We get
and .

Thus,