# Jordan form II: Computations

This post talks about computation of Jordan form and Jordan basis, along with a few examples.

In this post we focus on computing the Jordan form of a given matrix.

Proposition 1 (Calculating the number of Jordan blocks/sizes)For each eigenvalue , the number of corresponding Jordan blocks is . More generally, let

= the number of Jordan blocks of size

Then , i.e.

This further shows that the number of Jordan blocks of size is

This is obvious by looking at the Jordan form. A nice way to keep track of all these numbers is to use Young tableaux, which will be done later.

The next question would be how to find a Jordan basis. A possible procedure is outlined here,

1. Fix an eigenvalue . Find the smallest such that stabilizes, i.e. . This is the size of the largest Jordan block.

2. Find a basis of . Pull this basis back to; this set is clearly linearly independent. Let this pullback set be . Notice that the set defined by

is also linearly independent. Moreover, it spans a -invariant subspace, where the matrix of with respect to this basis is blocks of Jordan block of size .

3. Consider . We already have a linearly independent set . Extend this to a basis, say we concatenate it by a set . The set

is again linearly independent.

4. Iterate, the final set would be a Jordan basis, since it is a set of linearly independent elements with the right dimension (why?) such that has Jordan form with respect to it.

Example 1Find the Jordan form of .

The eigenvalues are clearly all 1. It is clear that , which means that there is only one Jordan block. Thus its Jordan form is .

To find a Jordan basis, notice that as there is only one Jordan block, the basis must be a cyclic one. Thus we only have to compute the image of .

Thus is one Jordan basis.

Example 2Find the Jordan form of .

After computations, the characteristic polynomial is . Therefore it suffices to check the number of Jordan blocks for the eigenvalue .

is clearly of rank 1. So the kernel has rank 2, meaning that there are 2 Jordan blocks, i.e. is diagonalizable and is similar to .

To find a Jordan basis, we first consider the eigenvalue 1.

After performing Gaussian elimination, we get

Then clearly the kernel is spanned by .

For the eigenvalue -1,

thus after Gaussian elimination, it becomes

yielding a basis of .

Example 3Find the Jordan form of .

After computations, the characteristic polynomial is . Therefore it suffices to check the number of Jordan blocks for the eigenvalue .

is clearly of rank 2. Thus the kernel is of 2 dimension, meaning that the Jordan form has to be

.

To find a Jordan basis, for the eigenvalue 2,

After some row reduction, we get

Therefore the kernel is spanned by .

For the eigenvalue 1,

1. By some computations, is spanned by .

The kernel contains . One can extend this to a basis by adding .

2. Compute

Then is the basis for the Jordan block of size 2.

3. Extend to a basis of by adding . Thus

spans .

Conclusion: A Jordan basis is .

**Young tableaux**

It is a convenient tool to track the number/size of Jordan blocks for each eigenvalue.

For example in the above case, we have exactly one Jordan block of size 3, we then give it a horizontal bar

If instead we have one Jordan block of size 2, and another of size 1, we give it a tableau

In general, each row represents one Jordan block, and the number of boxes represent the dimension. We list each block according to its size in a descending order. When we read this figure horizontally, we then obtain the numbers defined previously, i.e. the number of Jordan blocks with size .

Therefore if we merely wish to find the Jordan form, we may, for each eigenvalue ,

1. Calculate until it stabilizes.

2. Calculate , where is the number of Jordan blocks with size .

3. Draw boxes in the -th column, aligned to the top.

4. We then obtain a Young tableau, and we can just read off the number of Jordan blocks (the number of rows), and their sizes (the number of boxes in each row).

**Remark**

The last two examples are taken from the last page of http://www.math.uga.edu/~roy/rev.lin.alg.pdf .

Reblogged this on forsixthirty and commented:

nice.