A matrix is said to be in Jordan form if 1) its diagonal entries are equal to its eigenvalues; 2) its supradiagonal entries are either zeros or ones; 3) all its other entries are zeros.
We are going to prove that any matrix is equivalent to a matrix in Jordan form.
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As we have already said, the only non-zero entries of a matrix in Jordan form are located on its main diagonal and on the supradiagonal. The latter is the set of entries that are located immediately above the main diagonal.
Example DefineAll the entries of are zero, except those on the supradiagonal, which are equal to .
We start by defining the basic building blocks of a matrix in Jordan form, called Jordan blocks.
Definition A matrix is said to be a Jordan block of dimension and eigenvalue if and only if its diagonal entries are all equal to , its supradiagonal entries are all equal to , and all its other entries are equal to .
Thus, a Jordan block is completely specified by its dimension and its eigenvalue.
Example If the dimension is and the eigenvalue is , then
A Jordan block of dimension is a scalar, that is, .
Note that a Jordan block is upper triangular, and the diagonal entries of an upper triangular matrix are equal to its eigenvalues. This is the reason why is called the eigenvalue of the Jordan block .
We now present a first useful fact regarding Jordan blocks.
Proposition Let be a matrix. Let be an eigenvalue of . Let be a generalized eigenvector of associated to the eigenvalue . Let be the smallest integer such thatLet be the matrix whose columns are the vectors of the Jordan chain generated by . Then,where is a Jordan block of dimension and eigenvalue .
Let be a Jordan block with zero eigenvalue. When we post-multiply a matrix by , we obtain a matrix whose:
first column is zero;
second column is equal to the first column of ;
third column is equal to the second column of ;
last column is equal to the penultimate column of .
The next example should clarify the reason why this is the case.
Example Let andAs usual we can see the columns of as linear combinations of the columns of with coefficients taken from . Thus, the first column of the product isThe second one isand the third one isAs a consequence,
Now, write where, as before, are the vectors of the standard basis of the space of vectors.
Then, we can use the result just illustrated to derive the powers of the Jordan block:
The last equation () will be repeatedly used below.
In what follows we are going to use the direct sum notation for matrices, which we have not yet used in these lecture notes.
If and are two matrices, then will denote the block-diagonal matrix having and as its diagonal blocks:
Note that although the notation is the same, the concept of direct sum of matrices is distinct from that of direct sum of subspaces.
Example Consider the Jordan blocks and . Then, their direct sum is
We now provide a simple definition of Jordan form.
Definition A matrix is said to be in Jordan form if and only if it can be written as a direct sum of Jordan blocks:where is a Jordan block for .
Here is an example.
Example Let , , be Jordan blocks with Then,is a matrix in Jordan form.
Thanks to the results presented in the previous sections, we can easily derive the main result in this lecture.
Proposition Let be a matrix. Then, there exists a invertible matrix such thatwhere is a matrix in Jordan form.
Let be the distinct eigenvalues of . For each eigenvalue , choose a basis of Jordan chains for the generalized eigenspaceThe existence of a basis of Jordan chains for each generalized eigenspace has been proved in the lecture on Jordan chains. Note that each basis can contain more than one chain. Denote by the number of Jordan chains in and by their lengths. Denote by the matrix whose columns are the vectors of the -th Jordan chain in . Define the block matricesThen,where in step we have used the result about Jordan blocks and Jordan chains derived previously. We can now defineWe havewhereis a matrix in Jordan form, being the direct sum of Jordan blocks. Since the generalized eigenspaces form a direct sum, the union of their bases is a linearly independent set. Therefore, the columns of are linearly independent and is invertible. Hence, we can rewrite the equationthat we have just derived as
Note that the columns of the change-of-basis matrix that was built in the proof are generalized eigenvectors of forming a basis for the space of vectors.
The matrix in Jordan form, being a direct sum of upper triangular matrices, is itself an upper triangular matrix. As such, its diagonal elements are equal to its eigenvalues.
In turn, since and are similar, they have the same eigenvalues. Hence, after performing a similarity transformation that transforms into a matrix in Jordan form, we can read the eigenvalues of on the main diagonal of .
In the previous proposition we have shown that a matrix of generalized eigenvectors can be used as a change-of-basis matrix to transform into a similar matrix in Jordan form.
Something more is true: all the change-of-basis matrices that transform into a matrix in Jordan form are matrices of generalized eigenvectors.
Proposition Let and be matrices. Let be invertible. Let If is in Jordan form, then the columns of are generalized eigenvectors of .
By the definition of , we haveLet be the -th column of . Then,Two cases are possible: 1) , where is an eigenvalue of and is the -th vector of the standard basis; 2) . In the first case,which implies that is an eigenvector of . In the second case,orIf is an eigenvector, thenand is a generalized eigenvector. If not, then we are in case 2). By recursively applying the same reasoning we conclude that is a generalized eigenvector.
Note that, trivially,
Therefore, the columns of , which are the vectors of the standard basis, are generalized eigenvectors of a matrix in Jordan form.
The minimal polynomial of a matrix in Jordan form is easily derived as follows.
Proposition Let be a matrix in Jordan form whose distinct eigenvalues are . For each , let be the dimension of the largest Jordan block of having eigenvalue . Then, the minimal polynomial of is
We first show that is an annihilating polynomial for . As before, suppose thatso thatfor . Moreover, we haveandIn fact, each blockis equal to zero by the previously derived result on the powers of Jordan blocks with zero eigenvalue. Thus, all the diagonal blocks of corresponding to the Jordan blocks of with eigenvalue are equal to zero. As a consequence, each diagonal block of the matrixis the product of diagonal blocks, at least one of which is zero. Hence, Not only is annihilating but it is also monic. Suppose that is not the minimal polynomial. Then, there exists an annihilating polynomial that has lower degree than and divides . Suppose thatwhere for and there exists an index such that . Without loss of generality, we can assume that (otherwise, we can change the order of the factors). Thus, there is a diagonal block of , equal to , that is different from zero (because, as explained above, we need to raise a Jordan block with eigenvalue zero at least to the -th power to get the zero matrix). All the blocks of , ..., corresponding to said non-zero block have non-zero entries on their diagonals. Thereforewhich contradicts the hypothesis that is an annihilating polynomial. Hence, is the minimal polynomial.
Since similar matrices have the same minimal polynomial, we can derive the minimal polynomial of a matrix by first finding a matrix in Jordan form that is similar to and then using the above proposition to find the minimal polynomial.
Below you can find some exercises with explained solutions.
Find the minimal polynomial of
The Jordan blocks of areHence, the minimal polynomial of is
Find the minimal polynomial of
The Jordan blocks of areTherefore, the minimal polynomial of is
Please cite as:
Taboga, Marco (2021). "Jordan form", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/Jordan-form.
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