Jordan Normal Form - Example

Example

This example shows how to calculate the Jordan normal form of a given matrix. As the next section explains, it is important to do the computation exactly instead of rounding the results.

Consider the matrix

A = \begin{bmatrix} 5 & 4 & 2 & 1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & 3 & 0 \\ 1 & 1 & -1 & 2 \end{bmatrix}

which is mentioned in the beginning of the article.

The characteristic polynomial of A is

This shows that the eigenvalues are 1, 2, 4 and 4, according to algebraic multiplicity. The eigenspace corresponding to the eigenvalue 1 can be found by solving the equation Av = λ v. It is spanned by the column vector v = (−1, 1, 0, 0)T. Similarly, the eigenspace corresponding to the eigenvalue 2 is spanned by w = (1, −1, 0, 1)T. Finally, the eigenspace corresponding to the eigenvalue 4 is also one-dimensional (even though this is a double eigenvalue) and is spanned by x = (1, 0, −1, 1)T. So, the geometric multiplicity (i.e. dimension of the eigenspace of the given eigenvalue) of each of the three eigenvalues is one. Therefore, the two eigenvalues equal to 4 correspond to a single Jordan block, and the Jordan normal form of the matrix A is the direct sum

J = J_1(1) \oplus J_1(2) \oplus J_2(4) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}.

There are three chains. Two have length one: {v} and {w}, corresponding to the eigenvalues 1 and 2, respectively. There is one chain of length two corresponding to the eigenvalue 4. To find this chain, calculate

Pick a vector in the above span that is not in the kernel of A − 4I, e.g., y = (1,0,0,0)T. Now, (A − 4I)y = x and (A − 4I)x = 0, so {y, x} is a chain of length two corresponding to the eigenvalue 4.

The transition matrix P such that P−1AP = J is formed by putting these vectors next to each other as follows

P = \Big = \begin{bmatrix} -1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{bmatrix}.

A computation shows that the equation P−1AP = J indeed holds.

P^{-1}AP=J=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 4 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix}.

If we had interchanged the order of which the chain vectors appeared, that is, changing the order of v, w and {x, y} together, the Jordan blocks would be interchanged. However, the Jordan forms are equivalent Jordan forms.

Read more about this topic:  Jordan Normal Form

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