Calculation Results
What Is the Jordan Canonical Form?
The Jordan Canonical Form (JCF), also called the Jordan Normal Form, is a special block-diagonal matrix similar to a given matrix. It expresses a matrix in terms of its eigenvalues and generalized eigenvectors. When a matrix cannot be fully diagonalized, its Jordan form contains Jordan blocks.
Definition of a Jordan Block
A Jordan block of size n × n associated with eigenvalue λ is:
$$ J(\lambda)= \begin{pmatrix} \lambda & 1 & & 0 \\ & \lambda & 1 & \\ & & \ddots & 1\\ 0 & & & \lambda \end{pmatrix} $$The full Jordan form of a matrix \(A\) is defined as:
$$ J = P^{-1}AP $$where \(P\) is formed from eigenvectors and generalized eigenvectors.
Steps to Compute the Jordan Form
- Find eigenvalues and their algebraic multiplicities.
- Determine geometric multiplicities (number of independent eigenvectors).
- Decide the number and sizes of Jordan blocks.
- Compute generalized eigenvectors if the matrix is not diagonalizable.
- Build matrix \(P\) and compute \(J = P^{-1}AP\).
Example 1: A Diagonalizable Matrix
Given:
$$ A= \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $$Eigenvalues are 2 and 3. Each has algebraic multiplicity = geometric multiplicity = 1. Thus, the Jordan form is simply:
$$ J= \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} $$Example 2: A Non-Diagonalizable Matrix
Given:
$$ A= \begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix} $$Eigenvalue λ = 4 with algebraic multiplicity 2, but only one eigenvector exists (geometric multiplicity 1). Thus, the matrix is not diagonalizable. Its Jordan form is:
$$ J= \begin{pmatrix} 4 & 1 \\ 0 & 4 \end{pmatrix} $$Common Mistakes to Avoid
| Incorrect assumption | Correct understanding |
|---|---|
| Repeated eigenvalues always mean multiple blocks. | Number of blocks = geometric multiplicity. |
| Block size equals geometric multiplicity. | Block size depends on algebraic multiplicity + structure. |
| Only find standard eigenvectors. | Must find generalized eigenvectors if not diagonalizable. |
| All matrices are diagonalizable. | Only when algebraic = geometric multiplicity for each eigenvalue. |
Practice Exercises: Compute the Jordan Form
The exercises below require you to compute the Jordan Canonical Form of each matrix. Click "Show Answer" to reveal the solution.
Exercise 1:
$$ A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} $$Show Answer
Eigenvalue: λ = 2 (algebraic multiplicity 2, geometric multiplicity 1)
Jordan form:
$$
J = \begin{pmatrix}
2 & 1 \\
0 & 2
\end{pmatrix}
$$
Exercise 2:
$$ A = \begin{pmatrix} 3 & 0 & 0\\ 0 & 3 & 1\\ 0 & 0 & 3 \end{pmatrix} $$Show Answer
Eigenvalue: λ = 3 (algebraic multiplicity 3, geometric multiplicity 2)
Jordan form:
$$
J = \begin{pmatrix}
3 & 1 & 0\\
0 & 3 & 0\\
0 & 0 & 3
\end{pmatrix}
$$
Exercise 3:
$$ A = \begin{pmatrix} 4 & 1 & 0\\ 0 & 4 & 1\\ 0 & 0 & 4 \end{pmatrix} $$Show Answer
Eigenvalue: λ = 4 (algebraic multiplicity 3, geometric multiplicity 1)
Jordan form:
$$
J = \begin{pmatrix}
4 & 1 & 0\\
0 & 4 & 1\\
0 & 0 & 4
\end{pmatrix}
$$
Exercise 4:
$$ A = \begin{pmatrix} 5 & 0 & 0 & 0\\ 0 & 5 & 1 & 0\\ 0 & 0 & 5 & 0\\ 0 & 0 & 0 & 6 \end{pmatrix} $$Show Answer
Eigenvalues: λ = 5 (algebraic multiplicity 3, geometric multiplicity 2), λ = 6 (algebraic multiplicity 1)
Jordan form:
$$
J = \begin{pmatrix}
5 & 1 & 0 & 0\\
0 & 5 & 0 & 0\\
0 & 0 & 5 & 0\\
0 & 0 & 0 & 6
\end{pmatrix}
$$
Jordan form is easier to understand after exploring matrix multiplication, matrix inverse, and matrix rank.
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