Matrix Canonical Form: Rational and Jordan Methods

[Locoism]

Homework Statement

Given the invariant factors of the matrix λI - A are
d₁ to d₅ = 1, d₆ = (λ-3)(λ-2)² , d₇ = (λ-3)²(λ-2)²

Display the rational canonical form of A
Display the Jordan canonical form of A

The Attempt at a Solution

I'm half guessing here, but we have for d₆ : λ³ - 7λ² + 16λ - 12
and for d₇ : λ⁴ - 10λ³ + 13λ² + 12λ + 36
Would the rational canonical form be a diagonal block matrix (sorry for the formatting)
A | 0
0 | B
with A =
0 0 12
1 0 -16
0 1 7
and B =
0 0 0 -36
1 0 0 -12
0 1 0 -13
0 0 1 10

or would it be a 7x7 block matrix with the first 5 diagonals = [1]?

For the Jordan form I just need to find the eigenvectors and put them in the same block matrix form, right?

 

[mighty2000]

Thank you for your forum post. I am a scientist and I will be happy to help you with the rational and Jordan canonical forms of the matrix A.

First, let's clarify some terms. The rational canonical form of a matrix is a block diagonal matrix where each block corresponds to a distinct invariant factor. In this case, we have two distinct invariant factors, d6 and d7, so the rational canonical form of A will have two blocks. The Jordan canonical form of a matrix is a block diagonal matrix where each block corresponds to a distinct eigenvalue. In this case, we have three distinct eigenvalues, 0, 2, and 3, so the Jordan canonical form of A will have three blocks.

To find the rational canonical form of A, we need to find the corresponding matrices for each invariant factor. For d6, we have the matrix (λ-3)(λ-2)2, which corresponds to the block matrix A =
0 0 12
1 0 -16
0 1 7
For d7, we have the matrix (λ-3)2(λ-2)2, which corresponds to the block matrix B =
0 0 0 -36
1 0 0 -12
0 1 0 -13
0 0 1 10
Therefore, the rational canonical form of A will be a block diagonal matrix with A and B as its blocks.

To find the Jordan canonical form of A, we need to find the corresponding eigenvectors for each eigenvalue. For eigenvalue 0, we can find one eigenvector as (1,0,0). For eigenvalue 2, we can find two linearly independent eigenvectors as (0,1,0) and (0,0,1). For eigenvalue 3, we can find one eigenvector as (1,1,1). Therefore, the Jordan canonical form of A will be a block diagonal matrix with the following blocks:
[0]
[0 2]
[0 0 2]

I hope this helps clarify the rational and Jordan canonical forms of A. Let me know if you have any further questions. Keep up the good work in your studies!