Jordan Normal Form / Jordan basis

To find the Jordan normal form, we need to calculate the matrix M = B^-1AB. This will give us a diagonal matrix with the eigenvalues on the diagonal and 1's and 0's in the upper and lower triangles.M = [1 0 0; 0 1 0; 0 0 1]So, the Jordan normal form for matrix A is:J = [1 1 0; 0 1 1; 0 0 1]In summary, to find the Jordan normal form and a Jordan basis for the matrix A, we first found the eigenvalue t=1 and the corresponding eigenvector x=[3; 2;
  • #1
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Homework Statement



Determine the Jordan Normal form and find some Jordan basis of the matrix
3 -3 1
A = 2 -2 1
2 -3 2

Homework Equations



dim(A) = rk(A) + dimKer(A

The Attempt at a Solution



My problem here is that my lecturer seems to be doing completely different things with every question and it's getting confusing.

So, I calculated the characteristic polynomial of the matrix, and got one
eigenvalue of t = 1.

So I'm now dealing with the matrix A - tI, in this case A - I.

rk(A - I) = 1, so dimKer(A - I) = 2.

(A - I)^2 = 0 = B

So rk(B) = 0, then dimKer(B) = 3

Continually raising the powers of A - I will results in 0, so the kernels of powers stabilize at the second step, so we should expect a thread of length 2.

The kernel of A - I is spanned by the vectors ( 3/2, 1, 0) and (-1/2, 0, 1)
I should be using columns instead of rows, but I don't know latex so this was the easiest way to write it. Just imagine they were written as columns..Here's the main issue, when I find out what vectors span the kernel, and use those as columns of a new matrix, I then reduce that matrix to Reduced Column Echelon Form.

That's fine, but, in some of my lecturers examples he takes a vector corresponding to the missing leading one as a basis, and it others he takes one of the columns of the RCEF.

My question: why the differences and does it matter which of the columns are taken?I'm lost. Can you please explain what to do, step by step?

Thank you!
 
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  • #2


Hello,

First of all, great job on calculating the characteristic polynomial and finding the eigenvalue t=1. That's a great start!

Now, let's focus on finding the Jordan normal form and a Jordan basis for the matrix A.

Step 1: Find the eigenvalues and their corresponding eigenvectors.

You have already found one eigenvalue t=1, so let's focus on finding the corresponding eigenvectors. To do this, we need to solve the equation (A-I)x=0, where I is the identity matrix and x is the eigenvector.

Substituting t=1 into the matrix A-I, we get:

A-I = [2 -3 1; 2 -3 1; 2 -3 1]

Reducing this matrix to row echelon form, we get:

A-I = [1 -3/2 1/2; 0 0 0; 0 0 0]

This means that the only non-zero row in the reduced matrix corresponds to the eigenvector x = [3; 2; 2]. So, the eigenvector corresponding to the eigenvalue t=1 is x = [3; 2; 2].

Step 2: Find the generalized eigenvectors.

Since the rank of A-I is 1, we know that there will be one generalized eigenvector corresponding to the eigenvalue t=1. To find this eigenvector, we need to solve the equation (A-I)^2x=0.

Substituting t=1 into (A-I)^2, we get:

(A-I)^2 = [0 0 0; 0 0 0; 0 0 0]

Reducing this matrix to row echelon form, we get:

(A-I)^2 = [0 0 0; 0 0 0; 0 0 0]

This means that the only non-zero row in the reduced matrix corresponds to the generalized eigenvector y = [1; 0; -1].

Step 3: Constructing the Jordan basis.

To construct the Jordan basis, we need to combine the eigenvector x and the generalized eigenvector y into a single matrix. This matrix will be our basis for the Jordan normal form.

B = [3 1; 2 0; 2 -1]

Step 4: Finding the Jordan
 

What is Jordan Normal Form?

Jordan Normal Form is a way to represent a square matrix in a specific form, where the matrix is made up of diagonal blocks called Jordan blocks.

How is Jordan Normal Form different from diagonalization?

Jordan Normal Form is different from diagonalization because it allows for the presence of Jordan blocks, which are non-diagonalizable matrices. Diagonalization only works for diagonalizable matrices.

What is a Jordan basis?

A Jordan basis is a set of linearly independent vectors that form the columns of a matrix that is in Jordan Normal Form.

Can any square matrix be transformed into Jordan Normal Form?

No, not all square matrices can be transformed into Jordan Normal Form. Only matrices that have a complete set of eigenvectors can be transformed into Jordan Normal Form.

Why is Jordan Normal Form useful?

Jordan Normal Form is useful because it allows for a simpler representation of matrices, making it easier to perform calculations and understand the properties of the matrix. It is also useful in solving systems of differential equations and in other areas of mathematics and physics.

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