Finding the Jordan canonical form of a matrix

[Mr Davis 97]

Homework Statement

Find the Jordan canonical form of the matrix $\left( \begin{array}{ccc} 1 & 1 \\ -1 & 3 \\ \end{array} \right)$.

Homework Equations

The Attempt at a Solution

So my professor gave us the following procedure:

1. Find the eigenvalues for each matrix A. Your characteristic polynomial must definitely SPLIT over $\mathbb{C}$, but mostly likely you will be able to split it already over $\mathbb{R}$.

2. For each eigenvalue $\lambda$ of A set the matrix $B=A-\lambda I$

3. Calculate several consecutive powers of $B=A-\lambda I$, e.g., $B^1$,$B^2$,$B^3$, etc., and check the ranks of these matrices. Set $N$ to be the first exponent where the rank stabilizes, i.e., $\text{rk}(B^N)=\text{rk}(B^{N+1}).$.

4. Find a basis for the kernel of $B^N$; i.e., a basis for the generalized eigenspace $K_{\lambda}(T)=\text{Ker}(A-\lambda I)^N$.

5. In each part, put together the bases you found for all $K_{\lambda}(T)$, verify that you have obtained a basis for the whole space V, and write down the matrix of T in this basis.So, we have the matrix $\left( \begin{array}{ccc} 1 & 1 \\ -1 & 3 \\ \end{array} \right)$. It's characteristic polynomial is $f(\lambda) = (\lambda - 2)^2$. So we have an eigenvalue of 2 with multiplicity 2.
Now, let $B=A - \lambda I = \left( \begin{array}{ccc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)$. This matrix has rank 1. Now, $B^2 = \left( \begin{array}{ccc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)$. So it has rank 0. We can see that $\text{rk}(B^2) = \text{rk}(B^3)$, so we must find a basis for the kernel of $B^2$.

This is where I can get confused. We got the zero matrix, so the kernel of the zero matrix is all of $\mathbb{R}^2$. So would we just choose any two linearly independent vectors? Or am I completely off-track and making a big mistake somewhere? I feel like I am following the procedure...

 

[ehild]

Homework Statement

Find the Jordan canonical form of the matrix $\left( \begin{array}{ccc} 1 & 1 \\ -1 & 3 \\ \end{array} \right)$.

Homework Equations

The Attempt at a Solution

So my professor gave us the following procedure:

1. Find the eigenvalues for each matrix A. Your characteristic polynomial must definitely SPLIT over $\mathbb{C}$, but mostly likely you will be able to split it already over $\mathbb{R}$.

2. For each eigenvalue $\lambda$ of A set the matrix $B=A-\lambda I$

3. Calculate several consecutive powers of $B=A-\lambda I$, e.g., $B^1$,$B^2$,$B^3$, etc., and check the ranks of these matrices. Set $N$ to be the first exponent where the rank stabilizes, i.e., $\text{rk}(B^N)=\text{rk}(B^{N+1}).$.

4. Find a basis for the kernel of $B^N$; i.e., a basis for the generalized eigenspace $K_{\lambda}(T)=\text{Ker}(A-\lambda I)^N$.

5. In each part, put together the bases you found for all $K_{\lambda}(T)$, verify that you have obtained a basis for the whole space V, and write down the matrix of T in this basis.So, we have the matrix $\left( \begin{array}{ccc} 1 & 1 \\ -1 & 3 \\ \end{array} \right)$. It's characteristic polynomial is $f(\lambda) = (\lambda - 2)^2$. So we have an eigenvalue of 2 with multiplicity 2.
Now, let $B=A - \lambda I = \left( \begin{array}{ccc} -1 & 1 \\ -1 & 1 \\ \end{array} \right)$. This matrix has rank 1. Now, $B^2 = \left( \begin{array}{ccc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)$. So it has rank 0. We can see that $\text{rk}(B^2) = \text{rk}(B^3)$, so we must find a basis for the kernel of $B^2$.

This is where I can get confused. We got the zero matrix, so the kernel of the zero matrix is all of $\mathbb{R}^2$. So would we just choose any two linearly independent vectors? Or am I completely off-track and making a big mistake somewhere? I feel like I am following the procedure...

You make it too complicated. What can be the Jordan form of a 2x2 matrix?

 

[nuuskur]

Do take note of what your professor told you, but always start investigating simpler alternatives. What do you know about a 2x2 JCF ? Are there any necessary conditions you can think of that it must satisfy?

 

[Mr Davis 97]

You make it too complicated. What can be the Jordan form of a 2x2 matrix?

The Jordan form is $\left( \begin{array}{ccc} 2 & 1 \\ 0 & 2 \\ \end{array} \right)$, right? But I am still confused about how this fits into the given procedure.

 

[ehild]

The Jordan form is $\left( \begin{array}{ccc} 2 & 1 \\ 0 & 2 \\ \end{array} \right)$, right? But I am still confused about how this fits into the given procedure.

The original matrix A can be transformed into the Jordan form by choosing an appropriate matrix S: S^-1AS=J. The first column of S is v, the eigenvector of A(or B) , the second one is a "generalized eigenvector", eigenvector of B^2 but not eigenvector of B.

 

[Mr Davis 97]

The original matrix A can be transformed into the Jordan form by choosing an appropriate matrix S: S^-1AS=J. The first column of S is v, the eigenvector of A(or B) , the second one is a base factor which is not eigenvector of A (B).

That makes sense. However, I am still confused about the method that the professor gave us. Is the method not working? Obviously, a basis for the kernel of $B^2$ does not give the correct basis vectors which yield the JCF

 

[ehild]

That makes sense. However, I am still confused about the method that the professor gave us. Is the method not working? Obviously, a basis for the kernel of $B^2$ does not give the correct basis vectors which yield the JCF

As B²=0 you can choose any pair of independent vectors as basis. So this method does not work. You have to choose the other base vector so as you get the Jordan form.
Try https://math.berkeley.edu/~ogus/old/Math_54-05/webfoils/jordan.pdf

 

[vela]

That makes sense. However, I am still confused about the method that the professor gave us. Is the method not working? Obviously, a basis for the kernel of $B^2$ does not give the correct basis vectors which yield the JCF

The procedure in the original post doesn't quite work. You want to find two vectors $\vec{v}_1$ and $\vec{v}_2$ that satisfy
\begin{align*}
B\vec{v}_1 &= 0 \\
B\vec{v}_2 &= \vec{v}_1
\end{align*}
$\vec{v}_1$ is just a regular eigenvector. Note that $B^2\vec{v}_2 = B\vec{v}_1 = 0$, so $\vec{v}_2$ is in the kernel of $B^2$.

If you want to find $\vec{v}_2$ first, it can't be any vector in the kernel of $B^2$. It also has to satisfy $B\vec{v}_2 \ne 0$. Let $\vec{v}_2 = (x,y)$. Then
$$B\vec{v}_2 = \begin{pmatrix} -x + y \\ -x+y \end{pmatrix}.$$ You can choose $x$ and $y$ arbitrarily as long as $x \ne y$. Then $\vec{v}_1 = B\vec{v}_2$.