Jordan Canonical form problem

**tttcomrader** · April 23rd 2008, 02:02 PM

Find the Jordan Canonical form of the following matrice:

$\begin{bmatrix} 1 && 2 \\ 3 && 2 \end{bmatrix}$

For this one, here is what I have so far:

First, I found the eigenvalues are 1 and 2, and the basis of eigenvectors are $\beta = \{ (- \frac {1}{3} ,1 ) , ( 1 , ( \frac {1}{2} ) \}$

Now, I know I have to find $[T] _{ \beta }$ , but I'm bit rusty on this, I can't remember how to do it, any hint, please?

Thanks.

**Opalg** · April 24th 2008, 12:20 AM

Originally Posted by tttcomrader

Find the Jordan Canonical form of the following matrice:

$\begin{bmatrix} 1 && 2 \\ 3 && 2 \end{bmatrix}$

For this one, here is what I have so far:

First, I found the eigenvalues are 1 and 2, and the basis of eigenvectors are $\beta = \{ (- \frac {1}{3} ,1 ) , ( 1 , ( \frac {1}{2} ) \}$

Now, I know I have to find $[T] _{ \beta }$ , but I'm bit rusty on this, I can't remember how to do it, any hint, please?

Thanks.

I'm not sure where this has gone wrong, but the eigenvalues are 4 and –1, and the eigenvectors are (multiples of) (2,–3) and (2,3).

Since the eigenvalues are distinct, the Jordan form is the diagonal matrix whose diagonal entries are the eigenvalues: $\begin{bmatrix}4&0\\0&-1\end{bmatrix}$ .

**tttcomrader** · April 24th 2008, 09:01 AM

Thanks, I was being lazy and made careless mistake here.

The next one is: $A= \begin{bmatrix} 2 && 1 && 0 && 0 \\ 0 && 2 && 1 && 0 \\ 0 && 0 && 3 && 0 \\ 0 && 1 && -1 && 3 \end{bmatrix}$

So I have $f(t) = det (A - t I_{4} ) = (2-t)(3-t)[(2-t)(3-t)]$

So the eigenvalues are 2 and 3 with multiplty of 2 each. How would I compute that? thanks.

**Opalg** · April 24th 2008, 10:25 AM

Originally Posted by tttcomrader

Thanks, I was being lazy and made careless mistake here.

The next one is: $A= \begin{bmatrix} 2 && 1 && 0 && 0 \\ 0 && 2 && 1 && 0 \\ 0 && 0 && 3 && 0 \\ 0 && 1 && -1 && 3 \end{bmatrix}$

So I have $f(t) = det (A - t I_{4} ) = (2-t)(3-t)[(2-t)(3-t)]$

So the eigenvalues are 2 and 3 with multiplty of 2 each. How would I compute that? thanks.

The rule is that if an eigenvalue λ occurs with multiplicity 2, you have to find the dimension of the corresponding eigenspace (in other words, how many linearly independent eigenvectors are there for that eignvalue). The dimension of the eigenspace will be either 1 or 2. If it is 2, then the Jordan canonical form corresponding to that eigenvalue just consists of two λs on the diagonal. If it is 1 then the canonical form has a 2×2 block $\begin{bmatrix}\lambda&1\\0&\lambda\end{bmatrix}$ .

Fot the above matrix A, each eigenvalue has an eigenspace of dimension 1 (an eigenvector for the eigenvalue 2 must be a multiple of (1,0,0,0) and an eigenvector for the eigenvalue 3 must be a multiple of (0,0,0,1)). So the JCF looks like $\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}$ .

**tttcomrader** · April 25th 2008, 05:29 AM

Thanks, but I cannot understand why the the eigenvectors are such, let me show you how I worked mine, please correct me.

For a)

The eigenvalues are -1 and 4, so the eigenspace for -1 is:

$E_{ \lambda _{1}} = \{ \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in \mathbb {R} ^2 | \begin{bmatrix} 1 - (-1) && 2 \\ 3 && 2 - (-1) \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \}$

So I have to solve $\begin{bmatrix} 2 && 2 && 0 \\ 3 && 3 && 0 \end{bmatrix} = \begin{bmatrix} 1 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

Let $x_2 = t$ , we have $x_1 = -t$ , so the eigenvectors for this is $t(-1,1)$

But appearly it is not, what am I doing wrong here?

**Opalg** · April 25th 2008, 05:49 AM

Originally Posted by tttcomrader

Thanks, but I cannot understand why the the eigenvectors are such, let me show you how I worked mine, please correct me.

For a)

The eigenvalues are -1 and 4, so the eigenspace for -1 is:

$E_{ \lambda _{1}} = \{ \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} \in \mathbb {R} ^2 | \begin{bmatrix} 1 - (-1) && 2 \\ 3 && 2 - (-1) \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \}$

So I have to solve $\begin{bmatrix} 2 && 2 && 0 \\ 3 && 3 && 0 \end{bmatrix} = \begin{bmatrix} 1 && 1 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

Let $x_2 = t$ , we have $x_1 = -t$ , so the eigenvectors for this is $t(-1,1)$

But appearly it is not, what am I doing wrong here?

Sorry about that. You're right and I was wrong. The eigenvector for the eigenvalue –1 is (–1,1) (or any nonzero multiple of that). For the eigenvalue 4 you should get (2,3).

**tttcomrader** · April 25th 2008, 09:19 AM

For eigenvalue = 4, I ended up with this:

$\begin{bmatrix} -3 && 2 && 0 \\ 3 && -2 && 0 \end{bmatrix} = \begin{bmatrix} -3 && 2 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

Let $x_2 = s$ , then I have eigenvector $s( \frac {3}{2} , 1 )$

So did I do something wrong in this one? Thanks.

**Opalg** · April 25th 2008, 10:27 AM

Originally Posted by tttcomrader

For eigenvalue = 4, I ended up with this:

$\begin{bmatrix} -3 && 2 && 0 \\ 3 && -2 && 0 \end{bmatrix} = \begin{bmatrix} -3 && 2 && 0 \\ 0 && 0 && 0 \end{bmatrix}$

Let $x_2 = s$ , then I have eigenvector $s( \frac {3}{2} , 1 )$

So did I do something wrong in this one? Thanks.

You'd be surprised how many students make that mistake.

If $x_1=r$ and $x_2 = s$ , then the equation is $-3r+2s=0$ . So $r = \textstyle \frac23s$ , and the eigenvector is $s(\textstyle \frac23,1)$ . I prefer to take a scalar multiple of that, to avoid the fraction, which is why I gave the answer as $(2,3)$ .

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