# Matrix element of a field

• Sep 23rd 2011, 07:20 PM
Juneu436
Matrix element of a field
Hi, I am not 100% sure my working is correct for this question. Could someone be kind enough to check my working please.

$\displaystyle F= \{a+bi | a,b \in F_3 \}$

$\displaystyle A=\dbinom{2 \ \ 1}{1 \ \ 1} \in F^{2x2}$

The characteristic polynomial of A:

$\displaystyle det \dbinom{2-\lambda \ \ \ \ 1}{1 \ \ \ \ \ 1-\lambda}= \lambda -3\lambda +1$

So the eigenvalues are $\displaystyle \lambda_1 = \frac{-(\sqrt{5}-3)}{2}$ and $\displaystyle \lambda_2 = \frac{\sqrt{5}+3}{2}$.

To find the eigenvectors we sub in $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ into the equations:
$\displaystyle (2-\lambda)x+y=0$
$\displaystyle x+(1-\lambda)y=0$

to solve for x and y for both eigenvalues.

So for example, for $\displaystyle \lambda_2$ I get $\displaystyle y=\frac{3(\sqrt{5}+2)}{2}$ and $\displaystyle x=\frac{3(6\sqrt{5}+13)}{4}$.

Am I correct in my working out so far?

Thanks
• Sep 23rd 2011, 09:19 PM
FernandoRevilla
Re: Matrix element of a field
Quote:

Originally Posted by Juneu436
$\displaystyle F= \{a+bi | a,b \in F_3 \}$ $\displaystyle A=\dbinom{2 \ \ 1}{1 \ \ 1} \in F^{2x2}$ The characteristic polynomial of A: $\displaystyle det \dbinom{2-\lambda \ \ \ \ 1}{1 \ \ \ \ \ 1-\lambda}= \lambda -3\lambda +1$ So the eigenvalues are $\displaystyle \lambda_1 = \frac{-(\sqrt{5}-3)}{2}$ and $\displaystyle \lambda_2 = \frac{\sqrt{5}+3}{2}$.

I suppose $\displaystyle F_3=\mathbb{Z}/(3)=\{0,1,2\}$ . Then, the characteristic polynomial of $\displaystyle A$ is

$\displaystyle \chi(\lambda)=\lambda ^2-(2+1)\lambda +1=\lambda^2-0\lambda +1=\lambda^2+1$

so, the eigenvalues of $\displaystyle A$ are $\displaystyle 0+i,0-i$ or equivalently $\displaystyle i,2i$ .
• Sep 23rd 2011, 10:10 PM
Juneu436
Re: Matrix element of a field
Quote:

Originally Posted by FernandoRevilla
I suppose $\displaystyle F_3=\mathbb{Z}/(3)=\{0,1,2\}$ . Then, the characteristic polynomial of $\displaystyle A$ is

$\displaystyle \chi(\lambda)=\lambda ^2-(2+1)\lambda +1=\lambda^2-0\lambda +1=\lambda^2+1$

so, the eigenvalues of $\displaystyle A$ are $\displaystyle 0+i,0-i$ or equivalently $\displaystyle i,2i$ .

Oh yes, 3 isn't an element of $\displaystyle F_3$, thanks for that.

Why is -i equivalent to 2i? I know it is a stupid question.
Then how do I work out the eigenvectors?

Thanks
• Sep 24th 2011, 12:24 AM
FernandoRevilla
Re: Matrix element of a field
Quote:

Originally Posted by Juneu436
Why is -i equivalent to 2i? I know it is a stupid question.

Because in $\displaystyle F_3$ we have $\displaystyle 1+2=0$ so, $\displaystyle 2=-1$ .

Quote:

Then how do I work out the eigenvectors?
For example, if $\displaystyle \lambda=i$ then $\displaystyle \ker (A-iI)\equiv \begin{Bmatrix} (2-i)x_1+x_2=0\\x_1+(1-i)x_2=0\end{matrix}$

As $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .
• Sep 24th 2011, 05:51 PM
Juneu436
Re: Matrix element of a field
Quote:

Originally Posted by FernandoRevilla
For example, if $\displaystyle \lambda=i$ then $\displaystyle \ker (A-iI)\equiv \begin{Bmatrix} (2-i)x_1+x_2=0\\x_1+(1-i)x_2=0\end{matrix}$

As $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .

Thanks FernandoRevilla, but I don't really understand this. Could you please explain?
I can't see how to get the eigenvectors.

Thanks, and sorry for being a pain.
• Sep 24th 2011, 09:50 PM
FernandoRevilla
Re: Matrix element of a field
Quote:

Originally Posted by Juneu436
Thanks FernandoRevilla, but I don't really understand this. Could you please explain? I can't see how to get the eigenvectors.

Could you specify your doubts?. These things are just routine knowing the corresponding theory.
• Sep 24th 2011, 10:46 PM
Juneu436
Re: Matrix element of a field
Quote:

Originally Posted by FernandoRevilla
Could you specify your doubts?. These things are just routine knowing the corresponding theory.

This section:
Quote:

As $\displaystyle i$ is simple, $\displaystyle \dim \ker (A-iI)=1$ . Solving you'll find a basis, for example $\displaystyle B_i=\{(1,1+i)\}$ .
How does $\displaystyle \dim \ker(A-iI)=1$ and how do I go from there to find the eigenvectors?

• Sep 24th 2011, 11:00 PM
FernandoRevilla
Re: Matrix element of a field
Quote:

Originally Posted by Juneu436
How does $\displaystyle \dim \ker(A-iI)=1$ and how do I go from there to find the eigenvectors?

There is a well known theorem: if $\displaystyle \lambda$ is a simple eigenvalue then, the corresponding eigenspace $\displaystyle V_{\lambda}$ has dimension 1. This means that a non null vector of $\displaystyle V_{\lambda}$ form a basis of $\displaystyle V_{\lambda}$ (because is linearly independent). In our case, and choosing (for example) $\displaystyle x_1=1$ we obtain from the first equation that $\displaystyle x_2=1+i$ . You needn't verify the second equation because necessarily the rank of the system matrix is 1.
• Sep 25th 2011, 04:29 AM
Deveno
Re: Matrix element of a field
or you could just solve the system Av = λv for each eigenvalue λ.

for example, with λ = i, and writing a+bi = z, c+di = w this leads to:

$\displaystyle \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}z\\w\end{bmatrix} = \begin{bmatrix}iz\\iw\end{bmatrix}$

that is:

$\displaystyle \begin{bmatrix}2z+w\\z+w \end{bmatrix} = \begin{bmatrix}iz\\iw\end{bmatrix}$ so:

$\displaystyle z = iz - iw$
$\displaystyle z(1+2i) = 2iw$
$\displaystyle z(1+2i)(i) = -2w = w$
$\displaystyle w = (-2+i)z = (1+i)z$

so any eigenvector corresponding to i is of the form (z,(1+i)z)...z, of course, must be non-zero. it is convenient to choose z = 1+0i = 1.