Matrix element of a field

**Juneu436** · September 23rd 2011, 07:20 PM

Hi, I am not 100% sure my working is correct for this question. Could someone be kind enough to check my working please.

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

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

The characteristic polynomial of A:

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

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

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

to solve for x and y for both eigenvalues.

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

Am I correct in my working out so far?

Thanks

**FernandoRevilla** · September 23rd 2011, 09:19 PM

Originally Posted by Juneu436

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

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

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

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

**Juneu436** · September 23rd 2011, 10:10 PM

Originally Posted by FernandoRevilla

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

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

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

Oh yes, 3 isn't an element of $F_3$ , 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

**FernandoRevilla** · September 24th 2011, 12:24 AM

Originally Posted by Juneu436

Why is -i equivalent to 2i? I know it is a stupid question.

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

Then how do I work out the eigenvectors?

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

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

**Juneu436** · September 24th 2011, 05:51 PM

Originally Posted by FernandoRevilla

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

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

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.

**FernandoRevilla** · September 24th 2011, 09:50 PM

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.

**Juneu436** · September 24th 2011, 10:46 PM

Originally Posted by FernandoRevilla

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

This section:

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

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

Thanks again for your help.

**FernandoRevilla** · September 24th 2011, 11:00 PM

Originally Posted by Juneu436

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

There is a well known theorem: if $\lambda$ is a simple eigenvalue then, the corresponding eigenspace $V_{\lambda}$ has dimension 1. This means that a non null vector of $V_{\lambda}$ form a basis of $V_{\lambda}$ (because is linearly independent). In our case, and choosing (for example) $x_1=1$ we obtain from the first equation that $x_2=1+i$ . You needn't verify the second equation because necessarily the rank of the system matrix is 1.

**Deveno** · September 25th 2011, 04:29 AM

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:

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

that is:

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

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

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.

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