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We were talking about how the symmetric matrix

$$A=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

is not diagonalizable in the field consisting of only $0$ and $1$, since the eigenvalues are $0$ and $2$, but in this field $0=2$, and the two eigenvectors are the same

$$\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

and I was able to find the jordan normal form of the matrix being $$J = P^{-1} A P$$

$J=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, $P^{-1}=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, $A=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, $P=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$

since again in the field, $0=2=4=\dots$ and $-1=1=3=\dots$.

However I need to find another symmetric matrix in the same field that is also not diagonalizable, and also find its jordan normal form. I'm having a hard time with looking for this matrix.

The hint also said that $$ \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \quad\text{ or }\quad \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} $$ wouldnt work, even though they have no eigenvalues in ${\Bbb F}_2$; but they have two distinct eigenvalues in ${\Bbb F}_4$. so they are diagonalizable in ${\Bbb F}_4$.

What is ${\Bbb F}_4$? Is it a finite field that has only $0,1,2,3$? If that's the case, the eigenvalues for \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} are not even integers, how are they in ${\Bbb F}_4$?

Thanks in advance!

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$F_4$ is quite different from $\mathbb Z/4 \Bbb Z$, which is not a field. In $F_4$ we have elements $0$, $1$, $\alpha$, and $\alpha+1$, where $\alpha$ is a root of $f(t)=t^2+t+1$ (working over $F_2$), which just happens to be the characteristic polynomial of your matrix.

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In general, for the existence of the Jordan normal form you need all eigenvalues to be in the field you're working with.

For the case of $2\times2$ matrices we know that if the eigenvalues are different than the matrix is diagonalizable, so we want both eigenvalues to be the same.

Above the diagonal (or below it, depends on how you take the normal form to be) there is either $1$ or a $0$, since the normal form is already an upper triangular matrix, if the entry above the diagonal is $0$ you get a diagonal matrix - so the entry above the diagonal must be $1$.

On the diagonal you must have the eigenvalues, which are the same, hence both are $0$ or both are $1$.

You already gave an example where both are $0$ so this leaves you with the second option.

So: $$ J=\begin{pmatrix}1 & 1\\ 0 & 1 \end{pmatrix} $$

This matrix is similir to the matrix you want, find an invertiable $P$ s.t $P^{-1}JP$ is symmetric, and you are done

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As $A^2=0$, its JNF must be $\begin{pmatrix}0&1\\0&0\end{pmatrix}$.


Since symmetric matrices that aren't already in diagonal form look like $\begin{pmatrix}a&1\\1&d\end{pmatrix}$ and we've already had $a=d=1$ above and the hint says that $a=1, d=0$ and $a=0,d=1$ are no good choices, we are only left with $a=d=0$, i.e. $$ A=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$ It has $\begin{pmatrix}1\\1\end{pmatrix}$ as eigenvector with eigenvalue $\lambda=1$. And as $A-\lambda=\begin{pmatrix}1&1\\1&1\end{pmatrix}\ne0$, the JNF must be $$\begin{pmatrix}1&1\\0&1\end{pmatrix}.$$ Of course, we could have said right away that the only non-diagonal JNF's are $\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ with $\lambda\in\{0,1\}$.


$\mathbb F_4$ is a field with four elelements and contains $\mathbb F_2$ as a subfield. Especially, the equality $1+1=0$ that holds in $\mathbb F_2$ still holds in $\mathbb F_4$, which means that $\mathbb F_4$ is not simply $\mathbb Z/4\mathbb Z$. As a field, $\mathbb F_4$ must have an element $0$ and an element $1$. To be bigger than $\mathbb F_2$ it must contain some additional element $x$ and to allow addition, also an element $1+x$ (which is $\ne0$ because $x=x+0=x+(1+1)=(x+1)+1\ne1$, it is $\ne 1$ because $x\ne0$ and it is $\ne x$ becasue $1\ne0$). Then from $1+1=0$ we obtain $x+x=x(1+1)=x0=0$. What is $x\cdot x$? It can't be $0$ in a field. It can't be $1$, because the only solutions of $X^2-1=0$ in a field are $\pm1$ (which is just $1$ because $+1=-1$). It can't be $x$ because $x\cdot x=x$ implies $x(x-1)=0$, hence $x\in\{0,1\}$. We conclude that $x^2=x+1$. This simple rule $x^2=x+1$ allows you to write down the complete multiplication table for $\mathbb F_4=\{0,1,x,1+x\}$.

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