# Can the matrix $A=\begin{bmatrix} 0 & 1\\ 3 & 3 \end{bmatrix}$ be diagonalized over $\mathbb{Z}_5$?

Given matrix: $$A= \left[\begin{matrix} 0 & 1\\ 3 & 3 \end{matrix}\right]$$

whose entries are from $\mathbb{Z}_5 = \{0, 1, 2, 3, 4\}$, find, if possible, matrices $P$ and $D$ over $\mathbb{Z}_5$ such that $P^{−1} AP = D$.

I have found the characteristic polynomial: $x^2-3x-3=0$ Since its over $\mathbb{Z}_5$, $x^2-3x-3=x^2+2x+2=0$.

But from there I'm not sure how to find the eigenvalues, once I get the eigenvalues that are in the field it will be easy to find the eigenvectors and create the matrix $P$.

-
Your field has $5$ elements, just try everyone to check which ones are roots of the polynomial. Note, however, that the field in question isn't algebraically closed, so it might even not have roots there. –  Git Gud Apr 3 '13 at 20:30
Please see this about how to format equations on this site. In particular, I assume P-1AP=D is supposed to mean $P^{-1}AP=D$ but there is no indication of the power in what you wrote. –  Ross Millikan Apr 3 '13 at 20:32
@user547866: It really helps readability if you format your questions using MathJax (see FAQ). Regards –  Amzoti Apr 3 '13 at 20:33
@user547866 They don't have to be in the field and if they aren't, the matrix isn't diagonalizable over that field. –  Git Gud Apr 3 '13 at 20:57
@user547866 This might interest you. –  Git Gud Apr 3 '13 at 21:08

Hint: $x^2-3x-3\equiv x^2-3x+2$.

-
@user547866 What do you mean? The eigenvectors are $(1,1)^T$ (for $\lambda=1$) and $(1,2)^T$ (for $\lambda=2$). The matrix is happily diagonalisable. Actually, since the matrix has distinct eigenvalues, it must be diagonalisable. –  user1551 Apr 3 '13 at 21:52

yes over $\Bbb Z_5$ because: $\lambda^2 -3\lambda-3=o$ at Z_5 we will have $\Delta=9+12=4+2=6$ (9~4 and 12~2 at Z_5) so $\Delta=1$ and so $\lambda_1=\frac{3+1}{2}=2$ and $\lambda_2=\frac{3-1}{2}=1$

about: $\lambda_1$ we have :$( \left[\begin{matrix} 0 & 1\\ 3 &3 \end{matrix}\right]-\left[\begin{matrix} 2 & \\ 0 &2 \end{matrix}\right] )\left[\begin{matrix} x\\ y \end{matrix}\right]=0$

$$-2x+y=0$$ & $$( 3x+y=0 ~ -2x+y=0 )$$ and so $$y=2x$$ is our space of eigen value of $\lambda_1 =\{(2,4),(0,0)(1,2)\}$ => (dim =1) base={(1,2)}

about $\lambda_2$:

$( \left[\begin{matrix} 0 & 1\\ 3 &3 \end{matrix}\right]-\left[\begin{matrix} 1 & 0\\ 0 &1 \end{matrix}\right] )\left[\begin{matrix} x\\ y \end{matrix}\right]=0$ and so $y=x$ is our answer and eigenvector space of $\lambda_2=\{(0,0)(1,1)(2,2)(3,3)(4,4)\} \implies$ $(\dim=1)$ base ={(1,1)}

matrix at base of$\{(1,1),(1,2)\}$ will be diagonalizable $\left[\begin{matrix} 2 & 0\\ 0 &1 \end{matrix}\right]$

-