# Matrix of linear operator

I have troubles with following problem, can you please help me?

Matrix of linear operator $f$ over the field $\mathbf{Z}_5^3$ to a canonical base is $A$. $$A= \left( \begin{array}{ccc} 3 & 1 & 4\\ 3 & 0 & 2 \\ 4 & 4 & 3 \end{array} \right)$$

Find base $B$ over the field $\mathbf{Z}_5^3$ so the matrix $f$ to $B$ was $[f]_B^B =$ $\left( \begin{array}{ccc} c & 1 & 0 \\ 0 & c & 0 \\ 0 & 0 & c \end{array} \right)$ for some $c \in \mathbf{Z}_5$.

Thanks.

• You're being asked to diagonalize $A$. Where is your problem occurring? Is it working over $\Bbb Z_5$? or diagonalization in general? or worry about whether the final diagonalized form will match what $B$ has to be? What have you tried? – Greg Martin Jan 4 '15 at 22:40
• There is a reason why one writes \begin{array}{ccc} with "{ccc}". It has to do with the reason why one uses alignment tabs. But you didn't use any alignment tabs, the "{ccc}" didn't serve its purpose. I fixed it. – Michael Hardy Jan 4 '15 at 23:22

## 1 Answer

Hint: We calculate the characteristic polynomial over $\Bbb Z_5$ to be $$\det\pmatrix{3-t&1&4\\3&-t&2\\4&4&3-t} = \\ (3-t)[t(t-3) - (2)(4)] - 1[3(3-t) - (2)(4)] + 4[3(4) + 4t]$$ Your $c$ must be the unique zero of this polynomial. Modulo $5$, we have $$(3-t)[t(t-3) - (2)(4)] - 1[3(3-t) - (2)(4)] + 4[3(4) + 4t] = \\ (3-t)[t^2 - 3t + 2] - [1 - 3t] - [2 - t] = \\ -t^3 + t^2 - t + 1 - [1 - 3t] - [2 - t] =\\ -t^3 + t^2 - 2t - 2$$ Now, there are $5$ possibilities. Check them.