Check diagonalizability of a matrix without using eigen properties For the matrix 
$$
A=\begin{bmatrix}0 & 1 & 0\\
 0 & 0 & 1 \\
 0 & 0  & 0
\end{bmatrix}
$$
How can we determine if $A$ is diagonalizable over $(a) \mathbb{F}^2 (b) \mathbb{Q} (c) \mathbb{R} (d) \mathbb{C}$ ?
My professor has not yet covered eigenvalues but only provided the property that $A$ is diagonalizable if there exists an invertible matrix $P$ such that $AP = PD$. For the in-class examples, he would write out $P$ and $D$ using general terms and from the consistency/inconsistency, determine if $A$ is diagonalizable. Doing so for above yields
$$
\begin{bmatrix}0 & 1 & 0\\
 0 & 0 & 1 \\
 0 & 0  & 0
\end{bmatrix} \begin{bmatrix}p_1 & p_2 & p_3\\
 p_4 & p_5 & p_6 \\
 p_7 & p_8  & p_9
\end{bmatrix} =  \begin{bmatrix}p_1 & p_2 & p_3\\
 p_4 & p_5 & p_6 \\
 p_7 & p_8  & p_9
\end{bmatrix} \begin{bmatrix}d_1 & 0 & 0\\
 0 & d_2 & 0 \\
 0 & 0  & d_3
\end{bmatrix}$$
We obtain multiple equations which all seem linked. Is there an easier method without using eigenvalue properties or symmetry ?
Then there is the question of the fields as well. How does the answer change with each field ?
(This is not homework -- question from a practice set)
 A: Note that $A^3=\mathbf{0}$, so the matrix $A$ is nilpotent, and if $A$ were diagonalizable, then we could find an invertible matrix $P$ such that $A=PDP^{-1}$. But this implies that $$\mathbf{0}=A^3=PD^3P^{-1}$$ and hence $D=\mathbf{0}$. However, this is impossible, because it would imply that $A=PDP^{-1}=\mathbf{0}$ which is false. 
As you can see in this particular case, diagonalizability is field independent. 
A: There a nice little notion which precisely addresses this question.  It is expressed in the following
Proposition:  Let $\Bbb F$ be any field, and let $0 \ne N \in M_{n \times n}(\Bbb F)$ be an $n \times n$ matrix with entries in $\Bbb F$.  If $N$ is nilpotent, i.e., $N^m = 0$ for some positive integer $m$, then $N$ cannot be diagonalized.
Proof:  For it were possible to diagonalize $N$, for some diagonal matrix $D \in M_{n \times n} (\Bbb F)$ we could write
$D = S^{-1}NS, \tag{1}$
and hence
$N = SDS^{-1}, \tag{2}$
whence
$0 = N^m = (SDS^{-1})^m = SD^mS^{-1}. \tag{3}$
Since $S$ is nonsingular, we have from (3) that
$D^m = 0; \tag{4}$
since $D$ is diagonal, (4) implies that $d^m = 0$ for any diagonal entry $d$ of $D$.  But $d \in \Bbb F$, so $d^m = 0$ forces $d =0$, whence we must have $D = 0$ and thus $N = 0$ by (2).  This contradicts our assumption that $N \ne 0$.  Thus $N$ cannot be diagonalized.  QED.
Applying this proposition to the case at hand, viz.
$A= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \tag{5}$
we have
$A^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \tag{6} $
and
$A^3 = 0. \tag{7}$
We see that $A$, being nilpotent, cannot be diagonalized over any field.
Hope this helps!  Cheers,
and as always,
Fiat Lux!!!
A: It's very hard to talk about if a matrix is diagonalizable without
talking about eigenvalues and eigenvectors.
I will just say that generally the answer to whether a matrix is diagonalizable
or not depends on the field because eigenvalues are the roots of some
polynomial (called the characteristic polynomial of the matrix) and
so the number of roots depends on the field, there is a strong relation
between these roots and being diagonalizable.
Now, for this question, you might not have the tools right now, but
I'm sure you will learn them in a short time - if $A$ is upper triangular
then its eigenvalues are its diagonal entries, in your case all the
eigenvalues are $0$. 
The eigenvalues correspond to the diagonal of
$D$ when $A$ is diagonalizable so in your case if $A$ is diagonalizable
then there is an invertible $P$ s.t $$P^{-1}AP=D=0$$ but then $A=0$
which is a contradiction so this matrix is not diagonalizable over
any field
