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)

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.

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!!!

• This is essentially a copy of my post above...
– EPS
Sep 5 '14 at 3:20
• @Sam: I think your answer is a good one, so I upvoted it. Having said that, your comment above is simply NOT A TRUE STATEMENT! Certainly you are aware of the fact that a given mathematical idea may take on many forms of espression: do you say Rudin copies Ahlfors because both their books contain the Cauchy residue theorem? Think on it, if you would. Regards. Sep 5 '14 at 5:25
• Thanks for the upvote @Robert, but I just wanted to let you know that the residue theorem being contained in different books has nothing to do with my comment. If the same theorem is proved in the same book on different pages with essentially the same method, yes, I would say that those two pages are "copies" of each other. You might disagree; but that's a matter of personal opinion I guess. Peace.
– EPS
Sep 6 '14 at 2:14
• Peace backatcha, Sam, and sorry for being so grouchy. Cheers! Sep 6 '14 at 3:02

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