I would like to prove that four $2\times 2$ nilpotent matrices are always linearly dependent, using the Cayley-Hamilton theorem or the minimal polynomial in some way.

I think I have proved the statement using a "brute force" method, wherein I just squared every possible $2\times 2$ matrix (there's $16$ different kinds) to see if it vanished. I concluded that the only nilpotent $2\times 2$ matrices are $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$ and any other nilpotent $2\times 2$ matrix is a multiple of one of these. So actually, any three nilpotent matrices are always linearly dependent.

I would like to construct something more "sophisticated". I am also not quite sure my "brute force" approach is $100\%$ correct, anyway!

  • $\begingroup$ What about $\pmatrix{-1&1\\-1&1}$? $\endgroup$ – Omnomnomnom Dec 19 '14 at 14:13
  • $\begingroup$ Yes indeed, a good counter-example to my claims! $\endgroup$ – pow Dec 19 '14 at 15:27

Notice that the trace of any nilpotent matrix is $0$. Using that the map

$$\operatorname{tr}:\mathcal M_2(\Bbb F)\to \Bbb F$$ is a linear form then $\ker \operatorname{tr}$ is an hyperplan of $\mathcal M_2(\Bbb F)$ containing the set of nilpotent matrices. Hence $4$ nilpotent matrices doesn't form a basis of $\mathcal M_2(\Bbb F)$ hence they are linearly dependent.

  • $\begingroup$ Perhaps "form" is used some where, but I think the more common term is "linear functional" . Anyway, +1 $\endgroup$ – Timbuc Dec 19 '14 at 14:10
  • $\begingroup$ Thanks for your comment but I don't agree with it. See here $\endgroup$ – user63181 Dec 19 '14 at 14:13
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    $\begingroup$ For what is worth: the books in linear algebra by Treil, Golan, Kostrikin-Manin, Lang, Friedberg-Insel-Spence, Roman, Shilov, Hoffman-Kunze, etc. All thee agree with "linear functional'. Some other books call them "linear functions". I couldn't find one single book with the name "linear form" for these things. The only forms I could find are the usual ones: quadratic, bilinear ,symmetric and etc. $\endgroup$ – Timbuc Dec 19 '14 at 14:43
  • $\begingroup$ Do you insist on your opinion even I gave you the above link?! I didn't say that we can't call it linear functional but I said that we can call it also linear form and this is in fact the mostly used in linear algebra (at least in the vocabularies often used in French books). For the books you mentioned I just verified the book algebra of Serge Lang and I found (as I knew) that he used both vocabularies; see for example page 411 of this book. $\endgroup$ – user63181 Dec 19 '14 at 15:15
  • $\begingroup$ No, you didn't say anything about also , and I only remarked that the most usual term, imo, is functional. You said you don't agree with that, and I gave you several examples, some of them very well known linear algebra texts, where it is called "functional". Can you give me several examples where such a linear map is called "linear form" besides Wiki? So far functional is usued in several, form in none. Now, I have both Lang's Algebra 2nd edition and 3rd revised edition: in none I could find anything on page 411. Perhaps you can refer me to chapter, section, etc.? $\endgroup$ – Timbuc Dec 19 '14 at 15:21

For any nilpotent 2x2 matrix $P$ we have $P^2=0$.

Now consider the space $\Bbb R[X]/(X^2)$. It has dimension two and it isomorphic to the space of nilpotents 2x2 matrices.

  • $\begingroup$ If I am not mistaken, The $3$ matrices $$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \, \,\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\,\,\begin{bmatrix} -1 & 1 \\ -1 & 1 \end{bmatrix}$$ are nilpotent and linearly independant. $\endgroup$ – Kelenner Dec 19 '14 at 13:26
  • $\begingroup$ So what type of structure do you claim that the space of nilpotent matrices has? Do you claim that it is a ring? And ideal? $\endgroup$ – Omnomnomnom Dec 19 '14 at 14:11
  • $\begingroup$ The set of nilpotent matrices is not a subspace as it isn't closed under addition: $$\begin{pmatrix}0&1\\0&0\end{pmatrix}+\begin{pmatrix}1&-1\\1&-1\end{pmatrix}= \begin{pmatrix}1&0\\1&-1\end{pmatrix}$$ $\endgroup$ – Timbuc Dec 19 '14 at 14:14
  • $\begingroup$ Thank you for your comments and the counter-example! Sorry for the tardy answer. I am not well versed in abstract algebra and went to get a book on the subject. So $\Bbb R[X]/(X^2)$ is the space of polynomials in $x$, with coefficients in $\Bbb R$, modulo $X^2$. The basis is then on the form $a$, $bx$ for $a,b \in \Bbb R$. The significance of this space being isomorphic to the space of $2\times 2$ matrices is lost on me though. Does it mean, that as you add the nilpotent matrices, the minimal polynomial of the sum is not necessarily $X^2$ (as it could be $a$, $bx$, or $a+bx$)? $\endgroup$ – pow Dec 19 '14 at 15:21
  • $\begingroup$ @pow our comments were addressed at the answerer, ajotatxe, who should have explained his thoughts more clearly. It is not at all clear what is meant by the purported isomorphism. $\endgroup$ – Omnomnomnom Dec 19 '14 at 17:55

There are more nilpotent matrices than that. For example, if $A$ is nilpotent and $B$ is invertible, then $$\left(BAB^{-1}\right)^2=BAB^{-1}BAB=BA^2B^{-1}=0$$


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