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I can see that the standard base is not of nilpotent element (the elements on the diagonal are not), but can't prove that proposition, or be sure it is even true.

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What standard basis? –  Tobias Kildetoft Jan 16 '13 at 14:50
    
by standard basis I mean the matrices with 1 at index i,j, and 0 anywhere else –  cruvadom Jan 16 '13 at 14:51
    
I somehow missed the matrix part. Please edit the question so that the full question is there and not just in the title. –  Tobias Kildetoft Jan 16 '13 at 14:55
    
Do you mean $M_n(\mathbb{F})$, the algebra of all matrices over $\mathbb{F}$? Because "matrix algebra" usually allows subalgebras of these, and in some of these subalgebras (such as those consisting of strictly upper triangular matrices), all elements are nilpotent. –  Matt Pressland Jan 16 '13 at 14:55
    
@CalvinLin I think it's pretty clear: can you find $n^2$ linearly independent nilpotent matrices that span $M_n(F)$? –  rschwieb Jan 16 '13 at 14:57
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up vote 3 down vote accepted

"No" because of the following hints I'm giving you here.

The trace of a nilpotent matrix is zero. What do you know about the traces of $A+B$ and $\lambda A$ for matrices $A,B$ and scalar $\lambda$?

This should make it clear that you won't be able to generate just any matrix with a linear combination of nilpotent matrices.

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thanks. even easier: one element of the basis must have a non zero field element on the diagonal, so it's not nilpotent –  cruvadom Jan 17 '13 at 9:49
    
@user57076 No, that is false logic. The matrix $\begin{bmatrix}1&-1\\1&-1\end{bmatrix}$ is nilpotent. –  rschwieb Jan 17 '13 at 11:56
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