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Consider the vector space $M_{n\times n}(\mathbb F)$, the set for all $n\times n$ matrices. The basis for this vector space has dimension $n^2$. I read somewhere that the basis has $n$ matrices that have nonzero diagonal entries. I am trying to understand this intuitively, can anyone help clarify why there are $n$ of these elements?

Thanks.

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There is not just one single basis for $M_{n\times n}(\mathbb F)$, nor any distinguished one. There are infinitely many such bases. The one usually referred to as the standard basis consists of the matrices $E_{i,j}$ where $1\le i,j\le n$ and $E_{i,j}$ has $1$ at position $(i,j)$ and $0$ elsewhere. For this particular basis it is then obvious that precisely $n$ of these matrices have non-zero diagonal entries, namely the matrices $E_{i,i}$.

But, as said, there are plenty of other bases. In particular, there are bases in which every basis element has non-zero diagonal entries. One thing that can be said in general is that any basis must contain, for every position $(i,j)$, at least one matrix with non-zero $(i,j)$ entry.

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  • $\begingroup$ Yeah it did mention the standard basis. Thank you for clarifying! $\endgroup$ – Alti Nov 18 '13 at 6:14

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