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What would a basis of a space of $n \times n$ upper triangular matrices with trace 0 be? Is it trivial?

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A matrix does not have a basis; only a vector space has a basis. What do you really mean? –  Gerry Myerson Jun 25 '13 at 12:12
    
A vector space with all upper triangual matrices with trace 0. –  Ben Jun 25 '13 at 12:18
    
Ben, please edit your question so it says what you mean for it to say. People shouldn't have to go through the comments to understand the question. –  Gerry Myerson Jun 25 '13 at 12:21

1 Answer 1

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Hint: trace A=0 then $\sum_{i=1}^na_{ii}=0 $then we have $a_{nn=}-a_{11}-...-a_{n-1 n-1}$ $$ A=\begin{pmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n} \\ 0 & a_{22} & a_2^2 & \cdots & a_2^n \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & (-a_{11}-...-a_{n-1 n-1}) \end{pmatrix}$$

$$=\pmatrix{ \bullet&\bullet&\bullet&\bullet&\bullet\\ \color{white}\circ&\bullet&\bullet&\bullet&\bullet\\ \color{white}\circ&\color{white}\circ&\bullet&\bullet&\bullet\\ \color{white}\circ&\color{white}\circ&\color{white}\circ&\bullet&\bullet\\ \color{white}\circ&\color{white}\circ&\color{white}\circ&\color{white}\circ&\color{white}\circ\\ }$$

$B=\{E_{ij}:i\lt j\}$$\cup$$\{E_{ii}:i=1,2,..,n-1\}$

clearly $B$ is basis for an upper triangular matrix with trace 0 because $B$ generate it and $B$ is linear independent

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