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So, according to Humphreys:$\mathfrak{gl}_n$ is all $n \times n$ matrices, $\mathfrak{sl}_n$ is $n\times n$ matrices with sum of diagonal elements (trace) equal to zero and $\mathfrak{s}_n$ is set of all $\lambda I$ where $\lambda \in F$ (underlying field) and $I$ is the identity matrix (diagonal matrix having only 1 on the diagonal). He claims that $\mathfrak{gl}_n = \mathfrak{sl}_n+\mathfrak{s}_n$. This is just purely wrong, with his definitions. Consider

$$A = \begin{bmatrix} 1 & 2\\ 0 & 2 \end{bmatrix}$$

How is this a sum of scalar matrix and matrix having trace zero? What he should have said is : matrices in $\mathfrak{gl}_n$ that only have one eigenvalue, as those are known to be $\lambda I + N$ where $N$ is nilpotent matrix... or something... am I bonkers or right?

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  • $\begingroup$ Besides the mathematical lesson learned from the answer, you could also learn the life lesson to tone down your language. Since now according to you, you are bonkers. $\endgroup$ Apr 18 at 19:15
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    $\begingroup$ I identify as a clown.... so thankies Torsten! $\endgroup$
    – Dibidus
    Apr 19 at 4:43

1 Answer 1

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Well, you have$$\begin{bmatrix}1&2\\0&2\end{bmatrix}=\overbrace{\begin{bmatrix}-\frac12&2\\0&\frac12\end{bmatrix}}^{\phantom{\mathfrak{sl}_2}\in\mathfrak{sl}_2}+\overbrace{\begin{bmatrix}\frac32&0\\0&\frac32\end{bmatrix}}^{\phantom{\mathfrak{sl}_2}\in\mathfrak{s}_2}.$$More generally, if $A\in\mathfrak{gl}_n$, if $D=\frac{\operatorname{tr}(A)}n\operatorname{Id}_n$ and if $S=A-D$, then $A=S+D$, $S\in\mathfrak{sl}_n$, and $A\in\mathfrak{s}_n$.

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