# Problem 1.7 in Humphreys' Lie algebra book

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?

• 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. Apr 18 at 19:15
• I identify as a clown.... so thankies Torsten! Apr 19 at 4:43

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 $$A\in\mathfrak{s}_n$$.