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In field characteristic $p$, the identity matrix has trace $p=0$, and hence belongs to ${\mathfrak{sl}}_p$. Its span is a nonzero proper central ideal, so ${\mathfrak{sl}}_p$ is not a simple Lie algebra. In fact its center of a Lie algebra is precisely $k{I}_p$: commuting with every off-diagonal matrix unit forces a matrix to be scalar.

For $p>2$, the matrix-unit extraction lemma for special linear ideals still works because two is invertible. Any ideal containing a nonscalar matrix contains every off-diagonal matrix unit and every trace-zero diagonal matrix, hence equals ${\mathfrak{sl}}_p$. Therefore an ideal of the quotient Lie algebra pulls back either to the center or to the whole algebra. The quotient is nonabelian, since $[E_{12},E_{23}]=E_{13}$ remains nonzero modulo the center. ConsequentlyNo separation of diagonal root spaces is needed, so this proof also handles $p=3$ without assuming their weights are distinct.

For $p=2$, every trace-zero two-by-two matrix has the form $aI+b{E}_{12}+c{E}_{21}$. The only nonzero bracket between these basis vectors is $[E_{12},E_{21}]=I$, which is central. Thus

A simple Lie algebra is a nonabelian Lie algebra whose only Lie algebra ideals are zero and the whole algebra. Assume $n\ge 2$; ${\mathfrak{sl}}_1=0$ is not simple. We prove the assertion by the matrix-unit extraction lemma for special linear ideals, avoiding any classification theorem.

Let $I$ be a nonzero ideal in the special linear Lie algebra and $0\ne A\in I$. If an off-diagonal entry $A_{ji}$ is nonzero, two commutators with the matrix unit $E_{ij}$ giveso $E_{ij}\in I$. If $A$ is diagonal and nonscalar, choose unequal diagonal entries; $[E_{ij},A]=(A_{jj}-A_{ii})E_{ij}$ again supplies a matrix unit. A nonzero scalar matrix cannot have zero trace over $\mathbb{C}$, so these cases cover every nonzero $A$.

Once $E_{ij}\in I$, its bracket with $E_{ji}$ gives $H=E_{ii}-E_{jj}$, and $[H,E_{ji}]=-2E_{ji}$ puts the reverse unit in $I$. Bracketing with the other off-diagonal matrix units generates every off-diagonal unit: for a third index $k$, $[E_{ij},E_{jk}]=E_{ik}$ and $[E_{ki},E_{ij}]=E_{kj}$; using the reverse unit fills the remaining row and column, then $[E_{ki},E_{i\ell }]=E_{k\ell }$ fills the others. Opposite pairs generate every trace-zero diagonal matrix. These vectors span ${\mathfrak{sl}}_n$, so

Work over the complex numbers. The chain $S_0\subset S_1\subset \cdots \subset S_n$ has multiplicity-free restriction: each irreducible representation restricts to a direct sum of pairwise inequivalent irreducible representations. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted Olshanskii centralizer lemma makes $C_{\mathbb{C}{S}_r}(\mathbb{C}{S}_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative matrix algebra factor. Thus no identification of the branching graph with Young diagrams is being assumed here.

Successively restricting an irreducible representation $V$ gives one-dimensional spaces $V_T$ indexed by paths of irreducible representations from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a Gelfand–Tsetlin basis. Define the Gelfand–Tsetlin algebra as the subalgebra of $\mathbb{C}{S}_n$ acting diagonally in all these bases. The Artin–Wedderburn theorem identifiesUnder this identification the Gelfand–Tsetlin algebra is the direct sum of the full diagonal matrix algebras in the indicated bases, so it is commutative.

To see that it is actually available inside the group algebra, let $e_{\alpha }^{(r)}$ be the central primitive idempotent selecting the irreducible representation $\alpha$ of $S_r$. For a path $T=({\alpha }_0,\dots ,{\alpha }_n)$ the productis the projection onto $V_T$ in its final irreducible representation and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal matrix units, and therefore span the proposed algebra.

If an element of the group algebra commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the Gelfand–Tsetlin algebra. Thuswhich proves that it is a maximal commutative subalgebra.

The primitive idempotent products used above show that the Gelfand–Tsetlin algebra is generated by the centers at all levels: those products give every diagonal matrix unit, whereas each center acts diagonally. Hence

For a complex finite-dimensional simple Lie algebra, a Cartan subalgebra $\mathfrak{h}$ is a maximal commuting subalgebra of elements whose adjoint maps are semisimple. Equivalently in this setting it is a nilpotent self-normalizing subalgebra. Its dimension is the rank of a semisimple Lie algebra. Simultaneous diagonalization of its Adjoint representation gives the root-space decompositionA root of a root system is a nonzero linear functional $\alpha$ for which this root space is nonzero. For a complex semisimple algebra each root space is one-dimensional. A Cartan-Weyl basis consists of a basis $H_i$ of $\mathfrak{h}$ and one nonzero root vector $E_{\alpha }$ for every root.

The general Lie brackets have the formFor the opposite-root bracket, use the Killing form to define $h_{\alpha }$ by $\kappa (h_{\alpha },H)=\alpha (H)$. Its invariant bilinear form on a Lie algebra property givesOne may normalize the root vectors so that the pairing is one. If instead one uses a coroot as the opposite-root bracket, the root-vector normalization changes accordingly. In particular, root evaluation coordinates cannot simply be used as coefficients in a nonorthonormal Cartan basis.

For the matrix calculation take $N\ge 2$. The complexification of a Lie algebra of the special unitary group Lie algebra is the special linear Lie algebra ${\mathfrak{sl}}_N(\mathbb{C})$: traceless complex matrices. Its Cartan subalgebra consists of traceless diagonal matrices. Write $E_{jk}={\mathcal{T}}^{(j,k)}$ for the matrix units. The given Cartan basis is $H_i=E_{ii}-E_{i+1,i+1}$, $1\le i<N$, and the other basis elements are $E_{jk}$ with $j\ne k$.

The matrix-unit identity $E_{jk}{E}_{lm}={\delta }_{kl}{E}_{jm}$ givesThus all the roots, expressed as evaluation vectors in this precise Cartan basis, areThey are the functionals $e_j-e_k$ on traceless diagonal matrices; there are $N(N-1)$ of them. The corresponding root vector is $E_{jk}$. Together with $N-1$ Cartan generators, they give $N^2-1$ basis elements. The simple roots can be chosen as ${\alpha }_{i,i+1}$, whose evaluation vectors are the rows of the type-$A_{N-1}$ Cartan matrix, with $2$ on the diagonal and $-1$ on adjacent entries. These vectors are evaluations on $H_i$, not coordinates in an orthonormal realization of the root system.

To express every bracket strictly in the chosen basis, introduce the abbreviationThen all pairs are covered byHere both input root vectors have distinct row and column indices. The first case is the only one producing diagonal matrix units, and the displayed sum of $H_p$ resolves them completely into the chosen Cartan basis. Reversing the order gives the negative bracket. This also shows explicitly that the two nonzero non-Cartan cases have structure constants $+1$ and $-1$, and verifies the required root-addition rule.

Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are $H(t)\mapsto \pm t_i\pm t_j$, with the two signs independent. This is the D3 root system, with twelve roots.

For a direct check, the defining matrix identity gives the block form $X=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B^T=-B$ and $C^T=-C$. If $E_{ij}$ denotes a matrix unit, the root space for ${\epsilon }_i-{\epsilon }_j$ is spanned by $E_{ij}-E_{j+3,i+3}$; those for ${\epsilon }_i+{\epsilon }_j$ and $-{\epsilon }_i-{\epsilon }_j$ are spanned respectively by $E_{i,j+3}-E_{j,i+3}$ and $E_{i+3,j}-E_{j+3,i}$. Commuting these with $H(t)$ gives exactly the displayed maps.

For the left regular module, the radical series of a module is obtained by multiplying by powers of the Jacobson radical:Products of matrix units show thatwith $N^0=A$ and $N^n=0$. Thus each step removes one subdiagonal, and the Loewy length of ${}_{A}A$ is $n$.

For $1\le r\le n$, let $S_r=k$ with a lower triangular matrix $a=(a_{ij})$ acting by the scalar $a_{rr}$. These are pairwise nonisomorphic simple modules, since the diagonal matrix units $e_{ii}$ distinguish them.

Let $N$ be the ideal of strictly lower triangular matrices. It is nilpotent, andis a semisimple algebra. Hence $N$ is the Jacobson radical; explicitly,

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

Let $E_{rs}$ be the matrix unit with its only nonzero entry in row $r$, column $s$. Forone hasThereforeThe $n^2$ matrix units form an eigenbasis, with $E_{rs}$ having eigenvalue $r-s$. The zero eigenspace consists exactly of the diagonal matrices and has dimension $n$. The rank-nullity theorem now gives

For the symplectic Lie algebra $\mathfrak{sp}(4)$ preserving $J=\left(\begin{array}{cc}0& I_2\\ -I_2& 0\end{array}\right)$, write $h_1=E_{11}-E_{33}$ and $h_2=E_{22}-E_{44}$. Its positive roots of a root system have sl2 subalgebra associated with a root triplesThe matrix unit commutator identity verifies $[X_{\alpha },Y_{\alpha }]=H_{\alpha }$ and the two eigenvalue relations $[H_{\alpha },X_{\alpha }]=2X_{\alpha }$, $[H_{\alpha },Y_{\alpha }]=-2Y_{\alpha }$.

Over $\mathbb{R}$ or $\mathbb{C}$, every linear functional on all $n\times n$ matrices satisfying $f(AB)=f(BA)$ is a scalar multiple of the trace. The matrix units show that the functional vanishes off the diagonal and has equal values on every diagonal unit. The normalization $f(I)=n$ fixes the scalar to one.

Every commutator of square matrices is traceless by cyclicity of the trace. Conversely, off-diagonal matrix units and differences of diagonal units are themselves commutators and span all traceless matrices. This proves equality with the linear span of commutators, without needing a theorem about representation as one commutator.