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Geometric invariant theory constructs algebraic quotients of group actions by a reductive algebraic group using invariant sections and a chosen linearization of a line bundle. An affine quotient is the spectrum of the polynomial invariant ring; a projective quotient uses homogeneous invariants and excludes the points where all positive-degree invariants vanish. Stability separates the orbits which give well-behaved geometric quotient points.

Use the natural linearization of a line bundle on ${\mathcal{O}}_X(1)$ induced by the conjugation representation on $M_n\mathbb{C}$. The notation involving the dual does not change this matrix description: the nondegenerate pairing $(A,B)\mapsto \mathrm{tr}(AB)$ identifies the representation with its dual equivariantly. Let $[A]$ be a projective point, so $A\ne 0$.

The Hilbert-Mumford criterion for projective stability says that $[A]$ is not a semistable projective point in geometric invariant theory precisely when some algebraic one-parameter subgroup $\lambda :{\mathbb{G}}_m\to S{L}_n$ sends its affine representative to zero:Equivalently, all active weights for that subgroup are strictly positive. With the convention $\mu ([A],\lambda )=-\min \{\text{active weights}\}$, semistability requires $\mu \ge 0$ for every subgroup, including conjugates of diagonal ones.

Suppose such a zero limit exists. The coefficients $c_j(A)$ inare invariant under conjugation. Continuity along the zero limit gives $c_j(A)=c_j(0)=0$ for every $j$. By the Cayley-Hamilton theorem, $A^n=0$: the matrix is nilpotent.

Conversely, if $A$ is a nonzero nilpotent matrix, choose a basis in which it is strictly upper triangular, for example a Jordan normal form. The change of basis can be chosen in $S{L}_n$: multiply any invertible change-of-basis matrix by a scalar with the required $n$th power, which does not affect its conjugation. In this basis takeThe exponents sum to zero, so this is an algebraic one-parameter subgroup of the special linear group. The $(i,j)$ matrix entry has weightEvery potentially nonzero entry has $i<j$ and therefore positive weight. Thus the conjugated matrix tends to zero. Conjugate this subgroup back to the original basis to obtain the required destabilizing subgroup for $A$.

It follows that the nilpotent cone for conjugation of matrices is exactly the affine null cone, andEach $c_j$ is a homogeneous invariant of degree $j$, so the second description also gives invariant open sets witnessing semistability. Equivalently, a representative has at least one nonzero eigenvalue. Nonzero singular matrices can be semistable; invertibility is not the criterion. For example, a diagonal matrix with diagonal entries $1,0,\dots ,0$ is semistable, while a nonzero strictly upper triangular matrix is unstable.

For the natural linearization of a line bundle of the action on lines in $V$, a point is semistable when some positive-degree homogeneous invariant on $V$ is nonzero on a representative. This condition is independent of the representative. The nonsemistable vectors form the affine null cone.