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Let $U\subseteq W\subseteq {\mathrm{End}}_{\mathbb{C}}(V)$ be vector subspaces and $M=\{T:[T,W]\subseteq U\}$. If $\alpha \in M$ and $\mathrm{tr}(\alpha \beta )=0$ for all $\beta \in M$, then $\alpha$ is a nilpotent endomorphism.

To prove it, let $\beta$ act by $\overline{\lambda }$ on the generalized eigenspace of $\alpha$ of eigenvalue $\lambda$. Polynomial interpolation on eigenvalue differences, together with adjoint compatibility of additive Jordan decomposition, expresses $\mathrm{ad}\beta$ as a polynomial in $\mathrm{ad}\alpha$ with zero constant term. Since $\mathrm{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta \in M$. Taking the matrix trace on each generalized eigenspace yields $0={\sum }_{\lambda }\dim (V_{\lambda })|\lambda {|}^2$, so every eigenvalue is zero. This is the linear-algebra step behind the Cartan solvability criterion; $U,W$ need not be Lie subalgebras.

A smooth map $f:M\to N$ is transverse to an embedded submanifold $S\subset N$ whenThis is a sum of vector subspaces, not necessarily a direct sum. Equivalently the composition of $d{f}_p$ with the quotient to $T_{f(p)}N/T_{f(p)}S$ is surjective. In a slice chart for an embedded submanifold of codimension $k$, the $k$ normal coordinate functions composed with $f$ have surjective derivative on their common zero set. If the preimage is empty the condition holds vacuously.

Here is a short proof that uses only the row space. For a vector subspace $W\subseteq {\mathbb{F}}^n$, let $d_j$ be the dimension of its projection map onto the first $j$ coordinates, with $d_0=0$. In any reduced row echelon form with row space $W$, $d_j$ counts the pivot columns among those first $j$ columns: later-pivot rows project to zero, while earlier-pivot rows remain linearly independent. Thus the positions where $d_j-d_{j-1}=1$ determine all pivot positions from $W$ alone.

If those positions are $p_1,\dots ,p_r$, the map sending $w\in W$ to $(w_{p_1},\dots ,w_{p_r})$ is an isomorphism $W\to {\mathbb{F}}^r$: the reduced nonzero rows map to the standard basis. Its $i$th row is therefore the unique vector of $W$ with pivot-coordinate values equal to the $i$th standard basis vector. Any other reduced row echelon form with that row space has identical nonzero rows in identical order. At a fixed matrix size, the remaining rows must be zero. Together with existence by Gaussian elimination, this proves uniqueness.

A universe for spectra is a countably infinite-dimensional real inner product space $U$, used to index finite-dimensional suspension directions. A standard model is the algebraic direct sum $U={⨁}_{i\ge 1}\mathbb{R}{e}_i$ with an orthonormal basis, carrying its direct-limit topology. Its vectors have finite support; this universe is not the completed Hilbert space ${\ell }^2$. Indexed topological spectra assign spaces to finite-dimensional vector subspaces $V\subseteq U$ and use the corresponding orthogonal differences for structure maps.^[ref]

A family of indexing vector subspaces is cofinal when every finite-dimensional $W\subseteq U$ is contained in some member of the family. The flag $V_n=\mathrm{span}(e_1,\dots ,e_n)$ is cofinal: a finite basis of $W$ uses only finitely many of the $e_i$. In contrast, $W_n=\mathrm{span}(e_2,e_4,\dots ,e_{2n})$ has unbounded dimension but never contains $e_1$, so it is not cofinal. Cofinality ensures that restricting the index system does not discard any finite collection of suspension directions.

A vector subbundle $F\subseteq E$ is a subset whose fibers are vector subspaces $F_x\subseteq E_x$ of constant dimension and which admits local frames extending to local frames of $E$.

A vector subspace is a nonempty subset of a vector space closed under vector addition and scalar multiplication, and is therefore itself a vector space. Equivalently, whenever $u,v$ belong to the subset, so does $au+bv$ for all allowed scalars $a,b$. In particular it contains the zero vector.

For example, the plane $W=\{(x,y,z):x+y+z=0\}\subseteq {\mathbb{R}}^3$ is a vector subspace, with basis $(1,-1,0),(1,0,-1)$. Indeed every element is $(-y)(1,-1,0)+(-z)(1,0,-1)$. The parallel plane $x+y+z=1$ misses the zero vector and is an affine subspace instead. Homogeneous constraints produce vector subspaces: $W$ is the kernel of the linear map $(x,y,z)\mapsto x+y+z$. This lets us describe all solutions of a homogeneous linear equation through a basis of its kernel.

For a linear map $\alpha :U\to V$ between finite-dimensional vector spaces, take $\beta \in \mathrm{Hom}(V,U)$. The vector subspace $M^l(\alpha )=\{\beta :\beta \alpha =0\}$ consists of the linear maps factoring uniquely through $V/\mathrm{im}\alpha$, whereas $M^r(\alpha )=\{\beta :\alpha \beta =0\}$ consists of those whose image of a linear map lies in $\ker \alpha$. ThusWriting $u=\dim U$, $v=\dim V$ and $r=\mathrm{rank}\alpha$, their dimensions are $(v-r)u$ and $(u-r)v$. Passing to the dual map reverses composition and interchanges the corresponding left and right conditions. The finite-dimensional hypothesis makes these dimension differences ordinary integer subtraction.