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Let nonnegative partition-normalized B-splines $N_i$ form a basis, with ${\sum }_i{N}_i\le 1$ on the integration interval and $d_i=\int N_i>0$. The row-normalized mixed Gram matrix is $G_{ij}=\int N_i{N}_j/d_i$. The normal equations for the orthogonal projection onto their linear span become $Ga=b$, with $b_i=\int f{N}_i/d_i$ and $\Vert b{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$. The subpartition of unity for B-splines gives $\Vert {\sum }_i{a}_i{N}_i{\Vert }_{\infty }\le \Vert a{\Vert }_{{\ell }^{\infty }}$ and proves the bound. The ordinary symmetric Gram matrix omits this essential row normalization. The projection maps continuous functions into the spline space, not onto all continuous functions.

Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.

A family is linearly independent if and only if every vector in its linear span has unique coefficients as a linear combination of that family. For a short proof, subtract two representations to get ${\sum }_i(a_i-b_i)v_i=0$; linear independence forces each difference to vanish. Conversely, uniqueness applied to the zero vector forces every linear relation to use only zero coefficients. This explains the practical role of linear independence: a coordinate vector relative to an ordered basis is unambiguous.

The vector projection multiplies the scalar projection by the chosen unit vector, giving the orthogonal projection onto the linear span of a nonzero real Euclidean vector $\mathbf{c}$:For $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$, this gives $\mathbf{p}=(6/4)(2,0)=(3,0)$ and $\mathbf{r}=(0,4)$; the dot product $\mathbf{r}\cdot \mathbf{c}=0$ checks the orthogonal projection residual. Replacing $\mathbf{c}$ by $(-2,0)$ gives the same $\mathbf{p}=(-6/4)(-2,0)=(3,0)$: reversing direction changes the scalar projection, but preserves the vector projection onto the line.

For a short proof of the nearest-point property, any vector $\mathbf{z}$ on this line satisfies $\Vert \mathbf{a}-\mathbf{z}{\Vert }^2=\Vert \mathbf{r}{\Vert }^2+\Vert \mathbf{p}-\mathbf{z}{\Vert }^2$ by the Pythagorean identity. The second term is uniquely minimized at $\mathbf{z}=\mathbf{p}$. This connects a directional measurement to the orthogonal projection used in approximation.^[ref]

Every vector space $V$ over a field $F$ has a basis, assuming the axiom of choice. More generally, any linearly independent subset $A\subseteq V$ extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

Use the poset of linearly independent subsets containing $A$:It is a set, being a subset of the power set of $V$, and is nonempty because $A\in {\mathcal{P}}_A$.

A short proof outline supplies both essential steps:

Taking $A=\varnothing$ proves existence, including $V=\{0\}$, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: $\{v\}$ and $\{2v\}$ for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.

A positive measurable weight $w$ defines the displayed norm on the L2 space for the measure $w(x)dx$, with functions identified when equal almost everywhere. This is the norm induced by the weighted inner product. Changing $w$ changes the contribution of each region to approximation error. If $p_0,\dots ,p_N$ are nonzero mutually orthogonal real functions, completing squares shows that the unique best approximation to real $f$ in their linear span has coefficients $a_j=⟨f,p_j{⟩}_w/\Vert p_j{\Vert }_w^2$.