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An orthogonal matrix satisfies $R^{T}R=I$, so it preserves the Euclidean norm. For a nonzero real eigenvector $v$,Therefore $\boxed{\lambda =\pm 1}$. Geometrically, eigenvalue $1$ fixes every vector on the linear span of $v$; eigenvalue $-1$ reverses every vector on that line. In either case the line is preserved as a set.

For $M$, solving $(M-I)v=0$ gives $y=x$, $z=y$, $x=z$. For $N$, the equations are $-2y-2z=0$ and $-2z=0$, hence $y=z=0$. For $P$, let $a=(1,1,1{)}^T$ and $J=a{a}^T$, so that $P=I-\frac{2}{3}J$. Then $(P-I)v=0$ is equivalent to $x+y+z=0$. The resulting eigenspaces and their bases areSolving all the equations shows that these lists contain the maximum possible numbers of independent eigenvectors with eigenvalue $1$.

The columns of $M$ form an orthonormal basis, so $M$ is an orthogonal matrix. Also $\det M=1$. Its fixed axis is the line through $a$, and on the perpendicular plane it acts as a two-dimensional rotation. If $\theta$ denotes the rotation angle, its matrix trace is $1+2\cos \theta$. Since $\mathrm{tr}M=0$, we obtain $\cos \theta =-1/2$. To fix the orientation relative to $u=a/\sqrt{3}$, note that $M{e}_1=e_3$ andThe component of $e_1$ perpendicular to $u$ has squared length $2/3$, and this scalar triple product equals $(2/3)\sin \theta$. Hence $\sin \theta =-\sqrt{3}/2$. ThusEquivalently, its unsigned angle is $2\pi /3$, or its positive angle is $2\pi /3$ about the opposite oriented axis.

For $P$, the reflection in a hyperplane formula givesIt fixes the plane $a\cdot v=0$ pointwise and maps $a$ to $-a$, so $\boxed{P\text{ reflects in }x+y+z=0}$. These actions also show that $P$ is an orthogonal matrix. In contrast, the second column of $N$ has squared length $5$, so $N$ is not orthogonal and cannot represent either of these Euclidean motions.

A diagonalizable matrix must have a basis of eigenvectors over the field in question. For $M$, a direct determinant computation gives ${\chi }_M(t)=1-t^3$. Its eigenvalues are $1,\omega ,{\omega }^2$, where $\omega =e^{2\pi i/3}$. Only one is real, with one-dimensional eigenspace, so $M$ is not diagonalizable over $\mathbb{R}$. Over $\mathbb{C}$ the three eigenvalues are distinct, and the vectors $(1,\lambda ,{\lambda }^2{)}^T$ for $\lambda =1,\omega ,{\omega }^2$ give an eigenbasis. The upper triangular matrix $N$ has only eigenvalue $1$, of algebraic multiplicity $3$, but its eigenspace has dimension $1$ over either field. Hence $N$ is not diagonalizable over either field. For $P$, the two real plane vectors displayed above, together with $a$, form an eigenbasis, with eigenvalues $1,1,-1$. They remain independent over $\mathbb{C}$. Therefore

Here is the matrix form of Rado's theorem. A matrix $A$ with entries in the rational numbers and columns $a_1,\dots ,a_n$ is a partition regular matrix if every finite coloring of the positive integers admits a vector $x\in {\mathbb{N}}^n$ whose coordinates are monochromatic and for which $Ax=0$. Its columns condition asks for an ordered partition into nonempty blocks $B_1,\dots ,B_s$, withRado's theorem states that these two conditions are equivalent. Coordinates need not be distinct; positivity is required for every coordinate.

First prove necessity using the finite separating-functional proof of the columns condition. Clearing denominators does not change either the kernel or the columns condition, so assume that $A$ has integer entries. For each pair of disjoint index sets $I,B\subseteq \{1,\dots ,n\}$, with $B$ nonempty, for which $b_B={\sum }_{i\in B}{a}_i$ is outside the rational linear span of the $I$-columns, choose an integer linear functional ${\ell }_{I,B}$ such thatSuch a linear functional exists: extend a basis over the rational numbers of the indicated linear span by $b_B$, prescribe values zero on the former basis and one on $b_B$, extend to a basis of the ambient vector space over the rational numbers, and then clear the functional's denominators. There are only finitely many pairs $(I,B)$. Choose a prime number $p$ larger than the absolute values of all their nonzero evaluations; if there are none, choose any prime number.

Color $x>0$ by its last nonzero digit coloring value, namelywhere $v_p$ is the P-adic valuation. By partition regularity, choose a positive solution $Ax=0$ all of whose coordinates have the same nonzero digit $u$. Group its indices into nonempty blocks $B_1,\dots ,B_s$ according to increasing distinct P-adic valuations ${\nu }_1<\cdots <{\nu }_s$.

Suppose that a block sum $b_j$ is outside the rational linear span of the earlier columns, and put $I=B_1\cup \cdots \cup B_{j-1}$, taking $I=\varnothing$ for $j=1$. Apply the selected linear functional ${\ell }_{I,B_j}$ to $Ax=0$. The earlier terms vanish exactly. Divide the remaining integer equality by $p^{{\nu }_j}$ and reduce modulo $p$. Later blocks vanish modulo $p$, while each coordinate in $B_j$ contributes its common digit $u$. We obtainwhich is impossible by the choice of $p$ and because $u\not\equiv 0(\mathrm{mod}p)$. Thus each block sum lies in the earlier linear span. For the first block, this span is zero, so $b_1=0$. This proves the columns condition.

For sufficiency, assume the columns condition and choose rational coefficients ${\lambda }_{ij}$, for $i\in B_1\cup \cdots \cup B_{j-1}$, such thatChoose a positive integer $c$ clearing all these coefficients' denominators and a positive integer $p$ with $p\ge |c{\lambda }_{ij}|$ for every coefficient. Use the permitted monochromatic m-p-c set theorem, in the suffix convention for an m-p-c set, to obtain positive generators $z_1,\dots ,z_s$ such that the full setis positive and monochromatic. The positivity of the full m-p-c set is important: we do not discard expressions that happen to be negative. For $i\in B_h$, setAll these coordinates belong to $D$, so they are positive and monochromatic. On collecting coefficients of each $z_j$, their matrix product isThis explicit Rado solution inside an m-p-c set proves sufficiency and completes the proof of Rado's theorem.

To deduce the Finite sums theorem directly from the columns condition, introduce one positive variable $y_F$ for each nonempty $F\subseteq \{1,\dots ,k\}$ and imposeFor $k=1$ there are no equations and any positive integer works. For $k\ge 2$, write $a_F$ for the column belonging to $y_F$. Partition the columns into blocks $B_j=\{F:\min F=j\}$ in the order $j=1,\dots ,k$. Consider the row indexed by a set $H$ with $|H|\ge 2$. The sum of the columns in $B_j$ has row valueFor $j=1$ this is always zero. For $j>1$ it equals one exactly for those $H$ with $\min H<j$ and $j\in H$. Each such $a_H$ is the negative coordinate vector in row $H$, and it belongs to an earlier block. Thereforewhich verifies the columns condition, rather than assuming the desired Finite sums theorem. Apply Rado's theorem to this system. Its monochromatic positive solution satisfies $y_F={\sum }_{i\in F}{x}_i$ with $x_i=y_{\{i\}}$, and hence

Two nonzero vectors are linearly independent when $s\mathbf{x}+t\mathbf{y}=0$ forces $s=t=0$. Their linear span then has dimension two. If they are linearly dependent, one is a nonzero scalar multiple of the other, and their span has dimension one. Thus the two requested dimensions are $\boxed{2\text{ and }1}$.

The Euclidean scalar product and its norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_{j=1}^n{x}_j{y}_j$ and $\Vert \mathbf{x}\Vert =\sqrt{\mathbf{x}\cdot \mathbf{x}}$. To prove the Cauchy-Schwarz inequality, first handle $\mathbf{y}=0$ trivially. Otherwise the squared normgivesEquality holds exactly when the residual vector is zero, meaning the vectors are linearly dependent; this includes the zero-vector cases. Expanding $\Vert \mathbf{x}+\mathbf{y}{\Vert }^2$ and applying Cauchy-Schwarz then givesTaking nonnegative square roots proves the triangle inequality.

For the unit-vector optimization, let $\mathbf{w}=\mathbf{x}+\mathbf{y}$. Independence ensures $\mathbf{w}\ne 0$. The variable part of $S$ is $\mathbf{z}\cdot \mathbf{w}$, which Cauchy-Schwarz bounds below by $-\Vert \mathbf{w}\Vert$. Equality occurs only for the antiparallel unit vector. ThusThis is minimizing a linear functional on a sphere. The two consequences follow below.

The finite-dimensional implication again follows from equivalence of norms and the Heine-Borel theorem. In an infinite-dimensional Banach space, inductively apply Riesz lemma to the proper closed finite-dimensional linear span $Y_n=\mathrm{span}(f_1,\dots ,f_n)$ to obtain a unit vector $f_{n+1}$ with $\mathrm{dist}(f_{n+1},Y_n)>1/2$. Therefore $\Vert f_n-f_m\Vert >1/2$ for distinct indices, precluding a Cauchy subsequence.

For completeness, Riesz lemma here follows by taking $z\notin Y$, setting $d=\mathrm{dist}(z,Y)>0$, choosing $y\in Y$ with $\Vert z-y\Vert <2d$, and putting $f=(z-y)/\Vert z-y\Vert$. Its distance from $Y$ exceeds $1/2$. This proves the closed unit ball is norm compact if and only if the Banach space is finite-dimensional. The printed hint's equality of all pairwise distances is unnecessary; the separation inequality is what the argument supplies.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.

If the spectrum has one point, the preceding part gives $T=\lambda I$, and the linear span of any nonzero vector is a closed proper invariant subspace because $\dim H>1$.

Otherwise choose distinct points ${\lambda }_1,{\lambda }_2\in K$ and disjoint nonempty relatively open neighborhoods $U,V\subseteq K$. The open-set result gives $P(U)\ne 0$ and $P(V)\ne 0$, whileThus $P(U)\ne I$. Set $L=\mathrm{ran}P(U)$. As the range of an orthogonal projection, $L$ is closed, and the preceding observations give $\{0\}\ne L\ne H$. The Borel functional calculus for a normal operator gives $TP(U)=P(U)T$, soIn fact both $L$ and its orthogonal complement are invariant.

Rado's criterion. Let $A$ be a matrix over the rational numbers, with columns $a_1,\dots ,a_q$. A partition regular matrix is one for which every finite coloring of the positive integers gives a monochromatic positive vector $x\in {\mathbb{N}}^q$ with $Ax=0$; repeated coordinates are allowed. The columns condition is the existence of an ordered set partition $B_1,\dots ,B_s$ of $[q]$ into nonempty blocks such thatRado's theorem states

Necessity. Clear denominators so the columns are integer vectors. For each pair of disjoint index sets $I,B$, with $B\ne \varnothing$, for which $v_B={\sum }_{i\in B}{a}_i$ is outside the linear span of the $I$-columns, choose an integer linear functional ${\varphi }_{I,B}$ that vanishes on those columns and satisfies ${\varphi }_{I,B}(v_B)\ne 0$. Such a functional exists by extending a basis of the span to one also containing $v_B$, defining a functional on that basis, and clearing its rational denominators.

There are only finitely many such pairs. Choose a prime number $p$ dividing none of these nonzero integer evaluations. If there are no such pairs, any prime will do. Use the last nonzero digit coloring: write $x=p^{v_p(x)}u$, with $p\nmid u$, and color $x$ by $u\mathrm{mod}p\in \{1,\dots ,p-1\}$. Since $A$ is a partition regular matrix, this coloring supplies a monochromatic positive solution, so all its unit parts have a common nonzero residue $d$.

Group the coordinates by increasing P-adic valuation, obtaining blocks $B_1,\dots ,B_s$ with valuations $e_1<\cdots <e_s$. Suppose for some $j$ the sum $v_{B_j}$ lay outside the linear span of the earlier columns, with earlier index set $I$. Apply ${\varphi }_{I,B_j}$ to $Ax=0$. Earlier terms vanish exactly. Divide the remaining integer equation by $p^{e_j}$ and reduce modulo $p$. Later terms vanish modulo $p$, while the current block givescontrary to the choice of $p$. Thus every block sum lies in the earlier linear span, and for the first block this means it is zero. This proves the columns condition by the finite separating-functional proof of the columns condition.

Sufficiency. Suppose the columns condition holds. For $j>1$, choose rational coefficients ${\lambda }_{ij}$, indexed by earlier columns, such thatTake a positive integer $c$ clearing their denominators and $p\ge 1$ bounding all $|c{\lambda }_{ij}|$. We use the allowed monochromatic m-p-c set theorem with $s$ levels. Its M-p-c set is the full setwhere the generators make every displayed expression positive. Positivity is part of this convention, rather than a rule that discards nonpositive expressions. For $i\in B_h$, setAll these numbers lie in the monochromatic set $D$. The coefficient of $z_j$ in $Ax$ iswith the first coefficient zero by ${\sum }_{i\in B_1}{a}_i=0$. Hence $Ax=0$. This Rado solution inside an m-p-c set proves sufficiency.

The equality-or-inequality dichotomy. If every finite coloring gives a monochromatic solution with $x_1=x_2$, the first alternative holds. Otherwise fix a finite coloring $\chi$ admitting no such solution. Given any other finite coloring $\psi$, use the refinement of a finite coloring $n\mapsto (\chi (n),\psi (n))$. Since $A$ is a partition regular matrix, it supplies a solution monochromatic for both colorings. Its first two coordinates cannot be equal, by the defining property of $\chi$, so it is the required unequal solution for $\psi$. Thus one of the two universal alternatives always holds. This is the partition regularity dichotomy under an extra constraint; the alternatives are not asserted to be mutually exclusive.

Both restrictions can fail. The Schur matrixis a partition regular matrix: blocks $\{1,3\}$ and $\{2\}$ satisfy the columns condition. Color $n$ by $v_2(n)\mathrm{mod}2$, using the 2-adic valuation. An equal-coordinate solution would have $x_3=2x_1$, and $v_2(2x_1)=v_2(x_1)+1$, so its coordinates cannot be monochromatic. This exhibits failure of the equality restriction.

Conversely, $A=(1-1)$ satisfies the columns condition in one block. Every solution has $x_1=x_2$, so even the constant one-color finite coloring admits no unequal-coordinate solution. Consequently

Let $\Pi$ be the orthogonal projection onto the good linear subspace $\mathcal{G}$, and let $|\psi ⟩$ be a prepared unit vector with $p=⟨\psi |\Pi |\psi ⟩$. For $0<p<1$, defineThe amplitude amplification theorem states that, if the reflection operators about the initial state and the good subspace can be implemented, the iterationsatisfiesTo prove it, the two reflections preserve the linear span of $|g⟩,|b⟩$. In this ordered orthonormal basis, direct multiplication givesMultiplying this rotation matrix by $(\sin \alpha ,\cos \alpha {)}^T$ replaces $\alpha$ by $\alpha +2\theta$, proving the formula by mathematical induction.

For known $0<p\le 1/2$, choose $j$ to be a nearest nonnegative integer to $\pi /(4\theta )-1/2$. Then $|(2j+1)\theta -\pi /2|\le \theta$, so measurement finds the good subspace with probability at least $1-p$, using $O(p^{-1/2})$ iterations. For $p>1/2$, measurement of the initial state already has constant success probability. If $p=1$ success is certain; if $p=0$ these reflections cannot create any good component. If $|\psi ⟩=A|0⟩$, the initial-state reflection is $A(2|0⟩⟨0|-I)A^{†}$, implemented using quantum state preparation and its inverse quantum circuit.

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 $A$ be a rational matrix with columns $a_1,\dots ,a_n$. It is a partition regular matrix when every finite coloring of the positive integers has a monochromatic vector $x$ with $Ax=0$. It has the columns property when the column indices have an ordered partition $B_1,\dots ,B_r$ such thatand, for every $s>1$,Rado's theorem states that $A$ is partition regular if and only if it has the columns property.

First suppose $A$ is partition regular, and clear its denominators. For a prime number $p$, apply the last nonzero digit coloring in base $p$: if $x=p^{v_p(x)}u$ with $p\nmid u$, its color is $u\mathrm{mod}p$. Choose a monochromatic solution and group its coordinates into blocks $B_1,\dots ,B_r$ of equal $p$-adic valuation, in increasing order of valuation. Only finitely many ordered partitions are possible, so one partition occurs for infinitely many primes.

For any such prime, divide ${\sum }_i{x}_i{a}_i=0$ by the lowest power of $p$ and reduce modulo $p$. All coordinates in $B_1$ have the same nonzero leading digit $d$, while later blocks vanish, soThe integer vector ${\sum }_{i\in B_1}{a}_i$ is divisible by infinitely many primes and therefore equals zero.

Now fix $s>1$, and let $t$ be the common valuation on $B_s$. Reduction modulo $p^{t+1}$ givesIf the first sum of columns were outside the rational linear span of the earlier columns, an integer linear functional would vanish on every earlier column but not on that sum. Applying it to the congruence would say that infinitely many primes divide one fixed nonzero integer, a contradiction. Thus the displayed partition has the columns property.

Conversely, suppose $A$ has the columns property. For $s>1$, choose rational numbers $q_{is}$ such thatLet $d_{is}$ equal $1$ for $i\in B_s$, equal $-q_{is}$ in the earlier blocks, and equal zero in the later blocks. For $s=1$, let $d_{i1}$ be the indicator of $B_1$. Thenfor every $s$. Choose a positive integer $c$ clearing all denominators and a positive integer $p$ with $|c{d}_{is}|\le p$ whenever $s$ is later than the block containing $i$.

By the monochromatic m-p-c set theorem, the coloring contains a monochromatic $(r,p,c)$-set with generators $z_1,\dots ,z_r$. DefineIf $i\in B_t$, thenso every $y_i$ lies in that one monochromatic M-p-c set. Moreover,Thus $A$ is partition regular, proving Rado's theorem.

For the equation $4x+3y-6z=0$, a monochromatic solution in the last nonzero base-$p$ digit coloring would, at the least $p$-adic valuation among $x,y,z$, force a nonempty subset sum of $4,3,-6$ to vanish modulo $p$. The seven possible sums areNone is divisible by $5$, so base $5$ gives no monochromatic solution. The smaller primes do admit solutions: $(x,y,z)=(3,2,3)$ works for base $2$, whose coloring has one color, and $(3,4,4)$ is monochromatic in base $3$. Hence the smallest prime is

For $x+2y-7z=0$, use the last nonzero base-$11$ digit itself. The nonempty subset sums of $1,2,-7$ arenone zero modulo $11$. This is the required $10$-coloring.

For a $5$-coloring, identify each nonzero residue $d$ modulo $11$ with $-d$, giving the five colorsIf a monochromatic solution existed, reduction at the least $11$-adic valuation would give a signed nonempty subset sum of $1,2,-7$ equal to zero modulo $11$. Singles have absolute residues $1,2,7$; pairs have absolute residues $1,3,5,6,8,9$; and triples have absolute residues $4,6,8,10$. None is zero modulo $11$, which proves that this $5$-coloring works.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

The row space of a matrix is the linear span of its rows. It is the column space of the matrix transpose.

A subset $S$ of a vector space is a spanning set when its linear span is the whole vector space.

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.