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Use $\{Q_{\alpha }^A,Q_{\beta }^B\}={ϵ}_{\alpha \beta }{Z}^{AB}$. Unitary skew-diagonalization of an antisymmetric matrix gives two-by-two blocks with entries $Z_r=2z_r$. At rest, the paired supercharges have anticommutators $2(M\pm |z_r|)$. The squared norms of an operator and its adjoint sum to the expectation of their anticommutator, so positive norm requires $M\ge |z_r|$ for every block. A convention with $2Z$ in the algebra instead writes $M\ge \max |Z_r|$.

Keep the same origin and the same vector $p=(2,1{)}^T$, but use the nonperpendicular basis vectors $b_1=(1,0{)}^T$ and $b_2=(1,1{)}^T$. They form a basis because the column matrix $B$ has nonzero determinant:The reconstruction is $p=b_1+b_2$. Its coordinates are read along lines parallel to the other axis, producing a parallelogram. These lines need not be perpendicular to the axis being measured. In contrast, the dot products are $p\cdot b_1=2$ and $p\cdot b_2=3$; treating those as coordinates would reconstruct $2b_1+3b_2=(5,3{)}^T$, a different vector.

The dual basis supplies the right measuring rules for any basis.^[ref] Here these coordinate functionals are $b_1^{\ast }(x,y)=x-y$ and $b_2^{\ast }(x,y)=y$, the rows of $B^{-1}$. They give $b_1^{\ast }(p)=b_2^{\ast }(p)=1$. Equivalently, their representing vectors $d_1=(1,-1{)}^T$, $d_2=(0,1{)}^T$ satisfy $d_i\cdot b_j={\delta }_{ij}$ and extract the coordinates as $d_i\cdot p$. These are Riesz dual basis in an inner product space representatives, rather than the original basis vectors.

The Gram matrix records how the nonperpendicular directions interact:For any real basis represented by a square invertible matrix $B$, writing $p=Bc$ gives these identities immediately. One can recover the coordinates either as $c=B^{-1}p$ or by solving $Gc=B^{T}p$. In this example, if $c=(a,b{)}^T$, the squared norm is $a^2+2ab+2b^2$, not $a^2+b^2$. Oblique coordinates remain useful when natural directions are not perpendicular, but the Gram matrix must accompany them when computing lengths or dot products.

A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

Rotate the standard basis through $30^{\circ }$ using the rotation matrix whose columns are the new basis vectors:Here $u\cdot v=0$ and $u\cdot u=v\cdot v=1$, so $Q$ is an orthogonal matrix with $Q^{T}Q=I$, where $T$ denotes transpose and $I$ is the identity matrix. For the same vector $p=(2,1{)}^T$, its coordinate vector in this Cartesian basis isThus $\alpha \approx 2.232$ and $\beta \approx -0.134$: travel forwards along $u$ and a little backwards along $v$ to reconstruct $p=\alpha u+\beta v$. The orthogonal projections onto the two axes give these signed coordinates because the measuring vectors are both perpendicular and of unit length.

The squared norm remainsThis is useful when choosing axes aligned with a geometric feature: the coordinates change but distances and dot products retain the same sum-of-products formula. The two directions of the change of basis must be kept straight: $Q$ reconstructs the vector in old coordinates, $p=Q[p{]}_{\mathcal{U}}$, and $Q^T$ extracts its new coordinates. Applying $Q$ to the old coordinate vector $p$ instead would actively rotate the vector.

A short proof explains both formulas. If $p=Qc$, multiplication by $Q^T$ gives $c=Q^{T}p$. Then $\Vert p{\Vert }^2=c^T{Q}^{T}Qc=c^{T}c$. Perpendicular unit-length columns make the matrix inverse equal the transpose; this is the computational convenience supplied by a Cartesian basis.

Every algebra character of the Wiener algebra is evaluation at a unique point of the unit circle. Writing $u(w)=w$, both $u$ and $u^{-1}$ have norm one. Thus $z=\varphi (u)$ satisfies $|z|\le 1$ and $|z^{-1}|\le 1$, hence $|z|=1$. Continuity then gives $\varphi ({\sum }_{n\in \mathbb{Z}}{a}_n{u}^n)=\sum a_n{z}^n$. Conversely, evaluation defines a continuous algebra character. The resulting bijection is a homeomorphism for the Gelfand topology.

For $u\in H_0^2(U)$, ${\sum }_{i,j}\Vert {\partial }_{ij}u{\Vert }_2^2=\Vert \Delta u{\Vert }_2^2$. Two integrations by parts prove this for compactly supported test functions, and $H^2$ density extends it to the clamped second-order Sobolev space. Together with the Poincare-Wirtinger inequality applied to the mean-zero first derivatives and the zero-boundary Poincaré inequality, it controls the full $H^2$ norm by $\Vert \Delta u{\Vert }_2$. The boundary conditions are essential to this exact identity on bounded domains.

This is the closure of the compactly supported test functions $C_c^{\infty }(U)$ in the $H^2(U)$ norm. On a smooth bounded domain it consists precisely of the $H^2$ functions with zero value and zero normal derivative traces. Tangential first derivatives then vanish as well. The clamped Hessian identity makes the Laplacian norm equivalent to the full $H^2$ norm on this space, enabling a Lax-Milgram theorem treatment of the clamped biharmonic problem.

A vector subspace closed in the norm topology of a Hilbert space is complete in the inherited norm and inner product. Its orthogonal complement gives a decomposition of the ambient Hilbert space into an orthogonal direct sum. This closedness permits orthogonal projection and ensures limits used in range arguments remain in the subspace. Every finite-dimensional vector subspace is closed.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

For a nonnegative partition-normalized B-spline basis, the synthesis linear map $Tc={\sum }_j{c}_j{N}_j$ has operator norm one. Its condition number is consequently $\Vert T^{-1}\Vert$, with the supremum norm on the spline space and the maximum norm on its coefficients. The uniform-norm stability of a B-spline basis bounds this number independently of the spline knot sequence when the order is fixed. This is different from the condition number of a particular sampled B-spline collocation matrix.

In an ordinary least squares regression with an intercept and a nonconstant response, $R^2=1-\mathrm{RSS}/{\sum }_i(Y_i-\overline{Y}{)}^2$. It measures the fitted proportion of sample variation relative to an intercept-only baseline. It increases when regressors are added, even if those regressors are unhelpful for prediction. A high value does not establish the assumptions of a normal linear model or rule out a systematic regression residual pattern.

For the five-point example in linear regression, the total centered sum of squares is $10.8$ and the residual sum of squares is $0.8$, so $R^2=25/27\approx 0.926$. This summarizes the fitted reduction in squared error. It does not record the order or shape of the regression residuals, which are worth examining in a residual-versus-fitted plot.

Here is a fully constructed illustration. Use $x=(-2,-1,0,1,2)$ and the same fitted line $3+x$ for both datasets. Add respectively the residual vectorsBoth sum to zero, both have zero inner product with $x$, and both have squared norm $0.56$. Their ordinary least squares fit is therefore exactly the same line, with the same $R^2=10/(10+0.56)=125/132\approx 0.947$. But the first residual is $0.2(x^2-2)$, a systematic quadratic shape, while the second changes direction across the predictor values. These examples show directly which information the scalar $R^2$ discards; neither constructed pattern proves a random-error model.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

If a velocity operator satisfies $\mathrm{Re}⟨Lh,h⟩\ge \delta \Vert h{\Vert }_2^2$, then sufficiently regular solutions of ${\partial }_{t}f+v\cdot {\nabla }_{x}f+Lf=0$ obey the displayed estimate when spatial boundary fluxes vanish. The free-transport part contributes zero to the energy derivative. Integrating the resulting differential inequality avoids dividing by a possibly zero norm.