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A surjective isometry $u:V\to W$ between normed vector spaces preserves every distance:

The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

The Euclidean norm of $x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ is

The trace norm of a finite-dimensional operator $A$ is the sum of its singular values:For a Hermitian operator $A=A_{+}-A_{-}$, it equals $\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$.

For $x\in {\mathbb{R}}^d$, the L1 norm is

This applies normalization of an algebraic curve to the two local branches of a node.

Normalization separates a node's two branches; a rational nodal cubic has normalization ${\mathbb{P}}^1$ and two points above its node.

The geometric genus of an irreducible curve is the genus of its smooth projective normalization.

A unit normal is a normal vector of Euclidean norm one. Given a nonzero normal vector $n_0$, normalization gives $n=n_0/\Vert n_0\Vert$. This makes a dot product $w\cdot n$ measure the signed length of the normal component, without an extra factor from the length of $n_0$. On a hypersurface, the two opposite choices $n$ and $-n$ correspond to opposite normal orientations; a continuous choice on an orientable surface fixes its orientation of a surface.

A closed subset $A$ of a normal space is a countable intersection of open sets exactly when it is the zero set of some continuous function $f:X\to [0,1]$.

Under a surjective homomorphism, the image of a normal subgroup is normal.

The preimage of a normal subgroup under any group homomorphism is normal.

A weak shock has a small relative pressure jump. Its density jump agrees with the reversible adiabatic prediction through second order in the pressure jump, while entropy production first appears at third order.

In the rest frame of a one-dimensional shock, states with velocities $w_0,w_1$ satisfyand

These are the Rankine-Hugoniot conditions specialized to the mass, momentum and energy fluxes of a perfect gas.

The shock frame is an inertial reference frame in which the shock is stationary. A travelling shock then becomes a steady flow through a fixed discontinuity.

A one-way normal model assigns each factor level its own mean while assuming independent Gaussian errors with a common variance.

This Gaussian model is the setting for a one-way instance of analysis of variance.

For fixed coefficient dimension and $Y=X{\beta }_0+\epsilon$, the ordinary least squares error satisfiesConsequently convergence in probability of $X^{\mathsf{T}}X/n$ to a positive-definite matrix $Q$, together with $X^{\mathsf{T}}\epsilon /n\to 0$ in probability, proves statistical consistency by continuity of matrix inversion.

For independent and identically distributed random-design pairs, finite second moments and $\mathbb{E}[x_i{\epsilon }_i]=0$ imply these limits by the law of large numbers, provided $Q=\mathbb{E}[x_i{x}_i^{\mathsf{T}}]$ is nonsingular. Conditional mean zero, $\mathbb{E}[{\epsilon }_i\mid x_i]=0$, is a familiar sufficient exogeneity condition. For a deterministic design matrix, the vanishing normalized design-error product must instead be justified from its errors and weights.

Finite moments and a nonsingular design alone are insufficient. In an intercept-only model with errors identically one, the estimated intercept is always ${\beta }_0+1$. The missing zero-expected value condition changes the limiting target.

Least-squares differentiation gives $X^{T}X\hat{\beta }=X^{T}Y$, or its precision-weighted analogue.

Weighted least squares minimizes ${\sum }_i{w}_i(Y_i-X_i^T\beta {)}^2$; inverse-variance weights give the efficient estimator under known heteroscedasticity.

Let $Y=X\beta +\epsilon$ with a fixed full column rank design matrix, $\mathbb{E}\epsilon =0$ and $\mathrm{Cov}(\epsilon )={\sigma }^{2}I$. Then ordinary least squares is a best linear unbiased estimator: every response-linear unbiased estimator has at least as much variance in every coefficient direction. A normal distribution is unnecessary.^[ref]

For a proof put $A=(X^{\mathsf{T}}X{)}^{-1}{X}^{\mathsf{T}}$. A competing estimator $CY$ is unbiased for every $\beta$ exactly when $CX=I$. Write $C=A+D$; then $DX=0$, so the cross terms vanish andThis positive semidefinite matrix inequality proves the simultaneous directional comparison. The scope is linear unbiased estimators, not all conceivable estimators: a biased ridge regression estimate may improve mean squared error for some signals. With correlated or unequal-variance errors, the corresponding known-covariance result uses generalized least squares.

For a linear regression design matrix $X$ with full column rank, the ordinary least squares coefficient vector is $\hat{\beta }=(X^{T}X{)}^{-1}{X}^{T}Y$. In a normal linear model with error covariance matrix ${\sigma }^{2}I$, it is an unbiased estimator with covariance matrix ${\sigma }^2(X^{T}X{)}^{-1}$. Replacing ${\sigma }^2$ by the residual variance estimate gives the reported coefficient standard errors.

In simple linear regression with a nonconstant predictor,