Differentiating the ridge regression objective givesFor every nonzero $z$,so $X^{T}X+\lambda I_p$ is positive definite and hence invertible. The strictly convex objective therefore has the unique closed-form ridge regression estimator

The Gaussian likelihood is proportional toso maximum likelihood estimation is equivalent to ordinary least squares. If $p\le n$ and $X$ has rank $p$, then $X^{T}X$ is invertible and the normal equations giveIf $p>n$, then $\mathrm{rank}X<p$ and $\ker X\ne \{0\}$. A least-squares minimizer always exists, but it is not unique: by rank-deficient ordinary least squares, the complete set isan affine subspace of dimension $p-\mathrm{rank}X>0$. Thus there are uncountably infinitely many likelihood maximizers. In the generic full-row-rank case their dimension is $p-n$.

Let $\nu$ be an invariant distribution for the continuous-time Markov chain. Using $Q=D_g(P-I)$, its stationarity equation $\nu Q=0$ becomesThus the measure with weights ${\nu }_{i}g(i)$ is invariant for the irreducible positive-recurrent jump chain. Its invariant distribution $\pi$ is unique, so ${\nu }_{i}g(i)=c{\pi }_i$. Normalization gives the invariant-measure transfer between a jump chain and a CTMC formulaThe denominator is finite and nonzero because $\epsilon <g(j)<1/\epsilon$. Thus $\nu$ is a probability distribution and $(X_t)$ is positive recurrent.

A continuous-time Markov chain is irreducible if every state can reach every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the chain returns to $i$ with probability one; an irreducible chain is recurrent if one, and hence every, state is recurrent. A recurrent state is positive recurrent if its expected continuous first return time is finite, and an irreducible chain is positive recurrent when all its states are.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

For $d=1$ and $X\sim N(0,{\sigma }^2)$, the characteristic function isFor a multivariate normal distribution with covariance matrix ${\sigma }^2{I}_d$, the coordinates are uncorrelated and hence independent by independence of uncorrelated jointly normal variables. ThereforeThis is the characteristic function of an isotropic Gaussian vector.

Choose $Q\in SO(n+1)$ with $Q{e}_{n+1}=v$. Conjugation by $Q$ identifieswithIt follows that the unit-vector stabilizer in a special orthogonal group is an embedded submanifold diffeomorphic to $SO(n)$, andAt $R\in S_v$, its tangent space is

For any $v\ne w$, both $S_v$ and $S_w$ contain the identity. At the identity,Identify $\mathfrak{so}(n+1)$ with the exterior square ${\Lambda }^2{\mathbb{R}}^{n+1}$. The orthogonal complement of $T_I{S}_v={\Lambda }^2(v^{\perp })$ is $v\wedge v^{\perp }$. If $v$ and $w$ are linearly independent, the nonzero bivector $v\wedge w$ belongs to the orthogonal complements of both tangent spaces. Their sum is therefore a proper subspace of $T_{I}SO(n+1)$, so the intersection is not transverse. If $w=-v$, then $S_w=S_v$, whose tangent space is proper, so they are again not transverse. Consequentlyas summarized by distinct unit-vector stabilizers are not transverse.

Let ${\mathrm{Sym}}_n$ be the vector space of real symmetric matrices and considerAt $R\in O(n)$,This derivative is surjective: for $S\in {\mathrm{Sym}}_n$, choosing $H=RS/2$ gives $D_R\Phi (H)=S$. Thus $I$ is a regular value and the orthogonal group $O(n)={\Phi }^{-1}(I)$ is a submanifold of codimensionThe determinant takes only the values $\pm 1$ on $O(n)$, so the special orthogonal group $SO(n)=\{R\in O(n):\det R=1\}$ is an open-and-closed submanifold of $O(n)$. Therefore

The tangent space is the kernel of $D_R\Phi$:where the Special orthogonal Lie algebra $\mathfrak{so}(n)$ is the vector space of skew-symmetric matrices. This proves the special orthogonal group as a submanifold description.

Define the smooth functionAt $x\in F^{-1}(1)$ its derivative isThis linear map is surjective because $x\ne 0$; for example, $D_{x}F(x)=2$. Hence one is a regular value, and the regular level set theorem makesan embedded codimension-one submanifold of ${\mathbb{R}}^{n+1}$. Its tangent space is the kernel of the derivative:This is the unit sphere as a regular level set.

Choose a $K$-basis $e_1,\dots ,e_n$ of $L$ and writethereby identifying $L$ with ${\mathbb{A}}_K^n$. Let $M(x)$ be the $n\times n$ matrix whose $j$th column is the coordinate vector of ${\alpha }^{j-1}$ in this basis, for $1\le j\le n$.

Write the multiplication table aswith fixed structure constants $c_{ij}^l\in K$. Repeated multiplication shows that every entry of $M(x)$ is a polynomial in $x_1,\dots ,x_n$. HenceBy the criterion supplied in the question,Consequentlywhich is a distinguished open set, and therefore a Zariski-open set. This is the determinant construction showing that primitive elements form a principal Zariski-open set. No assertion that $D$ is nonzero is needed.