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For a unit vector $v$, a Householder reflection is $H=I-2v{v}^T$. Clearly $H^T=H$, and $H^{T}H=(I-2v{v}^T{)}^2=I$.

The vector $v$ has eigenvalue $-1$, while every vector in $v^{\perp }$ has eigenvalue $1$. Their multiplicities are one and $n-1$, respectively.

Let $S_{xx}=\sum X_i^2$. The likelihood equations giveThese are orthogonal projections of a Gaussian vector onto the span of $X$ and its orthogonal complement, hence are independent.

The reverse-regression estimate is $\hat{b}=\sum X_i{Y}_i/\sum Y_i^2$. Thereforeby Cauchy–Schwarz. Equality means the two data vectors are proportional, which is exactly when both fitted residual sums, and hence both variance MLEs, vanish.

For probability vectors $p,q$, they are optimal with value $v$ exactly whenand $p^{T}Aq=v$. These are the mutual best-response and security-level conditions.

For every probability vector $x$, antisymmetry gives $x^{T}Ax=0$. Applying the minimax inequalities with the same strategy on both sides shows the lower value is at most zero and the upper value at least zero; transposition also changes the value to its negative. Hence the value is zero.

Put $u=\sqrt{m}\bar{X}$ and $v=\sqrt{n}\bar{Y}$. Test (a) accepts inside the discwhereas $Z$ is the projection of $(u,v)$ onto the unit vectorIf $q_1=[{\Phi }^{-1}(1-\alpha /2){]}^2$, then $q_1$ is the corresponding ${\chi }_1^2$ quantile. Since a ${\chi }_2^2$ variable is stochastically larger than a ${\chi }_1^2$ variable, $q_2>q_1$. Choose a point on that unit-vector line with squared radius strictly between $q_1$ and $q_2$. A whole open neighborhood then makes (b) reject while (a) accepts. The joint normal density of $(\bar{X},\bar{Y})$ is strictly positive everywhere for every true $(\lambda ,\mu )$, so this neighborhood always has positive probability.

The symmetric bilinear form associated with a real quadratic form $Q$ is obtained by polarization identity:In coordinates $Q(x)=x^{T}Ax$, replacing $A$ by its symmetric part does not change $Q$, and the formula gives $phi(u,v)=u^{T}Av$; this proves existence. A symmetric form is positive semidefinite when $phi(v,v)\ge 0$ for every $v$, and positive definite when the inequality is strict for every $v\ne 0$.

The diagonalization theorem for real quadratic forms says that some basis puts any symmetric form intowith $r$ further zero coordinates. Sylvester's law of inertia says that $(p,q,r)$ is independent of the diagonalizing basis. To prove this, let $P$ be the span of the positive coordinate vectors and let $N\oplus Z$ be the span of the negative and zero vectors. If $P^{\prime }$ is positive definite for another diagonalization and $\dim P^{\prime }>p$, thenso the two spaces intersect nontrivially. A vector in the intersection would have both positive and nonpositive square, a contradiction. Thus $p^{\prime }\le p$; symmetry gives $p=p^{\prime }$. Applying the same argument to $-\varphi$ gives $q=q^{\prime }$, and then $r=r^{\prime }$.

For the nondegenerate form on $V$, write its inertia as $(p,q,0)$, so $p+q=2n$. The restriction vanishes identically on $E$: polarization gives $phi(u,v)=0$ for $u,v\in E$. Projection of $E$ to the positive coordinate space is injective, since a vector with zero positive projection cannot be isotropic unless it is zero. Hence $k\le p$; projection to the negative space likewise gives $k\le q$. Therefore $k\le \min (p,q)\le n$.

The matrix of $l^2$ is $l{l}^T$. If $l\ne 0$, it has rank one and inertia $(1,0,n-1)$, hence signature one; if $l=0$, its rank and signature are zero. The coefficientwise product $(l^2,s^2)$ has matrix $t{t}^T$, where $t_i=l_i{s}_i$, so it has the same rank-one conclusion when $t\ne 0$ and is zero otherwise.

Finally, diagonalization writes every positive semidefinite form as a sum of squares, say $f={\sum }_a{l}_a^2$ and $g={\sum }_b{s}_b^2$. Bilinearity of coefficientwise multiplication givesa sum of positive semidefinite rank-at-most-one forms. Thus $(f,g)$ is positive semidefinite.

For every product vector,Product vectors span the two-register state space, so the single $A\otimes B$ gate equals parallel application of $A$ to the first register and $B$ to the second. Unitarity guarantees that both descriptions are valid quantum circuits.

Write $Y_i=X_i^T\beta +{\epsilon }_i$, where conditionally $\mathbb{E}({\epsilon }_i|X_i)=0$ and $\mathrm{Var}({\epsilon }_i|X_i)={\sigma }^{2}v(X_i)$. ThenThe law of large numbers sends the matrix towhile the vector tends to zero, proving consistency. The central limit theorem and Slutsky's theorem give

Similarly, with $B=\mathbb{E}(X{X}^T)$ and $C=\mathbb{E}(v(X)X{X}^T)$,These conclusions require the displayed matrices to be finite and nonsingular.

Let $1<p<\infty$ and $q=p/(p-1)$. Every $y\in {\ell }^q$ definesHölder's inequality gives $\Vert f_y\Vert \le \Vert y{\Vert }_q$, and equality follows by using a normalized sequence proportional to $\mathrm{sgn}(y_n)|y_n{|}^{q-1}$.

Conversely, for $f\in ({\ell }^p{)}^{\ast }$ put $y_n=f(e_n)$. On the first $N$ coordinates, choose the same norming vector to obtainLetting $N\to \infty$ shows $y\in {\ell }^q$. On finitely supported sequences, linearity gives $f=f_y$; density of those sequences in ${\ell }^p$ extends this identity to all $x$. Thereforeisometrically.

If $P=P^T=P^2$ and $\mathrm{tr}P=1$, the spectral theorem gives eigenvalues $1,0,0$. Thus $P$ is orthogonal projection onto a line andfor a unit vector $X$; the only choices are $X$ and $-X$.

The smooth map $\pi :S^2\to S_3(\mathbb{R})$, $X\mapsto X{X}^T$, has image $G{r}_{1,3}(\mathbb{R})$ and fibers $\{X,-X\}$. Near $X_0$, restrict to the hemisphere $X\cdot X_0>0$; there $\pi$ is one-to-one and has smooth inverse obtained by choosing the positive local unit eigenvector. Combining this with a chart of $S^2$ supplies two-dimensional local parametrizations. Hence $G{r}_{1,3}(\mathbb{R})$ is a two-dimensional submanifold, diffeomorphic to $\mathbb{R}{P}^2$.

Yes. On $Q\le 1$ the same inequality gives $\dot{Q}\le 0$, so the closed ellipse is positively invariant. Equality in $\dot{Q}=0$ occurs only at the origin and at the boundary pointAt that point the vector field isso $\{p\}$ is not invariant. The largest invariant subset of $\{\dot{Q}=0\}$ is therefore the origin. LaSalle's invariance principle shows that every trajectory in the closed ellipse converges to the origin. The inequality may consequently be extended to $Q\le 1$.

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.

Let $E_2=(e_1,e_2)$ andBecause $Q_k{R}_k$ is the QR factorization of $H^{k+1}$, $S_k$ is the span of the first two columns $q_1^{(k)},q_2^{(k)}$ of $Q_k$.

The first row of the tridiagonal eigenvalue equation isIn the generic case ${\lambda }_{n-1}\ne {\lambda }_n$, the two dominant coefficient vectors are independent, sinceHere $h_{12}\ne 0$, since otherwise the displayed eigenvalue equation and the nonzero $b_j$ would force all eigenvalues to equal $h_{11}$. The strict spectral gap below the dominant pair now impliesThis is two-dimensional subspace iteration. If the dominant eigenvalue is repeated, $H$ acts as a scalar on $W$; the same QR argument either captures all of $W$ or separates its missing direction from the strictly smaller eigenspaces, and the conclusion below is unchanged.

Since $q_2^{(k)}\in S_k$ and $q_3^{(k)}\perp S_k$,The space $W$ is invariant under $H$, soShifting the index does not affect the limit, and thereforeThus the unshifted QR algorithm asymptotically deflates a $2\times 2$ block associated with the equal-modulus dominant pair.