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Consider $F(A)=A^{T}A$ with values in the symmetric matrices. At the identity,which is onto the $n(n+1)/2$-dimensional space of symmetric matrices. The same rank holds at every orthogonal matrix. Thus $O(n)=F^{-1}(I)$ is a manifold, and its determinant-one component, the special orthogonal group, has dimensionThe tangent equation at the identity is $X^T+X=0$, so

For $-j\le m<j$, the matrix element of $[O,J_{+}]$ between $|j,m+1⟩$ and $|j,m⟩$ givesThe ladder coefficient is nonzero, so adjacent diagonal entries agree. Induction makes the diagonal entry independent of $m$.

Taking a matrix element of $[O,J_3]=0$ givesThus the matrix element vanishes unless $m=m^{\prime }$.

Since $O$ commutes with every angular momentum operator, it commutes with $J^2$. Taking a matrix element of $[O,J^2]=0$ givesThe allowed nonnegative values of $j$ make $j(j+1)$ injective, so the matrix element vanishes unless $j=j^{\prime }$.

The coupled basis consists of the tripletand the singletThese equations list all nonzero Clebsch--Gordan coefficients, each coefficient being $1$, $1/\sqrt{2}$, or $-1/\sqrt{2}$ as displayed, andBecause each irreducible total-spin representation occurs once, an operator commuting with total angular momentum isIn the product basis, the nonzero matrix elements areand

For real $E$ and real $V$, choose a real fundamental pair of solutions. The corresponding Floquet matrix has real entries, so $\mathrm{tr}F(E)$ is real. The trace is basis independent, so this holds for every fundamental pair.

For two solutions of the Schrödinger equation, their Wronskian is constant because the equation has no first-derivative term. The determinant of the Floquet matrix is the ratio of the Wronskian at $x+a$ to that at $x$, hence

In a linear model containing an intercept, the normal equations make the residual vector $e$ orthogonal to every design column. In particular $e\perp 1$, soup to floating-point rounding.

The second design matrix has $49$ random Gaussian columns plus the intercept, hence is a random $50\times 50$ matrix. It has full rank with probability one because the determinant vanishes only on a measure-zero algebraic set. Its column space is then all of ${\mathbb{R}}^{50}$, so the fitted vector equals $Y$, the residual sum of squares is zero, and $R^2=1$.

The matrix $U_{\theta }$ is real orthogonal. Part (a) givesExpanding the left side in the computational basis yields

With $p_i=\mathbb{P}({\mathrm{dis}}_i=1)$, the fitted logistic model isRisk category 1 is the reference level, so its effect is included in the intercept; adding a separate coefficient for both levels would make the design matrix linearly dependent.

For $q=k^2$, the mode matrix is $J-q\mathrm{diag}(D_1,D_2)$. Its trace remains negative. Its determinant isIt is negative for some $q>0$ exactly whenTogether with homogeneous stability, the spatial-instability region isFor fixed $\beta$, this is the region below the last curve, above $\alpha =\beta$, and left of $\lambda =1/\beta$.

If $D_1=D_2=D$, the linearized matrix for wavenumber $k$ isIts eigenvalues are those of the homogeneous Jacobian shifted left by $D{k}^2$, so a stable homogeneous equilibrium remains stable for every spatial mode. Thus equal diffusivities cannot produce a Turing instability. In part (b), equal diffusivities mean $\lambda =1$; because homogeneous stability requires $\beta >1$, the necessary condition $1-\lambda \beta >0$ fails, consistently ruling out instability.

A strongly regular graph with parameters $(k,a,b)$ is $k$-regular, with every adjacent pair having $a$ common neighbours and every distinct nonadjacent pair having $b$ common neighbours. Its adjacency matrix satisfiesOn the orthogonal complement of the all-one vector, the two possible eigenvalues areUsing $1+m_r+m_s=n$ and $k+m_{r}r+m_{s}s=0$ gives exactly the two displayed expressions for $m_r,m_s$. They are eigenspace dimensions and hence integers, proving the rationality condition.

The Petersen graph has parameters $(3,0,1)$, so its spectrum isIf three Petersen graphs partitioned $E(K_{10})$, their adjacency matrices would satisfy $A+B+C=J-I$. The five-dimensional eigenvalue-one spaces of $A$ and $B$ inside the nine-dimensional space $1^{\perp }$ intersect nontrivially. For a nonzero common vector $v$, $Av=Bv=v$ and thereforecontradicting the Petersen spectrum. No such partition exists.

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

The matrixis the orthogonal projection onto the row space of $X$. Since $XP\beta =X\beta$, the empirical risks of $h_{\beta }$ and $h_{P\beta }$ are equal. Orthogonality givesso $q(\beta )\ge q(P\beta )$, with strict inequality unless $(I-P)\beta =0$. Therefore the minimizer lies in the row space:Because $X^T$ is injective when $X{X}^T$ is invertible, minimizing over that row space is equivalent to minimizing $r(\alpha )=q(X^T\alpha )$.

Since $A_k=Q_k{R}_k$,Thus every step is an orthogonal similarity, so it preserves eigenvalues and symmetry.

For a symmetric $r$-banded matrix, its subdiagonal entries can be eliminated by Givens rotations ordered along the band. Each rotation creates only the next local bulge; multiplication in reverse order chases that bulge out without creating entries beyond the original upper band. Symmetry supplies the corresponding lower band. This is symmetric bandwidth preservation under QR iteration, and proves that $A_{k+1}$ is again $r$-banded.

The unshifted QR algorithm computes the eigenvalues of $A$. For a real symmetric matrix it drives $A_k$ toward a diagonal or block-diagonal matrix; the accumulated orthogonal factors simultaneously approximate its eigenvectors.

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

Set $y=tz$ and $u=(x,z{)}^T$. The system becomesThe matrix has eigenpairs $1,(2,1{)}^T$ and $-2,(1,1{)}^T$. A particular solution is $(-t/3-2,-1{)}^T$. Therefore

The eigenvalues of the recurrence matrix have modulus $e^{-\kappa T}<1$, so transients decay and the coefficients approach the fixed pointThusFor $T=k\pi$ and $\kappa k\ll 1$, this isEven multiples deliver impulses in phase and produce resonance; odd multiples alternate the phase and largely cancel.