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The subspace $y^{\perp }$ is fixed by $M=I-x{y}^{†}$, while the remaining determinant factor is $1-y^{†}x$. Thus $\det M=1-y^{\ast }\cdot x$.

Apply the matrix determinant lemma to $HP-\lambda I=(H-\lambda I)-Hu,u^T$: its determinant is ${\chi }_H(\lambda )[1-u^T(H-\lambda I{)}^{-1}Hu]$.

$A=\left(\begin{array}{cc}{x}_3& x_1-i{x}_2\\ x_1+i{x}_2& -x_3\end{array}\right)$, so $\mathrm{tr}A=0$ and $\det A=-|x{|}^2$. Unitary conjugation preserves Hermiticity, trace, and determinant; every traceless Hermitian $2\times 2$ matrix is $x\prime \cdot \sigma$ with real $x\prime$, and $|x\prime |=|x|$.

The inverse function theorem says that a $C^1$ map with invertible derivative at $x$ restricts to a $C^1$ diffeomorphism between neighbourhoods of $x$ and $f(x)$. It makes every zero isolated. Compactness of $\bar{\Omega }$, together with separation from zero on the boundary, then makes the zero set finite and interior. Continuity of $Df$ gives the displayed radii. Choose disjoint balls around the zeros and a positive lower bound for $|f|$ on their complement and boundary. A sufficiently small $C^1$ perturbation has no zeros outside, while the quoted local lemma gives exactly one zero in each ball with the same determinant sign. Summing proves $N(g)=N(f)$.

Generators define a surjection $R^k\to M$, and conversely images of the standard basis generate any quotient. Writing $\varphi (m_i)={\sum }_j{a}_{ij}{m}_j$ with $a_{ij}\in I$ and applying the adjugate to $XI-A$ gives a monic annihilating polynomial whose lower coefficients lie in $I$. Taking $\varphi =1$ when $IM=M$ yields $xM=0$ with $x-1\in I$; for $M=\mathbb{Z}/3$ and $I=(2)$ take $x=3$. The Jacobson radical criterion follows by placing a nonunit $1-rs$ in a maximal ideal. It yields Nakayama’s lemma. Applying the determinant trick to preimages of a finite generating set constructs a polynomial right inverse to any surjective endomorphism, proving injectivity.

PutAfter ordering $I_1$ before $I_2$, the only nonzero off-diagonal entries between the two blocks occur at $(i,j)$ and $(j,i)$, where they equal $-a$. Thuswhere the coordinate vectors in the two blocks are understood. Expanding the determinant according to whether neither or both cross-block entries are selected givesThe minus sign is the sign of the transposition pairing the two cross-block entries. This formula remains valid when either diagonal block is singular, so no inverse or Schur complement is needed.

At the origin the Jacobian isThus it is a saddle and unstable. At the positive equilibrium the trace is negative. The product of the off-diagonal entries is $\nu \hat{\nu }$ by the nullcline identities, while the product of the magnitudes of the diagonal entries is strictly larger, so the determinant is positive. Both eigenvalues therefore have negative real part and the endemic equilibrium is stable.

Apply the formula to the complete quotient and use the determinant identity to express $b_n$. Eliminating the complete quotient from the consecutive expressions yields the stated quadratic for $q_{n+1}/q_n$. Taking the compatible signs for adjacent indices and using $c_n\le b_j$ gives

Fixed points have $y=0$ and $x^2-bx-a=0$. They coalesce on $a=-b^2/4$, the saddle-node curve. Away from it, $x_{\pm }=(b\pm \sqrt{b^2+4a})/2$; the Jacobian has trace $-1$ and determinant $2x-b$, so $x_{+}$ is stable and $x_{-}$ a saddle. At $(a,b)=(0,0)$ the branches $x=0$ and $x=b$ cross in a transcritical bifurcation.

The two nonzero proper invariant subspaces areBoth are preserved by the rotation. To show there are no others, note thatis the linear projection onto $U$, and $I-P$ is the linear projection onto $W$. Every invariant subspace $L$ is preserved by $I,T^2$, and hence by both of these linear projections. ThereforeBecause $U$ has dimension one, $L\cap U$ is either $0$ or $U$. If $L\cap W$ contains a nonzero $v=(a,b,0)$, it also contains $Tv=(-b,a,0)$. These two vectors are linearly independent, because their two-dimensional determinant is $a^2+b^2>0$. They span $W$, so $L\cap W$ is either $0$ or $W$. The only possibilities for $L$ are therefore $0,U,W,V$.

Finally every $(x,y,z)$ has the unique decomposition $(0,0,z)+(x,y,0)$, with one summand in each of $U,W$. HenceThis also illustrates how invariant subspaces of a quarter-turn in three dimensions can be classified by linear projections.

Over the complex numbers, the characteristic polynomial splits into three distinct roots of a polynomial, $1,i,-i$. Direct multiplication gives the eigenvectorsThey are linearly independent: the last two have two-dimensional determinant $2i\ne 0$ in the first two coordinates, and the first has nonzero third coordinate. ThusThis is a basis of eigenvectors of ${\mathbb{C}}^3$, proving that $T^{\prime }$ has a diagonalizable matrix.

The gradient of the given polynomial isIts second component vanishes when $y=0$ or $x=-1$. The first component then gives exactly three critical points:The Hessian matrix and its determinant areAt $(-1/2,0)$ the Hessian matrix is $\mathrm{diag}(2,1)$, whose positive definiteness proves a strict local minimum by the second-derivative test. At $(-1,\pm 1)$ its determinant is $-4$, so both points are saddle points of a scalar function. Their values and classifications areThere are no local maxima.

Suppose a line $T$ through $\mathbf{y}$ meets each $L_i$. Its intersection with $L_i$ is different from $\mathbf{y}$, so $T$ lies in ${\Pi }_i$. A nonzero direction vector $\mathbf{d}$ for $T$ must therefore satisfyA square matrix admitting a nonzero vector $\mathbf{d}$ with $M\mathbf{d}=0$ has zero determinant. Expanding givesThus the necessary condition isThe condition is not sufficient. Take $\mathbf{y}=(2,1,0)$, which is on none of the three original lines. It satisfies the condition, since $2+1-2=1$. The three plane equations becomeTheir common intersection is the single line $T=\{(s,1,0):s\in \mathbb{R}\}$. Any proposed line through $\mathbf{y}$ meeting all three original lines would have to be this $T$. But $T$ never meets $L_1=\{(s,0,1):s\in \mathbb{R}\}$: their second and third coordinates differ. The zero determinant guarantees a common direction for the three planes; it does not guarantee the required intersections. This is a parallel obstruction to a transversal of three skew lines.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Write $u=(a,b{)}^T$, $v=(c,d{)}^T$ and $\Delta =ad-bc$. If $xu+yv=0$, thenEliminating $y$ gives $\Delta x=0$, and eliminating $x$ gives $\Delta y=0$. Hence $\Delta \ne 0$ implies linear independence.

For the converse, suppose $\Delta =0$. If $u=0$, the family already contains the zero vector and is linearly dependent. If $a\ne 0$, the equality $ad=bc$ gives $v=(c/a)u$. If $a=0$ but $u\ne 0$, then $b\ne 0$ and $bc=0$ forces $c=0$, so $v=(d/b)u$. Both cases give a nontrivial linear relation. ThereforeThis is the two-dimensional determinant criterion, proved directly without assuming that criterion.

Choose an ordered basis $\mathcal{B}$ of $V$ and let $A=[T{]}_{\mathcal{B}}$. Define the characteristic polynomial byIf a different basis gives $B=P^{-1}AP$ for an invertible matrix $P$, multiplicativity of the determinant givesThus the definition is independent of the chosen basis. The alternate convention $\det (A-xI)$ differs by the constant factor $(-1{)}^{\dim V}$ and gives the same roots.

For vectors $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{R}}^3$, the scalar product, cross product and scalar triple product are respectivelyExpanding the dot product givesInterchanging $\mathbf{a}$ and $\mathbf{b}$ changes the sign of their cross product, so its dot product with $\mathbf{b}$ is also zero. Thus the cross product is orthogonal to both vectors, including when it is the zero vector.

The displayed scalar triple product is a determinant. A square matrix has nonzero determinant exactly when its rows are linearly independent. Hence

Subtracting the first row from the other two in the determinant givesThis vanishes exactly when the two displacements $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ are linearly dependent. If $\mathbf{a}\ne \mathbf{b}$, that means $\mathbf{c}-\mathbf{a}=t(\mathbf{b}-\mathbf{a})$ for some real number $t$. Thus all three points lie on the lineConversely, putting three points on any one line makes their displacements linearly dependent, so the determinant vanishes.

The equivalence remains true without distinctness. If exactly two points coincide, the two distinct locations determine a line containing all three, and two rows of the original determinant coincide. If all three points coincide, any line through that point works and the determinant is again zero. Existence persists, although uniqueness of the line can fail.

The value $f(\mathbf{a},\mathbf{b},\mathbf{c})$ is twice the signed area of the triangle traversed in that order; it is positive for counterclockwise orientation. Choose points on the parabolaFor $i<j<k$, subtracting the first row in the determinant and factoring givesEach of the three factors is positive, proving the required property for every integer $n\ge 3$.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.