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For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and define the sign of a permutation bySince $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$,so this is a homomorphism. A transposition interchanges two columns of the identity matrix and therefore has determinant $-1$. If $\sigma$ is expressed as a product of $r$ transpositions, then $\mathrm{sgn}(\sigma )=(-1{)}^r$; because the determinant depends only on $\sigma$, the parity of $r$ is independent of the chosen expression.

For $I\subseteq [m]$, the independent vectors $\{v_i:i\in I\}$ lie in the coordinate subspace supported onwhich has dimension $|S_I|$. Hence $|I|\le |S_I|$, and the stated form of Hall marriage theorem gives an injection $f$ with $f(i)\in \mathrm{supp}(v_i)$.

Let $V$ be the $n\times m$ matrix whose $i$th column is $v_i$. Its column rank is $m$. By equality of row and column rank, it has $m$ linearly independent rows. Let $B$ be their indices. The $m\times m$ submatrix $V_B$ is invertible, so its columns are independent. Those columns are exactly the nonzero coordinates of $B{v}_1,\dots ,B{v}_m$, and therefore these truncated vectors are linearly independent.

Expanding $\det V_B$ gives a permutation $f:[m]\to B$ for whichThus $f(i)\in \mathrm{supp}(v_i)$. Finally, order the coordinates with $B$ first. The matrix whose columns are the $v_i$ together with the $e_j$ for $j\notin B$ is block triangular, with diagonal blocks $V_B$ and an identity matrix. Its determinant is nonzero. Consequentlyis a basis of ${\mathbb{C}}^n$. This is the simultaneous basis exchange from a nonzero minor.

A B-spline $N_j$ is positive exactly in the interior of its support $(t_j,t_{j+k})$. If the B-spline collocation matrix $A_{\mathbf{x}}$ is invertible, its determinant contains a nonzero permutation term. Hence there is a permutation $\pi$ such that

If $x_i\le t_i$, then every one of the first $i$ points satisfies $x_r<t_i$. Any support containing such a point must have left endpoint $t_j<x_r$, hence $j<i$. Only $i-1$ B-splines are available to match these $i$ rows, contradicting that $\pi$ is a permutation. Thus $x_i>t_i$. Similarly, if $x_i\ge t_{i+k}$, each of the last $n-i+1$ points can only be matched to an index $j>i$, but only $n-i$ such indices exist. Therefore $x_i<t_{i+k}$. We have proved

The Schoenberg–Whitney theorem states, for strictly increasing knots and interpolation sites, thatIndeed its determinant is positive under these inequalities.

Put $A_j=\left(\begin{array}{cc}{a}_j& b_j\\ 1& 0\end{array}\right)$. The identityfollows by induction: multiplying the formula for $n-1$ by $A_n$ gives the two stated recurrences in the first column and $b_n(p_{n-1},q_{n-1}{)}^T$ in the second. The associated Möbius transformations $z\mapsto a_j+b_j/z$, applied from the right to $a_n$, show that the first-column ratio is the displayed positive generalized continued fraction.

Now set every $b_j=1$. Taking determinants in the matrix identity givesConsequently adjacent convergents differ byThe recurrence and $a_j\ge 1$ imply $q_{n+1}\ge q_n+q_{n-1}$, so $q_n\to \infty$. The determinant signs show that the even convergents increase, the odd convergents decrease, and every even one is below every odd one. Their adjacent separation tends to zero, so both subsequences have a common limit $x$. Since $x$ lies between each adjacent pair, the convergence of a positive simple continued fraction gives exactly

If all $a_j=1$ as well, the recurrence and initial values giveThe characteristic roots of $F_{n+2}=F_{n+1}+F_n$ are
$\varphi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2=-{\varphi }^{-1}$, henceTherefore $F_{n+2}/F_{n+1}\to \varphi$. Moreover,Using $F_n{\varphi }^{-n}\to 1/\sqrt{5}$ yieldsandThis is the all-one continued fraction and Fibonacci ratios case.

For a positive equilibrium, $Y^{\prime }=0$ gives $X^{\ast }=N_c$. The equations $Z^{\prime }=0$ and $X^{\ast }+Y^{\ast }+Z^{\ast }=N$ then give the Endemic equilibrium of the SIR model with demographywhich is strictly positive when $N>N_c$.

Use $Z=N-X-Y$ to reduce to the $(X,Y)$ system. At the endemic equilibrium its Jacobian isIts trace is negative and its determinant is $\beta Y^{\ast }(\mu +\nu )>0$. The two eigenvalues therefore have negative real part, so the endemic equilibrium is locally asymptotically stable.

The determinant is continuous, sois open in ${\mathbb{R}}^{n^2}$. It is therefore a smooth manifold of dimension $n^2$, and

The differential of determinant at $A$ isIt is nonzero at every $A\in SL(n,\mathbb{R})$, so the regular-level-set theorem makes $SL(n,\mathbb{R})$ a manifold of dimension $n^2-1$. At the identity,

Linearization about $X(t)$ givesOne Floquet multiplier of $\Phi (T)$ is $1$, corresponding to a phase shift along the orbit. Liouville's formula givesIn the plane this determinant is the other, transverse multiplier. The periodic orbit is transversely asymptotically stable when the displayed quantity is less than one, equivalently when the divergence integral is negative, and unstable when it is greater than one.

For two solutions of the Schrodinger 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 $R^2=1$.

Setand recursivelyThen $p_n/q_n=[a_0,ldots,a_n]$. The determinant identityfollows by induction. Writing the remaining complete quotient as ${\theta }_{n+1}>1$ givesso $\theta$ lies strictly between consecutive convergents and their order alternates. For odd $n$,Moreover $|\theta -p_n/q_n|<1/(q_n{q}_{n+1})$, and $q_n\to \infty$, proving convergence.

The continued-fraction algorithm gives

For a positive homogeneous equilibrium,sorequiring $\alpha >\beta$. The reaction Jacobian there isIts determinant is ${\beta }^2/(\alpha -\beta )>0$, while its trace is $\beta (1-\beta )/(\alpha -\beta )$. Stability therefore requiresThis is the region above the line $\alpha =\beta$ and to the right of $\beta =1$ in the positive quadrant.

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 $\lambda =1/\beta$.

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

Orthogonality and $\Vert \tilde{\varphi }(x)\Vert =\Vert x\Vert$ giveso ${\psi }_{\epsilon }$ maps each boundary sphere to itself. AlsoOn compact $A$, this derivative is uniformly close to the identity; for sufficiently small $\epsilon$ its determinant is positive and it is locally invertible. The map is properly homotopic to the identity and has degree one, so the local diffeomorphism has one sheet and is a global diffeomorphism $A\to A$.

Adding the two equations gives $\dot{x}+\dot{y}=b-x$, so a fixed point hasAt this point the Jacobian has determinant $a+b^2>0$ and traceThe fixed point is asymptotically stable exactly when $T<0$, orThe equality curve exists only for $a\le 1/8$ and isFor $0<a<1/8$ the fixed point is unstable between these two curves and stable outside them; for $a>1/8$ it is stable everywhere.

With $\eta =\mathrm{diag}(1,-1,-1)$, the defining condition for the $(1+2)$-dimensional Lorentz group isTaking determinants gives $(\det \Lambda {)}^2=1$, hence $\det \Lambda =\pm 1$. The norm of the zeroth column givesso $|{\Lambda }^0{}_0|\ge 1$. The Proper orthochronous Lorentz group is the component with $\det \Lambda =1$ and ${\Lambda }^0{}_0\ge 1$.

On $r=0$, the screen metric is $h_{ij}$ and $U={\partial }_{\lambda }$. Therefore the null expansion isThe derivative formula for the determinant givesand henceThus $\theta$ is the fractional rate of change of an infinitesimal transverse area carried along the generators: positive expansion enlarges the beam and negative expansion focuses it.

Let the conserved population be $S>0$. At equilibrium $Y^{\prime }=-\beta YN=0$, so $Y$ and $N$ cannot both be positive. The other equations then force $U=0$. The only equilibria are

Near $E_Y$, using $(N,U)$ as coordinates gives the linearizationIts determinant is $-\beta S\zeta <0$, so $E_Y$ is a saddle and is unstable. Near $E_N$, coordinates $(Y,U)$ give the triangular linearization with eigenvalues $-\beta S$ and $-\zeta$. Hence $E_N$ is asymptotically stable.

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