Suppose $S^b=0$. Use the increasing chain of vector subspacesChoose a basis successively, extending a basis of each kernel of a linear map to one of the next kernel of a linear map. If a basis vector first enters at level $j$, its image under $S$ belongs to $\ker S^{j-1}$, spanned entirely by earlier basis vectors. Thus every matrix column has nonzero entries only above its diagonal: the matrix is strictly upper triangular.

Conversely, for such a matrix and its ordered basis $e_1,\dots ,e_n$, one has $S⟨e_1,\dots ,e_j⟩\subseteq ⟨e_1,\dots ,e_{j-1}⟩$. Iterating $n$ times kills every vector. Hence $\boxed{S^n=0}$, establishing both directions without assuming a Jordan form.

Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne 0$, then $0=R^{a}v={\lambda }^{a}v$, hence $\lambda =0$. The characteristic polynomial splits over $\mathbb{C}$ and has degree $n$, soThere is indeed an eigenvector for the eigenvalue zero: a nonzero vector in the last nonzero image of $R$ lies in its kernel of a linear map. For the bound, choose $v$ with $R^{a-1}v\ne 0$, using minimality. The vectors $v,Rv,\dots ,R^{a-1}v$ are linearly independent. Otherwise choose the least $j$ with nonzero coefficient in a relation and apply $R^{a-1-j}$; only its nonzero multiple of $R^{a-1}v$ survives, a contradiction. Therefore $\boxed{a\le n}$.

The equation is hyperbolic at a point when $\boxed{b^2-ac>0}$. The principal form has one positive and one negative eigenvalue and two distinct real characteristic directions. In the given equation $a=x^2$, $b=0$, $c=-y^2$, so $b^2-ac=x^2{y}^2>0$ throughout the positive quadrant, consistent with the two canonical characteristic families used above.

The equation is parabolic at a point when $\boxed{b^2-ac=0}$ and at least one of $a,b,c$ is nonzero. The principal matrix then has rank one, with a repeated characteristic direction. If all three vanish, there is no second-order principal part there and the usual parabolic classification does not apply.

The principal matrix is $\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)$. The equation is elliptic at a point when $\boxed{b^2-ac<0}$, equivalently the principal form has eigenvalues of the same nonzero sign and no real characteristic directions.

Choose a small positively oriented circle $|w|=\rho$ in $U$ within the convergence disk of the inverse series. Cauchy's coefficient formula givesLet $\gamma =g(\{|w|=\rho \})$. The biholomorphism property makes it a simple positively oriented closed curve about zero. Under $w=f(z)$, $g(w)=z$ and $dw=f^{\prime }(z)dz$, soThe last equality is (a). Apply (b), givingThis is the local Lagrange inversion theorem; the simple zero ensures that the expression being differentiated has a removable singularity and a nonzero value at zero.

For a positively oriented simple closed piecewise smooth curve, a point $a$ inside it and a function $h$ holomorphic on a neighborhood of the curve and its interior, Cauchy's differentiated integral formula isThe printed denominator uses $n+1$ while the derivative and factorial use $m$. The standard formula requires these indices to agree; the intended exponent in this display is $m+1$. The private transcription preserves the mismatch exactly.

For the following task, write $f(z)=zq(z)$ near zero with $q(0)=f^{\prime }(0)\ne 0$. Thus $H(z)=(z/f(z){)}^n=q(z{)}^{-n}$ extends holomorphically at zero, and is holomorphic inside the specified contour because $f$ has no other zeros there. Now $f(z{)}^{-n}=H(z)/z^n$. Apply the correctly indexed formula with derivative order $n-1$ to getPositive winding one supplies the orientation and multiplicity used in this expression.

The complex fundamental theorem of calculus says that for a holomorphic antiderivative $F$ and a piecewise $C^1$ path $\gamma$ from $a$ to $b$, one has ${\int }_{\gamma }{F}^{\prime }(z)dz=F(b)-F(a)$. The integral is therefore zero on a closed path; the antiderivative need only be defined on a neighborhood of that path, rather than on all its interior.

The assumptions make $F(z)=z/f(z{)}^n$ holomorphic near $\gamma$. Differentiate:Integrate this derivative around the closed path to obtain

Reduce the matrix modulo two. It is $A=1\oplus \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, so $A\ne I$ but $(A-I{)}^2=0$. Thus the minimal polynomial is $\boxed{m=(X+1{)}^2}$. Its monic divisors are $1,X+1,(X+1{)}^2$, giving exactly $\boxed{3\text{ ideals}}$ in the quotient.

This ring has the four elements $0,I,A,I+A$. Coprimality with $X+1$ is equivalent to the representing polynomial taking value one at $1$. Thus precisely $I$ and $A$ are units: $\boxed{2\text{ invertible matrices}}$. Indeed $A^2=I$, whereas $(I+A{)}^2=0$.

The displayed matrix is block diagonal with a $1\times 1$ block $1$ and a $2\times 2$ block $B$ having characteristic polynomial $X^2-4X-5=(X-5)(X+1)$. The latter block is not scalar, so its minimal polynomial has degree two and is exactly that polynomial. Taking the least common multiple of the block minimal polynomials gives $m=(X-1)(X-5)(X+1)$ over $\mathbb{R}$.

Ideals of $K[X]/(m)$ correspond to ideals of $K[X]$ containing $(m)$, namely the principal ideals generated by the monic divisors of $m$. Every one of the three distinct linear factors may be included or omitted, soThis includes both the zero ideal and the whole ring. Equivalently the Chinese remainder theorem identifies this ring with ${\mathbb{R}}^3$.

The assumption $v\notin W^{\perp }$ means $p=Pv\ne 0$. For every unit vector $w\in W$, orthogonality and Cauchy–Schwarz give $⟨v,w⟩=⟨p,w⟩\le \Vert p\Vert$. Equality is attained atEquality in the real Cauchy–Schwarz inequality requires $w$ to be a scalar multiple of $p$, and attaining the positive supremum fixes its sign; the unit norm fixes its magnitude. This proves uniqueness.

For any $w\in W$, the decomposition $v-w=Tv+(Pv-w)$ is orthogonal. The Pythagorean identity givesEquality is attained at $w=Pv$. Thus $\boxed{\underset{w\in W}{\inf }\Vert v-w\Vert =\Vert Tv\Vert }$, and the minimizer in $W$ is uniquely $Pv$.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha >0,\beta \ne 0}$. The exponential factor expands while the imaginary part produces rotation, giving an unstable spiral.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha <0,\beta \ne 0}$. The factor $e^{\alpha t}$ contracts while the imaginary part produces rotation, giving a stable spiral.

For the linearized system a centre occurs when $\boxed{{\lambda }_{1,2}=\pm i\beta ,\beta \ne 0}$. Its real solutions are periodic ellipses after an invertible linear coordinate change.

As explained by a centre in the linearization need not be a nonlinear centre, for a nonlinear system these eigenvalues alone do not guarantee a centre. For example $\dot{X}=-Y-X(X^2+Y^2)$, $\dot{Y}=X-Y(X^2+Y^2)$ has eigenvalues $\pm i$ at zero but polar equations $\dot{r}=-r^3$, $\dot{\theta }=1$, so trajectories spiral inward. Removing the cubic terms gives a genuine centre with the same Jacobian. Thus there is no eigenvalue-only sufficient condition for a nonlinear centre in the stated generality.

The eigenvalues are real and have opposite signs: $\boxed{{\lambda }_1{\lambda }_2<0}$. One eigendirection contracts and the other expands, giving a saddle.

Both eigenvalues are real and positive: $\boxed{{\lambda }_1>0,{\lambda }_2>0}$. The real eigendirections carry exponentially growing factors, giving an unstable node.

Both eigenvalues are real and negative: $\boxed{{\lambda }_1<0,{\lambda }_2<0}$, with distinctness as assumed. The two independent real eigendirections carry exponentially decaying factors $e^{{\lambda }_{j}t}$, giving a stable node.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha >0,\beta \ne 0}$. The exponential factor expands while the imaginary part produces rotation, giving an unstable spiral.

The eigenvalues are a nonreal conjugate pair $\boxed{\alpha \pm i\beta ,\alpha <0,\beta \ne 0}$. The factor $e^{\alpha t}$ contracts while the imaginary part produces rotation, giving a stable spiral.