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The determinant of the Hessian matrix is $4\cdot 2-3^2=-1<0$. Hence it has one positive and one negative eigenvalue (explicitly $3\pm \sqrt{10}$), so its quadratic form is indefinite. The cubic terms are of higher order and cannot remove these opposite local directions of change. ThereforeFor example the quadratic term is positive along $y=0$ but negative along $y=-3x/2$, confirming the saddle point of a scalar function classification.

Three vectors form a basis of three-dimensional space exactly when their scalar triple product is nonzero:It is the determinant of the matrix with these columns, so this condition is equivalent to linear independence and spanning the whole space.

Take the dot product of the expansion of $\mathbf{y}$ with $\mathbf{v}\times \mathbf{w}$. The terms containing $\mathbf{v}$ and $\mathbf{w}$ vanish, leaving $\mathbf{y}\cdot (\mathbf{v}\times \mathbf{w})=\alpha \Delta$. Cycling the same argument yieldsThe three cross-product numerators use the reciprocal basis.

Subtract the first row from the third. Expansion along the resulting row $(a-1,0,0)$ gives the determinantHence

The coefficient matrix has determinant $-1$, so it is an invertible matrix. Its homogeneous linear system therefore has only the zero solution:One can also subtract the first equation from the second to get $z=0$, then subtract the first from the third to get $y+2z=0$, giving $y=0$ and finally $x=0$. Thus there are no nonzero solutions.

Expand the determinant along the first row:The cubic terms cancel in the expansion.

Completing the square in the determinant givesfor every real $\alpha$. A square matrix is invertible exactly when its determinant is nonzero, so $A_{\alpha }$ is nonsingular for every real $\alpha$.

Differentiate the first ordinary differential equation and substitute the second:The original first equation also gives $bv=u^{\prime }-au$, soTherefore the requested coefficients areThey are the trace and determinant of the coefficient matrix. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u^{\prime }=au$ must also be retained when reconstructing the coupled system.

The determinant of $P$ is $2\cdot 1-1(-1)=3$. The two-by-two inverse matrix formula yields

Apply the determinant-preserving row operation $R_3\leftarrow R_3+3R_2$. The new third row is $(0,0,9-3\alpha )$, so expansion of the determinant along it givesThereforeThe parameter $\beta$ does not affect the determinant.

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.

Move the right-hand side to the left. The resulting homogeneous linear system has coefficient matrixA nonzero solution exists exactly when its determinant vanishes:Here the upper-left entry of the original matrix is $-1$, as verified in the original PDF; that entry is corrupted in the local OCR.

The determinant condition reduces to $3-2b=0$, givingAs a check, the first linear equation gives $y=2x$, and at this value of $b$ the second linear equation is satisfied for every such pair, including nonzero pairs.

If $Av=\lambda v$ for a nonzero eigenvector, then $A^{2}v={\lambda }^{2}v$. Since $A^2=A$, $\lambda (\lambda -1)=0$: every eigenvalue is zero or one. Let $m$ be the number of ones, counted with algebraic multiplicity. The trace is $m$ and the determinant is the product of these eigenvalues. ThereforeEvery output $y=Ax$ is fixed by a second application:The image of a linear map is thus the eigenspace for the eigenvalue one. To establish its dimension without assuming diagonalizability, decompose every input asThe first term lies in the image of a linear map $A$ and the second in its kernel. Their intersection is zero, because a vector in the image is fixed by $A$ whereas a vector in the kernel is killed by $A$. Thus the space is their direct sum, and an adapted basis puts $A$ into the form $\mathrm{diag}(I_r,0)$, where $r=\mathrm{rank}A$. Comparing traces gives $r=m$:This is an idempotent linear map, and trace of an idempotent equals its rank. If $m=0$, no nonzero output exists. The dimension in the question refers to the image subspace, including its zero vector.

Commuting inverse. Assume $I-M$ is invertible, as required by the displayed inverse. Multiplication givesMultiply on both sides by $(I-M{)}^{-1}$ to obtainMore generally, a matrix commuting with an invertible matrix also commutes with its inverse.

Orthogonality. For a real skew-symmetric matrix $A$, $x^{T}Ax=0$. If $(I-A)x=0$ for a real vector $x$, multiply by $x^T$ to obtain $\Vert x{\Vert }^2=0$. Thus $I-A$ is invertible; the same argument applies to $I+A$. For the Cayley transform of a real skew-symmetric matrix,The factors commute by the first result, soTherefore $\boxed{N\text{ is an orthogonal matrix}.}$

Eigenvector mapping. If $Av=av$, then $(I-A{)}^{-1}v=(1-a{)}^{-1}v$. ConsequentlyThe denominator cannot vanish. In fact the eigenvalues of a real skew-symmetric matrix are purely imaginary: $v^{†}Av$ is the negative of its complex conjugate, so $a=-\overline{a}$. For $a=i\beta$ the corresponding eigenvalue of $N$ is $(1+i\beta )/(1-i\beta )=e^{2i\arctan \beta }$, consistent with orthogonality.

Spectrum in three dimensions. Let $Q$ be any real three-by-three orthogonal matrix. For a complex eigenvector $v$, $\Vert Qv\Vert =\Vert v\Vert$ implies $|\lambda |=1$. A real cubic characteristic polynomial has a real root, necessarily $+1$ or $-1$. Any nonreal roots form a complex-conjugate pair of unit modulus. If all three roots are real, two of the three signs agree and can be grouped as a pair with angle zero or $\pi$. Thus in every case the eigenvalues can be writtenTheir product is $\epsilon =\det Q$.

Determinant and fixed axis for the Cayley transform. Since $A^T=-A$,so $\boxed{\det N=1}$. This determinant argument works in every dimension. In dimension three the preceding spectrum result then guarantees an eigenvector with $Nx=x$. Set $z=(I-A{)}^{-1}x$. The fixed-vector equation says $(I+A)z=(I-A)z$, so $Az=0$ and $x=(I-A)z=z$. HenceThis is the axis fixed by the three-dimensional rotation. Equivalently, a real odd-dimensional skew-symmetric matrix has zero determinant, because $\det A=\det A^T=\det (-A)=-\det A$.

The kernel of a group homomorphism is $K=\{g:\theta (g)=1_H\}$. It is a subgroup, since the identity maps to the identity and $\theta (k_1{k}_2^{-1})=1_H$ for $k_1,k_2\in K$. Alsoso $gK{g}^{-1}\subseteq K$. Applying the same inclusion with $g^{-1}$ gives equality, proving that $K$ is a normal subgroup.

The map $gK\mapsto \theta (g)$ is well defined, multiplicative and injective, with image $\mathrm{im}\theta$. Thus the first isomorphism theorem givesThe quotient is isomorphic to all of $H$ precisely when the original homomorphism is surjective.

For the general linear group, the determinant gives a homomorphism $\det :\mathrm{GL}(n,\mathbb{R})\to {\mathbb{R}}^{\times }$, since $\det (AB)=\det A\det B$. Its kernel is the special linear group $\mathrm{SL}(n,\mathbb{R})$, so that subgroup is normal. Every nonzero real $t$ is the determinant of $\mathrm{diag}(t,1,\dots ,1)$ for $n\ge 1$, making the map surjective. Thereforewhere the operation on the right is multiplication.

A group homomorphism preserves multiplication: $\Phi (gh)=\Phi (g)\Phi (h)$. Its group kernel $K=\{g:\Phi (g)=e_H\}$ is a subgroup, and $\Phi (gk{g}^{-1})=e_H$ for $k\in K$. Thus $gK{g}^{-1}=K$, proving it is a normal subgroup. The map $gK\mapsto \Phi (g)$ is well-defined because two representatives of a coset differ by an element of $K$. It is a homomorphism, has trivial kernel, and is surjective when $\Phi$ is. This proves the first isomorphism theorem, $G/K\cong H$. This proof does not use finiteness and therefore also applies below.

The identity matrix belongs to $S{L}_2(\mathbb{C})$. The determinant is multiplicative, giving closure, and matrix multiplication is associative. The inverse of a determinant-one matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, also of determinant one. Thus this special linear group is a group.

The fractional linear maps must be interpreted on the Riemann sphere, including infinity: a finite pole is sent to infinity and infinity is sent to $a/c$ when $c\ne 0$. As literal everywhere-defined maps of the complex plane they would not form the asserted group. Substitution into the two rational expressions gives $f_A\circ f_B=f_{AB}$. Matrix identities and inverses therefore give the identity and inverse transformations, proving that the Möbius group is a group and that $\Phi :A\mapsto f_A$ is a surjective homomorphism.

For $f_A$ to be the identity, $(az+b)/(cz+d)=z$ on all finite points where defined. Hence $c{z}^2+(d-a)z-b$ is the zero polynomial, so $b=c=0$, $d=a$, and $a^2=1$ from the determinant. ThereforeThe kernel is central and hence normal. In particular $\Phi$ is not an isomorphism, since $A$ and $-A$ represent the same transformation. The quotient is the projective special linear group in this complex two-dimensional case.

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

Two elements are conjugate group elements if $h=xg{x}^{-1}$ for some $x\in G$. The conjugacy class of $g$ is $\{xg{x}^{-1}:x\in G\}$. Conjugacy is an equivalence relation: it is reflexive using $x=e$, symmetric by replacing $x$ with $x^{-1}$, and transitive becauseThus the classes partition the group: every element belongs to its own class, and any two classes that intersect are equal. This proves that every element belongs to exactly one class.

If $G$ is Abelian, then $xg{x}^{-1}=g$, so every class is a singleton. In general the elements whose classes are singletons are precisely those commuting with every group element. Thus the center of a group isIt contains $e$. If $a,b\in Z(G)$, then $a{b}^{-1}$ commutes with every $g$, since $b^{-1}$ does and $a$ does. The subgroup test therefore proves that $Z(G)$ is a subgroup. Any two of its elements commute with each other, so it is an abelian subgroup.

For the square centred at the origin with vertices $(\pm 1,\pm 1)$, putThe dihedral group $D_4$ consists of $R^j$ and $R^{j}S$, $j=0,1,2,3$. Its eight actions and their matrices areThese exhaust the symmetries: the image of one vertex has four choices and the image of an adjacent vertex has two choices, determining the entire square. For an explicit pair of nonconjugate elements take $R^2$ and $S$. Both have order two, but their determinants are $+1$ and $-1$. A similarity transformation preserves the determinant, so $R^2$ and $S$ are not conjugate.

To find the centre, use $SR=R^{-1}S$. A rotation $R^j$ commutes with $S$ exactly when $R^j=R^{-j}$, so $j$ is even. No reflection $R^{j}S$ commutes with $R$, since $(R^{j}S)R=R^{j-1}S$ while $R(R^{j}S)=R^{j+1}S$, and $R^2\ne I$. Therefore

A square matrix is invertible precisely when its determinant is nonzero. Equivalent conditions are full rank, a trivial kernel, and zero not being an eigenvalue.

For a lower triangular matrix, the first row has only the possibly nonzero entry $A_{11}$. A Laplace expansion along that row gives $\det A=A_{11}\det A^{(1)}$, where the remaining principal submatrix is again lower triangular. Induction from the one-by-one case yieldsThis proof includes singular triangular matrices; it makes no division by diagonal entries.

For the diagonal example, put $N=n(n+1)/2$. Its determinant is $b^{1+2+\cdots +n}=b^N$, and invertibility requires $b\ne 0$. Write the complex number as $b=r{e}^{i\theta }$ with $r>0$. A nonzero purely imaginary determinant requires $\cos (N\theta )=0$. ThusEquivalently $b^N\in i\mathbb{R}\setminus \{0\}$; the angle is understood modulo $2\pi$.

For skew-Hermitian matrices, reversing products under the conjugate transpose gives $(CD{)}^{†}=D^{†}{C}^{†}=DC$. ConsequentlyThe latter condition is vanishing of the anticommutator. A universal choice for the last request isIndeed $(CD-DC{)}^{†}=DC-CD=-(CD-DC)$, so the commutator of two skew-Hermitian matrices is skew-Hermitian.

For a positive integer $k$, cancellation of $U^{†}U=I$ gives $A^k=U{D}^k{U}^{†}$. The trace is invariant under matrix similarity, since $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, and the determinant is multiplicative. ThereforeThis includes zero eigenvalues without requiring $A$ to be invertible.