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From part (f), $PQ=\mathrm{diag}(-1,-1,1)$. To obtain $PQR=-I$, choose the Householder transformation that reverses only the third coordinate:Indeed $\mathbf{r}{\mathbf{r}}^T=\mathrm{diag}(0,0,2)$, so $R_{ij}={\delta }_{ij}-r_i{r}_j$, and direct matrix multiplication gives $PQR=-I$. The alternative vector $-\mathbf{r}$ produces the same matrix. Geometrically $R$ reflects in the plane $x_3=0$, completing the reversal of all three coordinates.

The symmetric and antisymmetric parts of a matrix areIndeed $S+A=M$, while the transpose identities $(M^T{)}^T=M$ give $S^T=S$ and $A^T=-A$. Thus $S$ is a symmetric matrix and $A$ is an antisymmetric matrix. The decomposition is also unique: a matrix that is both symmetric and antisymmetric must equal its own negative and hence vanish.

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.

By the definitions of matrix multiplication, transpose and trace,Relabelling the summed indices givesthe Frobenius inner product of these real matrices.

An eigenvalue makes the shifted matrix singular, so its characteristic polynomial vanishes:Thus the characteristic equation is

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 $\det A=\det A^T=\det (-A)=-\det A$.

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

A group is a nonempty set with a closed binary operation that is associative, has an identity element, and gives every element an inverse element. A subgroup is a subset that is itself a group under the inherited operation. The left coset of $H$ represented by $g$ is $gH=\{gh:h\in H\}$; right cosets are defined analogously. Left cosets partition a finite group, and $h\mapsto gh$ is a bijection from $H$ to $gH$. Therefore $|G|=[G:H]|H|$, proving Lagrange's theorem and the required divisibility.

For the given matrices, put $a=A$ and $c=AB=iI$. Since $B=iA$, the generators commute; $a^2=I$, $c^4=I$, and the generated elements areThey are distinct, and $(c^r{a}^s)(c^{r^{\prime }}{a}^{s^{\prime }})=c^{r+r^{\prime }}{a}^{s+s^{\prime }}$, with exponents reduced modulo four and two. The Cayley table isThe table proves closure and identifies $I$ as the identity. The inverse element of $c^r{a}^s$ is $c^{-r}{a}^s$, and associativity follows from matrix multiplication. Thus this is the direct product of cyclic groups of orders four and two, an abelian group of order eight.

Its subgroups are exactlyTo see the list is complete, Lagrange's theorem leaves orders one, two, four and eight. There are exactly three elements of order two, giving the three order-two subgroups. An order-four subgroup containing an element of order four must be one of the two cyclic groups displayed; without such an element it must be the four-element set of elements whose squares are $I$. This is the enumeration in subgroups of C4 times C2.

For example choose $H=\{I,c,c^2,c^3\}$. A complete set of cosets is $\boxed{H,aH=\{a,ca,c^{2}a,c^{3}a\}}$. Since $G$ is Abelian, the left and right cosets coincide.

A complex $3\times 3$ matrix is diagonalizable exactly when it has a basis of three linearly independent eigenvectors. If all its eigenvalues are one, a diagonalization would have $D=I$ and hence $A=PI{P}^{-1}=I$. Therefore the specified matrix, which is not a scalar identity matrix, is not diagonalizable.

For the displayed real symmetric matrix, choose the orthogonal matrixIts columns form an orthonormal eigenbasis, andIn the orthogonal coordinates $u=(x+y)/\sqrt{2}$, $v=(x-y)/\sqrt{2}$, the quadric surface has equation $u^2+3v^2+3z^2=1$. It is a prolate spheroid, with long semiaxis one in direction $(1,1,0)$ and two equal short semiaxes $1/\sqrt{3}$. The figure gives its principal-axis sketch.

A diagonalizable matrix can certainly have exactly two distinct eigenvalues: $\mathrm{diag}(1,1,2)$ is already diagonal and has an eigenbasis despite its repeated eigenvalue.

Let $J=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Since $J^2=-I_2$ and $J^4=I_2$, its powers define the representation of the cyclic group of order four:For the quaternion group, denote the three displayed complex matrices by $A,B,C$. Direct multiplication gives $A^2=B^2=C^2=-I_2$, $AB=C$, $BA=-C$, and $ABC=-I_2$. Together with the images of $\pm 1$, these are exactly the quaternion multiplication relations, proving that the assignment defines a group homomorphism into $\mathrm{GL}(2,\mathbb{C})$.

Use the realification of a complex representation: for $M=X+iY$, define $\mathcal{R}(M)=\left(\begin{array}{cc}X& -Y\\ Y& X\end{array}\right)$. It acts on grouped real and imaginary coordinates and satisfies $\mathcal{R}(MN)=\mathcal{R}(M)\mathcal{R}(N)$. The explicit generators areSet $D(1)=I_4$ and $D(-q)=-D(q)$ to specify all eight elements.

For an invertible real $S$, the equivalent representation $\tilde{D}(q)=SD(q)S^{-1}$ obeys $\tilde{D}(q_1)\tilde{D}(q_2)=SD(q_1{q}_2)S^{-1}=\tilde{D}(q_1{q}_2)$. The trace is invariant under similarity, so their characters agree. Explicitly,

The Klein four-group has multiplication tableSince it is abelian, every conjugacy class is a singleton: $\{1\},\{a\},\{b\},\{c\}$. An irreducible representation is a nonzero representation whose only invariant subspaces are zero and the whole representation space. For a finite group, the number of inequivalent complex irreducible representations equals the number of conjugacy classes, so there are four here.

A one-dimensional representation sends the involutions $a,b$ to signs, independently. All four choices areThey are distinct irreducible representations, exhausting the four possibilities. The character table of the Klein four-group isThe specified diagonal matrices obey $D(a{)}^2=D(b{)}^2=I$, commute, and satisfy $D(a)D(b)=D(c)$, so the group relations prove that $D$ is a group representation. Its coordinate axes are invariant lines carrying respectively ${\chi }_{01},{\chi }_{10},{\chi }_{11}$. ThereforeThe trivial representation ${\chi }_{00}$ is absent. If $H$ denotes the four-row character matrix above, direct row products give $H{H}^T=4I$. This verifies character orthogonality:The character of $D$ is $(3,-1,-1,-1)$, whose inner products with the four rows are $0,1,1,1$, independently confirming the decomposition. All these irreducibles are real as well: commuting real involutions can be simultaneously diagonalized into their real sign eigenspaces.

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.

An eigenvalue $\lambda$ and associated eigenvector $e\ne 0$ satisfy $Ae=\lambda e$. The characteristic polynomial $\det (A-\lambda I)$ has degree equal to the positive matrix size. The fundamental theorem of algebra gives a complex root, at which $A-\lambda I$ is singular and therefore has a nontrivial kernel. Hence every nonempty square complex matrix has an eigenvector.

A nondiagonalizable example is the Jordan block $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: its only eigenvalue is zero, and its eigenspace is one-dimensional. A Hermitian matrix satisfies $A^{†}=A$ and has an orthonormal eigenbasis by the spectral theorem. One proof finds an eigenvector, notes that its orthogonal complement is invariant, and repeats. Its eigenvalues are real becausewhere $e^{†}e>0$.

With all eigenvalues negative, $A$ is invertible. Its same complete eigenbasis givesSome powered eigenvalues can coincide, but the original eigenvectors remain a complete basis. Writing $A=U\mathrm{diag}({\lambda }_i)U^{†}$, the convergent matrix exponential series givesThus the matrix inverse from a decaying matrix exponential isThe identity printed in the original PDF is missing a minus sign. Already for the one-by-one matrix $A=(-1)$ its integral is $1$, whereas its inverse is $-1$. Equivalently, integrating $d(e^{tA})/dt=A{e}^{tA}$ gives $A{\int }_0^{\infty }{e}^{tA}dt=-I$, confirming the correction without diagonalization.

The trace of a square matrix is $\mathrm{tr}A={\sum }_i{A}_{ii}$. Expanding a product and interchanging its finite sums givesHence every commutator has trace zero. Taking the trace of $AB-BA=cI$ gives $0=cn$. Over the complex numbers, with $n\ge 1$, this forcesFor that value, choose $A=B=0$, or any commuting pair. For every nonzero $c$ the trace contradiction proves nonexistence. The finite matrix size is essential to this trace argument.

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.

Over the chosen scalar field, an $n$-by-$n$ matrix is diagonalizable if and only if it has $n$ linearly independent eigenvectors. Repeated eigenvalues are allowed provided all eigenspaces together supply a basis.

For the given matrix, the characteristic polynomial is ${\lambda }^2-4\lambda -5=(\lambda -5)(\lambda +1)$. One may chooseTaking these as columns gives the change of basisThis is not a rotation matrix. The PDF's terminology cannot hold literally: $A$ is real and nonsymmetric, whereas a real matrix orthogonally similar to a real diagonal matrix is symmetric. Also $V_1\cdot V_2=-4\ne 0$, so scaling these eigenvectors cannot make an orthonormal rotation basis. A general invertible similarity is the valid construction.

Let $D=P^{-1}AP$. Inserting $P{P}^{-1}$ between successive factors gives $A^n=P{D}^n{P}^{-1}$, and henceTo write $A^n={\alpha }_{n}A+{\beta }_{n}I$, match its action on the two eigenspaces:Solving gives the matrix powers for a diagonalizable matrix with two eigenvaluesThe cases $n=0,1$ recover $I,A$, respectively. Equivalently the spectral projections of a diagonalizable matrix are $(A+I)/6$ and $(5I-A)/6$, and the power multiplies them by $5^n$ and $(-1{)}^n$.

Let $v_1,\dots ,v_n$ be the linearly independent eigenvectors, and form $V=(v_1\cdots v_n)$. This matrix is invertible and $AV=VD$, where $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Hencewhich proves diagonalizability.

If $B=P^{-1}AP$, then $P^{-1}{v}_j$ are linearly independent eigenvectors of $B$ with the same eigenvalues. Explicitly, with $W=P^{-1}V$,Thus matrix similarity preserves diagonalizability.

Using part (i) and $A^{†}=-A$ in the Cayley transform with argument $-A$ givesThe factors commute because $I-A$ and $I+A$ are polynomials in the same matrix, and an invertible commuting matrix also commutes with the other factor's matrix inverse. Both inverses exist by part (ii). Thus $U$ is unitary.

A Hermitian matrix satisfies $H^{†}=H$, where $†$ denotes the conjugate transpose. For a nonzero eigenvector $v$ with $Hv=\lambda v$, the scalar $v^{†}Hv$ equals its own complex conjugate, andThus every Hermitian eigenvalue is real.

A unitary matrix satisfies $U^{†}U=U{U}^{†}=I$. If $Uv=\mu v$, preservation of the squared norm givesSince $v\ne 0$, $\boxed{|\mu |=1}$. These statements concern complex eigenvectors, even when the matrix entries happen to be real.

The matrix is $M=2I-e{e}^T$, where $e=(1,1,1{)}^T$. Along $\hat{e}=e/\sqrt{3}$, its eigenvalue is $2-3=-1$. Every vector perpendicular to $e$ has eigenvalue two. For example, use the orthonormal basisThe principal-coordinate equation isIt depends only on distance from the $u_3$ axis and position along it, so every rotation about that axis preserves the surface. ThereforeThis is a one-sheet hyperboloid with waist radius $\sqrt{3/2}$. The repeated positive eigenvalue makes the two transverse principal axes nonunique.