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Direct matrix multiplication givesConsequently $\boxed{A^2-6A+5I=0}$, the Cayley-Hamilton theorem for its characteristic polynomial $t^2-6t+5$. Rearranging gives $A(6I-A)=5I$, so

For the specified coordinate vectors,Choose the orthogonal matrix representing a quarter-turn,Then $S^{T}S=I$ and direct matrix multiplication gives $S^{T}LS=\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)=M$. Thus these two shear mappings are orthogonally similar.

Direct matrix multiplication gives $MN=\left(\begin{array}{cc}5& 18\\ 7& 25\end{array}\right)$. The three determinants are $\det M=8-9=-1$, $\det N=4-3=1$, and $\det (MN)=125-126=-1$. Hence $\boxed{\det (MN)=(-1)(1)=\det M\det N}$, as required.

Direct matrix multiplication givesSubtracting verifies $\boxed{A^2-4A-5I=0}$. This is also the Cayley-Hamilton theorem for the characteristic polynomial ${\lambda }^2-4\lambda -5$.

A vector perpendicular to ${\mathbf{e}}_1$ has the form $\mathbf{v}=(s,t,-s)$. Direct matrix multiplication givesThus every nonzero such vector is an eigenvector with $\boxed{\lambda =-2}$. The zero vector must be excluded from the literal claim about “any” perpendicular vector, since an eigenvector is nonzero. The eigenspace is two-dimensional. Equivalently, $A=-2I+6{\mathbf{e}}_1{\mathbf{e}}_1^{\mathsf{T}}$ exhibits the spectrum of a scalar matrix plus a rank-one projector.

The determinant is multiplicative under matrix multiplication:This also holds when $B$ is singular, in which case both sides vanish.

Use Einstein notation, with repeated suffixes summed. The matrix multiplication and transpose giveInterchange the free indices, then relabel the dummy indices $k,\ell$:The negative sign uses $B_{k\ell }=-B_{\ell k}$. Therefore $C$ is skew-symmetric. No orthogonality of $A$ is needed for this part: matrix congruence preserves skew-symmetry.

Adding the diagonal entries gives $\mathrm{tr}A=4+4+2=10$. Expansion along the first row gives $\det A=4(8-4)+2(0-8)=0$. For the remaining trace, direct matrix multiplication givesThus $\mathrm{tr}(A^2)=20+20+12=52$. By multiplicativity of the determinant, $\det (A^2)=(\det A{)}^2=0$. Altogether,

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

Scalar entries commute, so $B_{jk}{A}_{ij}=A_{ij}{B}_{jk}$ is exactly the summed entry formula for matrix multiplication. Hence

The matrix multiplication $A\mathbf{c}$ adds each row. Equating it to $\mathbf{b}$ gives the linear systemThe last equation is the sum of the first two, so it supplies no further restriction. Subtracting the first from the second gives $2\beta =2$, hence $\beta =1$, and then $\gamma =1-\alpha$. Thus all parameter choices are

Direct matrix multiplication givesThus $D$ is diagonal. The columns of $P$ are eigenvectors of $A$ with eigenvalues one and four.

Subtracting the second row from the third gives a third row $(0,2,0)$, so expansion along it shows $\det B=-2\ne 0$. Thus the matrix inverse exists and every right-hand side has a unique solution.

To calculate it, temporarily solve $B\mathbf{x}=\mathbf{b}$ with general $\mathbf{b}=(b_1,b_2,b_3{)}^T$. Subtracting the second equation from the third and from the first givesBack-substitution into the second equation givesThereforeFor the given right-hand side, matrix multiplication yields the unique answer

By the definition of matrix multiplication and trace, with all indices summed from $1$ to $D$,The entries are scalars, so their multiplication commutes. Reordering them and renaming the summed indices givesEquivalently, in Einstein notation, $P_{ij}{Q}_{jk}{R}_{ki}=Q_{jk}{R}_{ki}{P}_{ij}$. This proves the cyclic property of the trace.

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

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.

Choose the unnormalized eigenvectorsTheir linear combination giving the specified vector isApplying the reciprocal eigenvalues of the inverse matrix therefore yieldsIn Cartesian coordinates, this is $(-1/3,2/3{)}^T$, and direct matrix multiplication gives $S\mathbf{y}=(1,0{)}^T$.

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

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

Cosets and Lagrange. A left coset is $gH=\{gh:h\in H\}$. The left cosets of a subgroup partition a group, each with $|H|$ elements. Lagrange's theorem states $\boxed{|G|=[G:H]|H|}$ for a finite group, so every subgroup order divides the group order.

The 64-element group. Work in the commutative ring $R=\mathbb{Z}/4\mathbb{Z}$, which has zero divisors. Denote the displayed lower triangular matrix by $A(a,b,c)$, where $a,b,c$ are its three entries below the diagonal. Direct matrix multiplication givesAll arithmetic is modulo four, so the set is closed. Associativity comes from matrix multiplication, the identity element is $A(0,0,0)$, andThe formula works in either multiplication order and requires no division. Hence these matrices form a group of order $\boxed{|G|=4^3=64}$. This is the lower triangular version of the Heisenberg group over a commutative ring.

The abelian subgroup. If $a=c$, write $h(a,b)=A(a,b,a)$. ThenThe identity is $h(0,0)$ and the inverse is $h(-a,a^2-b)$, proving an abelian subgroup. Its two independent entries give $\boxed{|H|=16}$ and $\boxed{[G:H]=4}$, in accordance with Lagrange's theorem. This is the abelian plane subgroup of the Heisenberg group modulo four.

More explicitly, $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of $R$, with kernel $H$. The four left cosets are exactly the four sets with fixed $a-c=t$, $t=0,1,2,3$; representatives are $A(t,0,0)$. Thus $H$ is also a normal subgroup.

Matrices squaring to the identity. The product formula givesThe first and third coordinates vanish only when $a,c\in \{0,2\}$. Then $ca=0$ modulo four, so the middle coordinate vanishes exactly when $b\in \{0,2\}$. Therefore the complete set isIt is a subgroup: multiplying two such elements keeps all three entries even, and their cross term $cd$ vanishes modulo four, so the product simply adds the three even coordinates. The identity is included and each element is its own inverse. In fact $K$ is isomorphic to the direct product of groups $C_2\times C_2\times C_2$. The seven nonidentity elements are involutions.

Finally, the last subset in the PDF is defined by $a\ne 0\ne b$, with no restriction on $c$. It has $3\cdot 3\cdot 4=36$ elements, which cannot be a subgroup because $36$ does not divide $64$. It also excludes the identity element. Thus