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Apply trace orthogonality of symmetric and antisymmetric matrices. In detail, invariance of the trace under transpose and its cyclic property giveTherefore $\boxed{\mathrm{tr}(SA)=0}$. The negative sign comes from $A^T=-A$, while $S^T=S$.

The transpose reverses a matrix product. Since $S^T=S$, we have $(S^2{)}^T=S^T{S}^T=S^2$. Therefore $S^2$ is a symmetric matrix.

For a matrix proportional to its transpose, applying the transpose twice gives $A^T=5A$ and therefore $A=25A$. Over the real numbers or complex numbers, $24A=0$ implies

Write $P=I-B$, $Q=I+B$, so $P^T=Q$ and $Q^T=P$. The factors commute because both are polynomials in $B$. Also $P$ has a matrix inverse since it is the transpose of the invertible $Q$. Thus the Cayley transform of a real skew-symmetric matrix, with its parameter replaced by $-B$, satisfiesHence $C$ is an orthogonal matrix. Its inverse is $Q{P}^{-1}$; this lets us distinguish the symmetric and skew-symmetric Cayley transforms below.

The transpose rule gives $(BC{)}^{\mathsf{T}}=C^{\mathsf{T}}{B}^{\mathsf{T}}$. If $C$ is orthogonal, $C^{-1}=C^{\mathsf{T}}$, and the identity becomesMultiplying on the left by $C$ yields $\boxed{B^{\mathsf{T}}=-B}$. This is precisely the defining condition for an antisymmetric matrix.

The transpose reverses product order, so $(CB{)}^{\mathsf{T}}=B^{\mathsf{T}}{C}^{\mathsf{T}}$. If $B$ is an orthogonal matrix, then $B^{-1}=B^{\mathsf{T}}$ and the given identity becomesMultiply on the left by $B$ to obtain $C^{\mathsf{T}}+C=0$. Therefore $C$ is antisymmetric, so it is an antisymmetric matrix, namely $\boxed{C^{\mathsf{T}}=-C}$.

The $(i,j)$ entry of $B^{\mathsf{T}}B$ is ${\mathbf{e}}_i\cdot {\mathbf{e}}_j={\delta }_{ij}$. Hence $B$ is an orthogonal matrix, whose matrix inverse equals its transpose:This obtains the inverse without unnecessary cofactor calculations.

Use the order-reversal rule for the transpose of a product:Therefore the product of two orthogonal matrices is orthogonal. The argument does not require $A$ and $B$ to commute.

The middle row is $-1/2$ times the sum of the first and third rows, so the rows are linearly dependent and the determinant is zero. The matrix trace is the sum of diagonal entries, $1+4+1=6$. A transpose preserves the determinant, while the determinant of a product is the product of its determinants. Consequently

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.

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

For a skew-symmetric matrix, $M^T=-M$. Invariance of the determinant under transpose givesFor odd $n$, $2\det M=0$, so $\det M=0$ over the real or complex numbers.

Taking determinants and using invariance under transpose,This determinant identity actually holds in every dimension. In dimension three, skew symmetry additionally givesso $\det A=0$. Hence there is a real nonzero vector $x$ in the kernel:The eigenvalue relation then gives $Nx=x$. Thus the Cayley transform fixes an axis and is a proper three-dimensional rotation, including the identity as a degenerate case.

First, $I-A$ is invertible. If $(I-A)x=0$ for a real vector, then $Ax=x$, whereas the skew-symmetric matrix identity gives $x^{T}Ax=0$; hence $x^{T}x=0$ and $x=0$. Thus its real kernel is trivial, which suffices for invertibility. The same argument applies to $I+A$.

Using $A^T=-A$ and the rule for the transpose of a product,The factors $I+A$ and $I-A$ commute, as do the required inverses. ConsequentlyThus $\boxed{N\text{ is orthogonal}}$. This is the Cayley transform of a real skew-symmetric matrix.

The adjoint operator $A^{†}$, when it exists, is defined byfor all vectors in the relevant domains. This convention takes the inner product to be conjugate-linear in its first argument. Nondegeneracy makes the adjoint unique.

For the standard complex inner product $⟨x,y⟩={\overline{x}}^{T}y$, the equality becomes ${\overline{x}}^T{\overline{A}}^{T}y={\overline{x}}^T{A}^{†}y$ for every $x,y$, soThus the matrix adjoint is the Hermitian conjugate, not merely the transpose unless the entries are real. In a basis with positive Hermitian Gram matrix $G$, the same definition instead gives $A_G^{†}=G^{-1}{\overline{A}}^{T}G$, making the dependence on the chosen inner product explicit.

The entries of $M^{T}M$ are real, and the transpose rule gives $(M^{T}M{)}^T=M^{T}M$. Thus it is a real symmetric matrix, hence also a Hermitian matrix.

For any complex vector $v$,In particular, if $M^{T}Mv=\lambda v$ with $v\ne 0$,Thus $M^{T}M$ is positive semidefinite. Its zero eigenspace is precisely the kernel of $M$, since $\Vert Mv{\Vert }^2=0$ is equivalent to $Mv=0$. This is the Gram-matrix structure behind the nonnegative eigenvalues.

Both factors and their product are real, so the Hermitian conjugate equals the transpose. Part (i) therefore givesHence $\boxed{R_1\text{ is unitary for every }\theta ,\varphi .}$ A real orthogonal matrix is always a unitary matrix.

The Hermitian conjugate of a real matrix is its transpose, soTherefore $\boxed{M\text{ is anti-Hermitian for all }\theta ,\varphi .}$ It is also a real skew-symmetric matrix. It is Hermitian as well only in the exceptional case $M=0$, precisely when $R_1$ is symmetric. This includes the identity and half-turn cases from part (d), so the zero-matrix overlap should not be excluded.

The contraction is $\mathrm{tr}(A^3)$. Since $A^T=-A$, transpose and invariance of the trace giveThis is the odd-power case of vanishing odd complete contractions of an antisymmetric matrix.

The determinant is invariant under transpose. Therefore the characteristic polynomials coincide:Their polynomial roots, including their algebraic multiplicities, are the same, so $A$ and $A^T$ have the same eigenvalues. This allows column sums as well as row sums to supply Gershgorin discs for the same spectrum.