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Regard the real skew-symmetric matrix as acting on ${\mathbb{C}}^n$. Its Hermitian adjoint is $A^{\ast }=A^T=-A$. Parts (i) and (iii) show that its eigenvalues are purely imaginary and its complex minimal polynomial has no repeated roots.

Since $A$ has real entries, nonreal eigenvalues occur in conjugate pairs. A nonzero pair has the form $i\alpha ,-i\alpha$, with $\alpha >0$, and contributes the real quadratic polynomialZero, if present, contributes one factor $x$. Hence the real minimal polynomial isThe product annihilates $A$ after complexification of a real vector space, so it also annihilates the original real linear map; no factor can be removed because each displayed complex root is an eigenvalue. This proves both the asserted form and its minimality. The case $A=0$ has $\epsilon =1$ and an empty product.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

The change of variables in b diagonalizes the complexified action into $z\mapsto Uz$ and $\bar{z}\mapsto \bar{U}\bar{z}$. Consequently the conventional complex branching rule isPrecisely, this is a statement about the complexification of a real vector space carrying the real defining representation. Over the real numbers, the six-dimensional space is the realification of a complex representation $3$; the conjugate summands above are not separate three-dimensional real invariant subspaces.

The block embedding $SO(6)\hookrightarrow SO(7)$ fixes the seventh coordinate. It adds a trivial representation, givingThe dimensions are complex dimensions after complexification of a real vector space, or equivalently $6+1$ as real dimensions before it.

A complex group representation $\rho :G\to \mathrm{GL}(n,\mathbb{C})$ defines a real representation on ${\mathbb{R}}^{2n}$ by viewing each complex coordinate as its real and imaginary parts. In grouped coordinates its matrices are the realification of a complex matrix shown in the title. Direct block multiplication gives $\mathcal{R}(AB)=\mathcal{R}(A)\mathcal{R}(B)$ and $\mathcal{R}(I_n)=I_{2n}$, proving the representation property and preservation of invertibility.

Taking the trace gives its character ${\chi }_{\mathbb{R}}(g)=2\mathrm{Re}{\chi }_{\mathbb{C}}(g)$. After complexification of a real vector space, this realification splits as the original representation and its conjugate. Realification need not preserve irreducibility: the answer depends on whether the irreducible complex representation has real, complex or quaternionic representation type. The standard two-dimensional complex representation of the quaternion group gives a four-dimensional real example with character four at the identity, minus four at its central negative identity, and zero at the other six elements.