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A real-linear map between complex vector spaces that commutes with multiplication by $i$. On $\mathbb{C}$, its real matrix is $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, representing multiplication by $a+ib$. On $\mathbb{C}$, such a map with real determinant one is a rotation. In higher complex dimension, determinant one does not force the map to be a unitary operator.

A Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

On the complex vector space $\mathrm{CF}(G)$ of class functions, define the linear operatorThis is a complex-linear map and satisfies $T^2=I$. It is well-defined on class functions, because inversion of conjugacy classes takes an entire conjugacy class $K$ to another conjugacy class $K^{-1}$.

Use first the basis consisting of the indicator functions $1_K$ of the conjugacy classes. The linear operator $T$ permutes this basis by $1_K\mapsto 1_{K^{-1}}$. A permutation matrix has trace equal to the number of fixed basis vectors, soThe right-hand side counts the real conjugacy classes.

By character orthogonality and completeness, the irreducible characters are another basis of $\mathrm{CF}(G)$. Part (a) gives $T\chi =\overline{\chi }$. The pointwise complex conjugate of an irreducible character is again an irreducible character: conjugate the representing matrices, and conjugate any hypothetical proper invariant subspace back to contradict irreducibility. Thus $T$ also permutes this basis, andHere $\chi =\overline{\chi }$ means exactly that $\chi$ is a real-valued character. Equality of the trace in these two bases provesThis is the real character and real conjugacy class counting theorem. The linear operator used in the proof is $f\mapsto f\circ \mathrm{inv}$; on the irreducible character basis it agrees with pointwise complex conjugation, although pointwise complex conjugation itself is not a complex-linear map on all class functions.

In the monomial basis $\mathcal{B}=(1,x,x^2,x^3)$ of the complex vector space $V=\mathbb{C}[x{]}_{\le 3}$, the binomial theorem givesThe characteristic polynomial of this upper triangular matrix is the product of $t$ minus its diagonal entries, so ${\chi }_T(t)=(t-1{)}^4$.

For the minimal polynomial, set $\Delta =T-I$, the forward difference with step one. For a nonconstant polynomial of polynomial degree $d$, the binomial theorem shows that $f(x+1)-f(x)$ has polynomial degree $d-1$. Constant polynomials have zero forward difference. Consequently ${\Delta }^4=0$ on $V$. To prove that the nilpotency index is exactly four, computeIt follows thatIf $T$ were diagonalizable, its only eigenvalue would be $1$, and its diagonal matrix in an eigenbasis would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus $T$ is not diagonalizable. Equivalently, its minimal polynomial has a repeated root.

We prove by mathematical induction on $n=\dim V$ that every linear map on an $n$-dimensional complex vector space admits both triangular forms. For $n=0$ or $1$ this is immediate. When $n>1$, the fundamental theorem of algebra supplies an eigenvector $v$; its span $U=⟨v⟩$ is an invariant subspace.

By parts (ii) and (iii), $U^0$ has dimension $n-1$ and is an invariant subspace of the dual map $T^{\prime }$. Apply the induction hypothesis to $T^{\prime }{|}_{U^0}$ to choose a basis ${\phi }_1,\dots ,{\phi }_{n-1}$ with an upper triangular matrix. Extend it by a linear functional ${\phi }_n$ satisfying ${\phi }_n(v)=1$. Such a functional exists by basis extension, and cannot belong to $U^0$. In this basis of $V^{\prime }$, the matrix of $T^{\prime }$ has the formwhere $B$ is an upper triangular matrix; here the last diagonal entry follows by evaluating $T^{\prime }{\phi }_n$ at $v$, where $Tv=\lambda v$.

There is a basis $e_1,\dots ,e_n$ of $V$ with ${\phi }_i(e_j)={\delta }_{ij}$, since the dual basis construction gives a finite-dimensional bidual evaluation map isomorphism. In particular $e_n=v$. The matrix transpose relationship for a dual map givesa lower triangular matrix. Reversing this basis turns a lower triangular matrix into an upper triangular matrix. If $R$ is the permutation matrix reversing the order, we have provedThis completes the induction by triangularization using dual spaces.

Each endomorphism has three distinct eigenvalues in a three-dimensional complex vector space. Choose a nonzero eigenvector for each eigenvalue; these vectors are linearly independent and hence form an eigenbasis. Both operators are therefore diagonalizable with diagonal matrix $D=\mathrm{diag}(1,2,3)$.

Their characteristic polynomials areIf a polynomial $p$ annihilates either operator, applying it to each eigenvector gives $p(1)=p(2)=p(3)=0$. Thus its minimal polynomial must contain all three factors. Conversely, their product annihilates the diagonal matrix, soFinally, send an ordered eigenbasis of $S$ to the correspondingly ordered eigenbasis of $T$. The resulting invertible map $C$ satisfies $CS=TC$, and hence $T=CS{C}^{-1}$: the two endomorphisms are conjugate.

A traceless skew-Hermitian matrix has the unique formThis real-linear parametrization is bijective, proving $\boxed{V\cong {\mathbb{R}}^3}$ as real vector spaces. It is not a complex vector space: multiplication by $i$ generally turns a skew-Hermitian matrix into a Hermitian one.

An antilinear endomorphism $J$ with square $-1$ equips a complex vector space with a quaternionic structure: complex multiplication supplies the quaternionic generator $i$ and $J$ supplies $j$, satisfying $Ji=-iJ$. Such a space has even complex dimension. There is no invariant complex line, because $Jv=cv$ would imply $-v=|c{|}^{2}v$. In dimension four the invariant complex two-planes are quaternionic lines, forming ${\mathbb{HP}}^1\cong S^4$. This is the Euclidean reality structure on twistor space used in twistor theory.

For a finite group, the number of irreducible characters which are real-valued characters equals the number of real conjugacy classes. On the complex vector space of class functions, the linear operator $Tf(g)=f(g^{-1})$ permutes the basis of class indicators by inversion and the basis of irreducible characters by pointwise complex conjugation, using the character inversion identity. The trace of a basis permutation is its number of fixed members. Equality of the trace in the two bases proves the theorem. The linear operator $T$ is complex-linear; pointwise complex conjugation on arbitrary class functions is not.

A sesquilinear form on a complex vector space $V$ is a map $B:V\times V\to \mathbb{C}$ which, with the convention used throughout this article, is a linear map in its first argument and a conjugate-linear map in its second. For vectors $u,v,w$ and complex numbers $\alpha ,\beta$, the defining identities areThe bars denote complex conjugation. No symmetry or positivity is required. Some authors choose the opposite convention; every formula here keeps the first argument linear.^[ref]

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space, write $u={\sum }_i{x}_i{e}_i$, $v={\sum }_j{y}_j{e}_j$, and $A_{ij}=B(e_i,e_j)$. Expanding by the two identities givesEvery complex matrix $A$ gives a sesquilinear form this way. The superscript $T$ means transpose, not conjugate transpose. Equivalently, the scalar is $y^{\ast }{A}^{T}x$, where $y^{\ast }={\overline{y}}^T$. This is the convention used for the matrix of a Hermitian form when symmetry is imposed. The coordinate formula follows directly by expanding the basis coefficients.^[ref]

For a concrete example on ${\mathbb{C}}^2$, let $B(z,w)=z_1\overline{w_2}$. Its matrix is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$: $B(e_1,e_2)=1$ but $B(e_2,e_1)=0$, so it is not a Hermitian form. Also $B((1,i),(1,i))=-i$, so even real diagonal values cannot be assumed for an arbitrary sesquilinear form. By comparison, the bilinear form $F(z,w)=z_1{w}_1+z_2{w}_2$ satisfies $F(iz,w)=F(z,iw)=iF(z,w)$ and has $F((i,0),(i,0))=-1$; it cannot supply positive squared lengths on this complex vector space.

The effect of changing the complex argument of a multiplier is particularly revealing. For real numbers $\theta ,\varphi$, sesquilinear forms satisfyThus a common change of complex argument cancels from the diagonal, whereas the two arguments rotate the output in opposite directions. Conjugating one argument is what replaces the ${\alpha }^2$ scaling of a bilinear form by the nonnegative $|\alpha {|}^2$ scaling needed for a Hermitian inner product. This mechanism also works in function spaces: for complex continuous functions on $[0,1]$, $B(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ obeys the same identities, letting integration measure overlap rather than merely multiply coefficients.