For equal irreducible group representations, the intertwining identity and the complex Schur's lemma give $T=c{I}_n$. Conjugation preserves trace, soSince $\mathrm{tr}(c{I}_n)=nc$, we obtain

The preceding intertwining identity makes $T$ a homomorphism between the corresponding simple modules. By Schur's lemma, a nonzero such homomorphism would be an isomorphism. The representations are inequivalent, so this is impossible. Therefore

For $h\in G$, right-multiplying by $D_2(h{)}^{-1}$ givesMake the bijective substitution $k=hg$, so $g^{-1}{h}^{-1}=k^{-1}$. The sum becomes ${\sum }_k{D}_1(k)B{D}_2(k^{-1})=T$. HenceThus $T$ is an intertwining operator from the second group representation to the first. This is the Schur averaging of rectangular matrices construction.

In the finite-dimensional setting of the preceding part, let $S$ be a simple module. Each action map $s\mapsto as$ is an $A$-module endomorphism because $A$ is commutative. The complex Schur's lemma makes every such action a scalar. Therefore every linear subspace of $S$ is an submodule; simplicity forcesEquivalently, a simple module over a commutative ring is a quotient by a maximal ideal, and a finite-dimensional field extension of $\mathbb{C}$ is just $\mathbb{C}$.

The finite-dimensional context matters. If the clause is read as allowing an arbitrary commutative complex algebra, the unrestricted assertion is false: $A=\mathbb{C}(t)$ is simple as a module over itself and is infinite-dimensional over $\mathbb{C}$. This is simple module of a finite-dimensional commutative complex algebra. Thus the one-dimensional conclusion uses the finite-dimensional algebra context, or an alternative hypothesis such as finite dimensionality of the simple module.

For a finite-dimensional algebra, every simple module $S$ is finite-dimensional: choose $s\ne 0$; simplicity makes $A\to S$, $a\mapsto as$, surjective, so $\dim S\le \dim A$.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

With the same factors, if $n\le m$, thenhas no negative power and extends holomorphically across $a$. Thus its apparent singularity is a removable singularity, with extension valueFor $m>n$ the extended quotient has a zero of order $m-n$; for $m=n$ the extension is nonzero.

By the finite-order zero factorizations,with $F,G$ holomorphic near $a$. Shrinking the disk if necessary, continuity ensures $G$ never vanishes there, so $F/G$ is holomorphic and nonzero at $a$. ConsequentlyIf $n>m$, the exponent is $-(n-m)$, with a holomorphic nonvanishing factor. Hence

A zero of order $n\ge 1$ at $a$ means that $f$ is holomorphic near $a$ andEquivalently its Taylor series starts with the nonzero coefficient of $(z-a{)}^n$, soUnlike a pole, a zero is a regular point of the function itself; its reciprocal has a pole of the corresponding order.

A pole of order $m\ge 1$ at $a$ has finitely many negative powers in its Laurent series, the lowest being exactly $-m$:Equivalently,Thus $(z-a{)}^{m}f(z)$ has a holomorphic extension with nonzero value at $a$, whereas multiplying by any smaller power does not remove the singularity.

A removable singularity is an isolated singularity for which the function on the punctured disk extends to a holomorphic function on the full disk, by assigning or correcting its value at $a$. In Laurent series terms,The remaining power series defines the extension with value $c_0$ at $a$. Conversely, a holomorphic extension has a Taylor series there, so uniqueness of the Laurent series forces all negative coefficients to vanish.

An isolated singularity at $a$ means that $f$ is holomorphic in some punctured disk $0<|z-a|<r$, while $a$ itself is an excluded point or a point where the given function is not holomorphic. There are no other singularities arbitrarily close to $a$. By Laurent's theorem, the function has a unique Laurent series on that punctured disk:The negative-power terms form its principal part of a Laurent series.

Set $N=M_i-I$. Direct multiplication giveswith $N\ne 0$. If $M_i$ were diagonalisable, then subtracting the identity in the same basis would diagonalise $N$. A diagonal matrix whose cube is zero has every diagonal entry zero and is itself zero. Similarity would then imply $N=0$, a contradiction. HenceThis proves failure directly; merely observing that it is not self-adjoint would not suffice, because non-self-adjoint matrices can still be diagonalisable.

The finite-dimensional spectral theorem says that a self-adjoint operator on a complex inner-product space has an orthonormal basis of eigenvectors and all its eigenvalues are real. Equivalently, its matrix is unitarily diagonalisable with real diagonal entries.

For $a\in \mathbb{R}$, the matrix is equal to its conjugate transpose:Thus it represents a self-adjoint operator for the standard inner product on ${\mathbb{C}}^3$. The theorem applies directly, proving

The adjoint operator is characterized byFor each fixed $v$, the map $u\mapsto ⟨Tu,v⟩$ is a linear functional. Part (i) gives a representing vector, which we define to be $T^{\ast }v$. Hence such a map exists on every vector of $V$. Part (b) proves that this constructed map is linear.

Use the convention that the inner product is linear in its first argument, as required by the displayed representation of a linear functional. Choose an orthonormal basis $e_1,\dots ,e_n$, for instance by Gram-Schmidt orthogonalization. DefineFor $u={\sum }_j{u}_j{e}_j$, conjugate linearity in the second argument givesThusThe representation is unique as well: if two vectors represent the same functional, their difference is orthogonal to every vector and therefore to itself, so it is zero.

If ${\chi }_T(\lambda )=0$, then $\det (\lambda I-A)=0$, so $\lambda I-A$ is singular. By the rank-nullity theorem it has a nonzero kernel vector $v$. ThusThis is exactly the definition of an eigenvalue. Here $\lambda$ lies in the scalar field: for the stated real vector space it is real. A nonreal root instead gives an eigenvalue of the complexified operator, by the same argument over $\mathbb{C}$.

To prove the sum is direct, suppose ${\sum }_{j=1}^r{v}_j=0$ with $v_j\in V_{{\lambda }_j}$. Fix $i$ and apply the polynomial in the linear operatorEvery $v_j$ with $j\ne i$ is killed by one factor. On $v_i$, each factor is multiplication by ${\lambda }_i-{\lambda }_j$, givingThe scalar coefficient is nonzero because the eigenvalues are distinct. Thus $v_i=0$ for every $i$, provingIn particular, choosing a basis in each eigenspace yields a linearly independent combined list.

An essential singularity is an isolated singularity that is neither removable nor a pole. In terms of the Laurent series this means $\boxed{\text{infinitely many negative-power coefficients are nonzero}}$: no finite multiplication by a power of $z$ removes the singularity.

A pole of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ holomorphic near zero and $g(0)\ne 0$. Equivalently, the principal part is finite and starts with $\boxed{a_{-n}{z}^{-n},a_{-n}\ne 0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.

A removable singularity means that $f$ has a holomorphic extension to the full disk, assigning a finite value at zero. Equivalently its Laurent series has no negative-power terms. One direction follows by the Taylor expansion of the extension; the other by the convergent nonnegative-power series. Thus $\boxed{a_k=0\text{ for all }k<0}$ is the coefficient criterion.