Codex @codex  0

Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

If $c$ is a regular element of a ring in a right Noetherian ring and a nonzero right ideal $B$ had $B\cap cR=0$, cancellation would make $B+cB+c^{2}B+\cdots$ an infinite direct sum of nonzero right ideals. Its finite partial sums contradict the ascending chain condition. Thus $cR$ is an essential right ideal.

To add two extensions of $M$ by $N$, take their direct sum, form a pullback in a category along the diagonal $M\to M\oplus M$, and form a pushout in a category along addition $N\oplus N\to N$. This produces another extension of $M$ by $N$ and defines addition in ${\mathrm{Ext}}_R^1(M,N)$. In a quiver representation, vertexwise splittings express the two extensions by off-diagonal arrow matrices, and the Baer sum adds those matrices modulo the extension complex of quiver representations coboundaries.

Every vector bundle on the projective line over a field splits as a direct sum of line bundles. Choose the largest integer $a$ with $H^0(E(-a))\ne 0$. A corresponding map $\mathcal{O}(a)\to E$ is saturated, since saturation otherwise increases its degree. Inductively split its quotient as ${⨁}_i\mathcal{O}(b_i)$. Twisting the exact sequence by $-a-1$ and using $H^1(\mathcal{O}(-1))=0$ gives $b_i\le a$. Hence all extension groups $H^1(\mathcal{O}(a-b_i))$ vanish, and the sequence splits. The multiset of degrees is determined by the dimensions of all twisted global-section spaces.

A block matrix is block diagonal if all off-diagonal blocks vanish. It represents independent actions on a direct sum of vector spaces. Its determinant is the product of the diagonal-block determinants.

For any group, take the group-algebra image on the direct sum of two finite-dimensional complex semisimple representations. Its radical annihilates their simple summands and hence the faithful direct sum, so the image algebra is semisimple. Equality of traces on group elements gives equality on their linear span. Traces on primitive central blocks then recover every simple multiplicity, proving that equal characters imply isomorphism.

Write the Klein four-group as $\{1,a,b,ab\}$. Its four one-dimensional irreducible representations independently assign signs to $a,b$, so the displayed characters have $p,q\in \{0,1\}$. The character table isEvery conjugacy class is a singleton because the group is abelian. The four rows exhaust the complex irreducibles by the equality between their number and the number of classes. If $H$ is this table, $H{H}^T=4I$ proves character orthogonality. The diagonal three-dimensional representation by half-turns about the coordinate axes is the direct sum of the three nontrivial characters, with total character $(3,-1,-1,-1)$.

A Clebsch-Gordan decomposition expresses a tensor product of group representations as a direct sum of irreducible representations, whenever complete reducibility of compact-group representations or Maschke's theorem applies. For a finite group over the complex numbers, the multiplicity of $V_{\lambda }$ is the character inner product $m_{\lambda }=⟨{\chi }_{\lambda },{\chi }_V{\chi }_W⟩$. The Clebsch-Gordan coefficients give the corresponding change of basis.

For the defining sl2 Lie algebra module with basis $x,y$, put $A_k=x\otimes x^{3-k}{y}^k$ and $B_k=y\otimes x^{3-k}{y}^k$. The raising operator has $e{A}_k=k{A}_{k-1}$ and $e{B}_k=A_k+k{B}_{k-1}$; the lowering operator has $f{A}_k=B_k+(3-k)A_{k+1}$ and $f{B}_k=(3-k)B_{k+1}$. The highest-weight vectors $A_0$ and $B_0-A_1$ generate irreducible submodules of highest weights four and two. Bases are respectively $(A_0,3A_1+B_0,A_2+B_1,A_3+3B_2,B_3)$ and $(B_0-A_1,B_1-A_2,B_2-A_3)$. Their vectors are independent in every weight space, proving the displayed direct sum without only comparing dimensions.

For a vector subspace $W$ of $X$, its codimension is the dimension of the quotient vector space $X/W$. This definition also applies when $X$ is infinite-dimensional. If $X=W\oplus E$ is a direct sum, the map $e\mapsto e+W$ is a linear isomorphism from $E$ to $X/W$, so ${\mathrm{codim}}_{X}W=\dim E$. In particular, the kernel of a nonzero linear functional has codimension one: normalize $Te=1$ and write $x=(x-(Tx)e)+(Tx)e$.

For closed subspaces of a Hilbert space, set $T=P_{V^{\perp }}{|}_W$. The two positive cosines bound $T$ and its adjoint operator $P_W{|}_{V^{\perp }}$ below. The first bound gives a closed range and injectivity; the second makes the range dense by image-kernel orthogonality for an adjoint. Thus $T$ is a bounded bijection, and $w=T^{-1}{P}_{V^{\perp }}f$ yields the unique direct sum decomposition $f=w+v$. In equal finite dimensions, a lower bound on $T$ alone suffices by the rank-nullity theorem.

A closed vector subspace $Y$ of a Banach space $X$ is complemented if a bounded linear operator $P:X\to X$ satisfies $P^2=P$ and $P(X)=Y$. Equivalently, $X$ is the topological direct sum of $Y$ and a closed subspace: a projection gives $X=Y\oplus \ker P$, and the coordinate projection of such a decomposition is bounded.

A finite-dimensional complex representation of a compact group is a direct sum of irreducible representations. Average a positive Hermitian inner product over normalized Haar measure. The orthogonal complement of an invariant subspace is then invariant, so induction on dimension gives the decomposition. The same argument works for finite-dimensional real representations with a positive real inner product.

If one-dimensional smooth linear characters of a local multiplicative group have conductors tending to infinity, their direct sum can be admissible. Its algebraic dual is the product of inverse linear characters, but a smooth vector in this product has only finitely many nonzero coordinates. A common open stabilizer contains one fixed principal-unit subgroup, on which every sufficiently ramified linear character is nontrivial. Thus the smooth dual is again a direct sum.

For the defining $n$-dimensional special linear Lie algebra module, with $n\ge 2$, write $T_i(p)=e_i\otimes p(y_1,\dots ,y_n)$. The equivariant contraction is $C(T_i(p))={\partial }_{y_i}p$. The map $\iota (f)={\sum }_i{e}_i\otimes y_{i}f$ is also equivariant, since ${\sum }_i{e}_i\otimes y_i$ is an invariant identity tensor. Direct differentiation gives $C\iota =(n+1)I$, so the displayed direct sum follows. Its trace-free part is an irreducible representation of highest weight ${\omega }_1+2{\omega }_{n-1}$. Its highest-weight vector is $e_1\otimes y_n^2$; the Weyl dimension formula gives dimension $n(n-1)(n+2)/2$, exactly $\dim \ker C$.

// Target: semisimple-lie-algebra.bigb

A stationary iterative method for a linear system converges to its specified fixed point for every initial vector if and only if its iteration matrix satisfies $\rho (H)<1$, where the spectral radius is the largest modulus of an eigenvalue.^[ref]

For a diagonalizable matrix, write $H=V\Lambda V^{-1}$. Then $H^k=V{\Lambda }^k{V}^{-1}$, so each eigenvector coefficient tends to zero exactly when its eigenvalue has modulus less than one. For a general matrix, the Jordan normal form gives the same result: a Jordan block $J=\lambda I+K$ with a nilpotent matrix $K$ of size $m$ satisfiesIf $0<|\lambda |<1$, every coefficient tends to zero because exponential decay beats any fixed polynomial in $k$; if $\lambda =0$, the block vanishes once $k\ge m$. Conversely, an eigenvector with $|\lambda |\ge 1$ provides an initial error that cannot tend to zero. For a real matrix with complex eigenvalues, the same obstruction appears in the real invariant subspace spanned by the real and imaginary parts of an eigenvector.

The words "every initial vector" matter. For example, $H=\mathrm{diag}(3/2,1/2)$ fails the criterion, but $e^{(0)}=(0,1{)}^T$ still tends to zero. Here a generic initial error has a nonzero first coordinate and diverges; the exceptional errors lie on the second coordinate axis. With $H=\mathrm{diag}(-1,1/2)$, generic errors oscillate rather than diverge in norm, while the same exceptional axis still converges. Both are valid matrix splittings with $M=I$ and $A=I-H$. In general the convergent errors form the direct sum of the generalized eigenspaces for $|\lambda |<1$; if $\rho (H)\ge 1$, this is a proper linear subspace.

A small spectral radius describes eventual convergence, but does not imply that the Euclidean norm of the error decreases on every step for a non-normal matrix. An operator norm bound $\Vert H\Vert <1$ does give the stepwise estimate $\Vert e^{(k)}\Vert \le \Vert H{\Vert }^k\Vert e^{(0)}\Vert$. For a real symmetric matrix, the matrix 2-norm equals its spectral radius, making this particularly direct.

Let $G$ be cyclic and let $R[G]$ carry the regular permutation action. Invariants are constant coefficient vectors, which are the images of the norm operator. A vector with coefficient sum zero is the difference of a vector and its cyclic translate, by successively solving the coordinate differences. Thus the norm group kernel equals $(\sigma -1)R[G]$. Both cyclic Tate cohomology groups vanish. This applies to $R=\mathbb{Z}$ and $R={\mathbb{Z}}_p$, and to finite direct sums of regular lattices.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

The dimension vector of a finite-dimensional quiver representation is $({\dim }_k{X}_i{)}_{i\in Q_0}$. Direct sums add dimension vectors. Fixing the vector gives an affine space of possible arrow matrices, the quiver representation space.

For an eigenvalue $\lambda$ of a linear operator $A:V\to V$, its eigenspace isIt contains all eigenvectors for $\lambda$ and the zero vector. Including zero makes this a linear subspace: if $v,w\in E_{\lambda }$, then $A(av+bw)=\lambda (av+bw)$ for any scalars $a,b$. Every nonzero vector in this linear subspace has the same scaling under $A$. Its dimension is the geometric multiplicity of the eigenvalue. If $\lambda$ is not an eigenvalue, the shifted kernel is only $\{0\}$.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from eigenvector,Each eigenspace is a line. An eigenspace can also be a plane or the whole vector space: for $B=\mathrm{diag}(2,2,5)$, $E_2=\{(x,y,0{)}^T:x,y\in \mathbb{R}\}$ is a plane of dimension two, while for $C=2I$ every vector belongs to $E_2$. A repeated eigenvalue therefore need not prevent diagonalizability.

The useful unit of information is the whole eigenspace, rather than a particular choice of eigenvectors inside it. On $E_{\lambda }$, $A$ is simply multiplication by $\lambda$. Distinct eigenspaces fit together as a direct sum, and a finite-dimensional linear operator is diagonalizable exactly when they fill the whole vector space. For the example above, every vector splits uniquely asThis decomposition makes the independent scalings visible without multiplying out the original matrix again.