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For a linear, fixed-frequency beam splitter, choose power-normalized complex amplitudes $a_1,a_2$ at its input optical ports and $b_1,b_2$ at its output optical ports: $|a_j{|}^2$ and $|b_j{|}^2$ are powers. The scalar description treats one specified electromagnetic polarization and matching transverse profiles; other fields require additional components. With the geometrical ordering in the diagram,Here $t,t^{\prime }$ are transmission complex amplitudes per unit input and $r,r^{\prime }$ are reflection complex amplitudes per unit input. In this power normalization, a single input gives $T=|t{|}^2$ and $R=|r{|}^2$. Raw electric field amplitudes in different media generally need additional power-normalization factors.

If these two output optical ports account for all outgoing power and the device has no loss or gain, $S$ is a unitary matrix:The dagger is the Hermitian transpose and a star is complex conjugation. To prove the condition, substitute $\mathbf{b}=S\mathbf{a}$ into conservation of total power for every input vector: ${\mathbf{a}}^{†}(S^{†}S-I)\mathbf{a}=0$. Inputs $(1,0)$ and $(0,1)$ fix the diagonal entries; $(1,1)$ and $(1,i)$ fix the real and imaginary parts of the off-diagonal entry. A lossy two-output description instead has $S^{†}S\le I$ for a passive device.^[ref]

One consistent ideal symmetric 50:50 choice isEach reflection carries $+\pi /2$ relative wave phase in this convention. The columns have unit vector norm and zero inner product, directly verifying that $S$ is a unitary matrix. Choosing a different phase convention for complex amplitudes at the optical ports replaces $S$ by $D_{\mathrm{out}}S{D}_{\mathrm{in}}^{-1}$, where the $D$ are diagonal matrices whose entries have modulus one; the input complex amplitudes must be transformed consistently. Observable powers are unchanged. The displayed $i$ is a convention for this ideal model, not a universal reflection rule for every coating or incidence geometry.

For a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$ over a field, matrix multiplication gives $(BD-DB{)}_{ij}=(d_j-d_i)b_{ij}$. Hence a matrix $B$ commutes with $D$ exactly when $b_{ij}=0$ wherever $d_i\ne d_j$. If all diagonal entries are distinct, the centralizer of a diagonal matrix consists exactly of diagonal matrices. When entries repeat, the allowed entries form unrestricted blocks on the corresponding eigenspaces.

Fix the convention: the columns of the change-of-basis matrix $P$ are the new basis vectors written in the old basis. Thus $P$ converts new coordinate vectors into old ones; its matrix inverse converts old coordinate vectors into new ones. For the preceding basis,Its determinant is $-2$, so $P$ is an invertible matrix. Checking which direction $P$ converts is essential: different books may call either $P$ or $P^{-1}$ the change-of-basis matrix.

For a linear operator, convert the input into the old basis, apply the old matrix, and convert the output back. Reading that sequence from right to left givesFor example, let the old diagonal matrix stretch horizontal components by $2$ and vertical components by $3$. ThenThe output represents $2u+2v=(4,0)$, just as $A_{\mathrm{old}}(2,0{)}^T=(4,0{)}^T$. The matrices look different because their bases differ, but they encode the same linear operator. This relation is called matrix similarity; it preserves the underlying action.^[ref]

Over an algebraically closed field of characteristic zero, any two Cartan subalgebras of the symplectic Lie algebra are conjugate by the symplectic group. A Cartan subalgebra $\mathfrak{h}$ is an abelian Lie algebra and equals its centralizer of a Lie subalgebra. The symplectic centralizer decomposition therefore forces all its weight spaces in $V$ to be one-dimensional, nonzero, and paired by opposite weights of a representation. Choose a basis vector in each space, normalizing the perfect pairing with its opposite to one. These vectors form a symplectic basis $\mathcal{B}$. In this basis, $\mathfrak{h}$ is contained in the Lie subalgebra of diagonal matrices with entries $(t_1,\dots ,t_n,-t_1,\dots ,-t_n)$. All those diagonal matrices commute with $\mathfrak{h}$, so equality with its centralizer of a Lie subalgebra forces equality with the entire diagonal Lie subalgebra. Repeating for ${\mathfrak{h}}^{\prime }$ gives a second symplectic basis. The linear map sending one symplectic basis to the other belongs to the symplectic group and conjugates the two diagonal Lie subalgebras, proving the claim.

If a matrix with orthonormal columns is divided into two row blocks, each having at least as many rows as there are columns, it admits simultaneous thin factorizations $A_1=U_{1}C{V}^T$, $A_2=U_{2}S{V}^T$, with orthonormal $U_1,U_2$, an orthogonal matrix $V$, and nonnegative diagonal matrices $C,S$ satisfying $C^2+S^2=I$. To prove this, apply the spectral theorem to $A_1^T{A}_1$; the same orthonormal eigenbasis diagonalizes $A_2^T{A}_2=I-A_1^T{A}_1$. Normalize the nonzero columns of $A_{i}V$ and complete the zero columns to orthonormal sets. The entries of $C,S$ can be written as cosines and sines of angles in $[0,\pi /2]$.

The degree matrix is the diagonal matrix whose $i$th diagonal entry is the degree of a vertex $i$. For a finite simple graph, $d_i={\sum }_j{A}_{ij}$, where $A$ is its adjacency matrix. For a Weighted graph Laplacian, use the sum of incident edge weights instead of the number of incident edges.

The determinant assigns a scalar to a square matrix. For a linear map on the real plane, it measures signed area scaling: its absolute value tells how much area grows, and its sign records whether the orientation of a vector space is preserved or reversed. In three dimensions, the same idea measures signed volume. Over a field, a square matrix is an invertible matrix exactly when its determinant is nonzero. These are two views of one phenomenon: losing an independent direction collapses full-dimensional volume and makes the original input impossible to recover uniquely.^[ref]

Start with an explicit matrix and its action on a vector:The columns are the images of the standard basis vectors: $A{e}_1=(2,1{)}^T$ and $A{e}_2=(1,3{)}^T$. Consequently the unit square with corners $(0,0),(1,0),(1,1),(0,1)$ becomes the parallelogram with corners $(0,0),(2,1),(3,4),(1,3)$. Its area is five. Every small patch of this square undergoes the same linear map, so the factor describes the whole deformation, rather than just one specially chosen boundary. The matrix mixes horizontal and vertical components, but its determinant compresses that combined effect into one area factor.^[ref]

Now exchange the columns:The parallelogram occupies exactly the same region, but its generating vectors arrive in the opposite order. Moving from the first column to the second now turns clockwise rather than counterclockwise. Ordinary area remains five; the oriented volume in two dimensions changes sign. A negative determinant therefore signals reversed orientation of a vector space, not a negative amount of material. Keeping the sign makes successive reversals calculable: two reversals restore the original orientation of a vector space.

Compare a singular matrix:Its columns are linearly dependent: the second is twice the first. The unit square collapses onto the segment from $(0,0)$ to $(3,6)$, which has zero area despite nonzero length. Since $C(-2,1{)}^T=0$, the kernel of a linear map contains a nonzero vector; inputs differing by that vector have identical outputs. The collapsed picture explains why an inverse matrix cannot exist.

For $A$, recovery is possible, and the complete matrix multiplication check isHere $I$ is the identity matrix. The inverse matrix restores the original square, shrinking its area by $1/5$. Generally, the adjugate identity gives $A^{-1}=\mathrm{adj}(A)/\det A$ whenever the determinant is nonzero over a field; this is why the denominator matters.^[ref]

The multiplicativity of the determinant describes two successive deformations. Matrix multiplication means that $AB$ applies $B$ first, then $A$. TakeThe first linear map doubles area; the second multiplies the resulting area by five. Thus the unit square passes through areas $1,2,10$, with the last parallelogram bounded by $(0,0),(2,1),(4,7),(2,6)$. Direct calculation agrees:This explains physically why area factors multiply instead of adding. The general identity holds for square matrices of the same size, including singular matrices. If either stage loses a direction, later processing cannot recover the discarded information.^[ref]

The same interpretation works in three dimensions without drawing a new picture. The diagonal matrixsends the unit cube to a rectangular box with side lengths two, three, and one. Its ordinary volume is six. The minus sign comes from reversing the third direction, so the oriented volume is minus six. Replacing the last diagonal entry by zero flattens the cube into a planar region: its area may be positive, but its three-dimensional volume is zero. This distinction matters whenever a linear map changes shape: the determinant measures size in the full input dimension, rather than the size of whichever lower-dimensional object remains.

The determinant measures neither trace nor perimeter. For $A$, the trace happens also to be five, but the identity matrix in two dimensions has trace two and determinant one. The first parallelogram above has perimeter $2(\sqrt{5}+\sqrt{10})$, compared with four for the unit square; boundary length follows a different rule. Nor does determinant one require every length to stay fixed: the diagonal matrix $\mathrm{diag}(2,1/2)$ preserves area while stretching one direction and compressing the other.

These properties make the determinant useful when coordinates obscure what a deformation does. A nonzero determinant certifies that linearly independent input directions remain distinguishable; its magnitude counts the area or volume represented by a transformed cell. That second role reappears in the change of variables formula, where the Jacobian determinant corrects the sizes of small cells. Signed volume also separates a deformation that reverses the orientation of a vector space from one that merely stretches it, even when both give identical ordinary volume factors.^[ref]

For general $n$, regard the columns $v_1,\dots ,v_n$ as ordered arguments. The determinant is the unique alternating multilinear map to the scalar field normalized by $\det I=1$. The multilinearity of the determinant means linearity in each column separately, with all others held fixed:Alternation means equal columns give zero; exchanging columns negates the result. Normalization sets the oriented volume of the standard basis to one. These properties also hold in rows. For real three-dimensional columns, their determinant is the scalar triple product, and its absolute value is the enclosed volume. Over other fields, the algebraic definition still works even without a geometric notion of positive volume.^[ref]

Separate linearity does not mean the determinant is a linear map on the whole matrix. Scaling every column of an $n$-by-$n$ matrix by $t$ gives $\det (tA)=t^n\det A$. For our two-dimensional example, $\det (2A)=4\det A=20$: doubling both directions quadruples area. The power records how many independent directions are being scaled together.

The properties suggest practical computations. Adding a multiple of one column to another preserves the determinant: the extra term has repeated columns and vanishes. This is the algebra behind an area-preserving shear mapping. Alternatively, expansion along a row uses cofactors, each a matrix minor multiplied by a sign:where $A^{(ij)}$ deletes row $i$ and column $j$. The Leibniz formula for determinants supplies the general entry formula. Both calculations encode the same geometric invariant; the defining properties explain their structure without requiring memorization of a separate recipe for every size.^[ref]

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]

A positive diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with $d_i>0$, is an invertible matrix and changes the coordinate scales through the similarity transformation $B=D^{-1}AD$. Its diagonal entries are still $a_{ii}$, while $b_{ij}=a_{ij}{d}_j/d_i$. Thus the Gershgorin discs have the same centres but different radii. Matrix similarity preserves the characteristic polynomial and eigenvalues, so the original and scaled disk unions are both valid bounds and may be intersected. Suitable scaling can compensate for badly unbalanced off-diagonal entries; it need not shrink every radius.^[ref]

For example,The original Gershgorin discs are $|z-2|\le 100$ and $|z-3|\le 0.01$. Their union is simply the huge first disk; the small second disk contains no eigenvalue. After scaling, both radii are $1$, giving $|z-2|\le 1$ or $|z-3|\le 1$. The two Gershgorin discs overlap, so their individual eigenvalue counts are not certified. The scaled matrix is a real symmetric matrix, however, so matrix similarity now proves that the original matrix has real eigenvalues in $[1,4]$ and is an invertible matrix. Directly, both matrices have characteristic polynomial ${\lambda }^2-5\lambda +5$, henceThe scaling changes the available bounds, while these eigenvalues stay fixed.

A scalar $\lambda$ is an eigenvalue of a linear operator $A:V\to V$ whenfor some nonzero vector $v$, called an eigenvector. The scalar must belong to the field over which $V$ is a vector space. Geometrically, a real eigenvalue tells us what $A$ does along a special line: every vector on that line is multiplied by the same number. The eigenvector gives the line and the eigenvalue gives its scaling.

For a real eigenvalue, $\lambda >0$ preserves the orientation along the line, $\lambda <0$ reverses it, and $\lambda =0$ collapses the line to zero. The length is multiplied by the modulus $|\lambda |$, so $|\lambda |>1$ stretches and $0<|\lambda |<1$ contracts. For example, the diagonal matrix $\mathrm{diag}(2,-1/2,0)$ doubles one coordinate direction, reverses and halves another, and annihilates the third. Its eigenvalues are $2,-1/2,0$; the last direction lies in its kernel.

For a finite square matrix, the eigenvalues are the roots of a polynomial $p_A$, its characteristic polynomial:Here $I$ is the identity matrix. The reason is that $Av=\lambda v$ means $(A-\lambda I)v=0$ with $v\ne 0$: the shifted matrix must have a nontrivial kernel, equivalently a zero determinant. The worked symmetric matrix example in eigenvector shows both stages: find the eigenvalues, then find their eigenspaces.

Eigenvalues are especially useful for repeated application of a linear map. If $Av=\lambda v$, thenOne multiplication proves the next step, so a matrix power becomes an ordinary power on each eigenvector. This explains growth, decay, and alternating signs along these directions. Gilbert Strang develops this connection, and its extension to the matrix exponential, in his lecture "Eigenvalues and Eigenvectors" (MIT 18.065, 2018).^[ref]

A documented historical connection is C. G. J. Jacobi's "Über ein leichtes Verfahren die in der Theorie der Säcularstörungen vorkommenden Gleichungen numerisch aufzulösen" (1846, Journal für die reine und angewandte Mathematik 30, pp. 51–94). Its opening discusses symmetric matrices arising from small oscillations, with a common unknown subtracted from every diagonal entry. In modern notation this is $(A-\lambda I)v=0$; Jacobi seeks the parameter and the ratios of the components of $v$. This is evidence of an early physical and computational setting for the eigenvalue problem, without assigning its discovery to a single person.^[ref][ref]

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

With $A=D+L+U$, the Gauss-Seidel method solves $(D+L)x^{(k+1)}=b-U{x}^{(k)}$. It uses each newly computed coordinate immediately in the remaining triangular solve. It is successive over-relaxation with parameter one, and converges from every start when $A$ is a symmetric positive-definite matrix.

To solve a linear system $Ax=b$, split its matrix into the diagonal matrix $D$ and the strictly lower and upper parts $L,U$, retaining their actual signs. Every diagonal entry must be nonzero. A sweep updates coordinates in order:The Jacobi method uses the old vector throughout; the Gauss-Seidel method reuses each fresh coordinate. Thus the update order is part of the method.^[ref]

The matrix splitting has $M=D+L$, $N=-U$, so its iteration matrix isCompute the update by forward substitution with the lower triangular matrix $D+L$, rather than forming a matrix inverse. For a sparse matrix, a sweep only visits stored entries and can overwrite the working vector. Dependencies on fresh coordinates constrain parallel execution. This makes the Gauss-Seidel method an inexpensive solver component, including in a multigrid method.^[ref]

Let $C$ be a connected component of the union of the row Gershgorin discs of an $n\times n$ matrix. If $C$ comprises $k$ of the indexed Gershgorin discs, then it contains exactly $k$ eigenvalues, counted with algebraic multiplicity. Coincident disks still count separately, once for each row. In particular, a Gershgorin disc disjoint from all the others contains one simple eigenvalue. More generally, any union of $k$ indexed Gershgorin discs disjoint from the remaining $n-k$ contains exactly $k$ eigenvalues, whether or not that union is connected.^[ref]

The separation is essential: overlapping Gershgorin discs do not individually promise one eigenvalue each. Discs that touch at a boundary point belong to the same connected component. For example, the matrix $\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 0\\ 0& 0& 8\end{array}\right)$ has Gershgorin discs of radii $1,0,0$ with centres $2,2,8$. The component around $2$ comprises two indexed disks and contains the eigenvalue $2$ with algebraic multiplicity two; the isolated point at $8$ contains the remaining eigenvalue.

For intuition, continuously turn on the off-diagonal entries using the diagonal matrix $E=\mathrm{diag}(a_{11},\dots ,a_{nn})$ and $A(t)=E+t(A-E)$, $0\le t\le 1$. At $t=0$ the eigenvalues are the disk centres; at every later time the Gershgorin discs stay inside their final union. The continuity of the eigenvalues as a collection means that no eigenvalue can cross a gap between final components. The count gives a useful certificate that an eigenvalue cluster is complete.^[ref]

When a ground state exists, its ground-state degeneracy is the dimension of the ground-state subspace:Here $I$ is the identity operator, and the kernel of a linear map is taken within the operator domain of the Hamiltonian operator $H$. The count concerns linearly independent energy eigenstates, rather than every normalized quantum superposition or every choice of global phase. Even a single energy eigenspace contains infinitely many representing vectors; its dimension measures independent freedom. Ground-state degeneracy is the lowest-energy instance of degenerate energy levels.

For an exact finite model with energy unit $\epsilon >0$, let the Hamiltonian operator be the diagonal matrixIts ground-state energy is zero and $g_0=2$: the first two coordinate unit vectors $e_1,e_2$ span the ground-state subspace. Every normalized $a{e}_1+b{e}_2$, with $|a{|}^2+|b{|}^2=1$, is a ground state, including $(e_1+e_2)/\sqrt{2}$. Its energy measurement is certain to return zero, even though a measurement distinguishing $e_1$ from $e_2$ need not have a definite outcome. Under the unchanged Hamiltonian operator, this quantum superposition is a stationary state because both components acquire the same global phase.

Ground-state degeneracy matters when an interaction selects or splits previously equal energy eigenvalues. In the finite model, adding $\epsilon \mathrm{diag}(\delta ,-\delta ,0,0)$ with $0<\delta <2$ gives the formerly lowest pair the distinct energies $\pm \epsilon \delta$, leaving a unique ground state. This direct calculation illustrates why degenerate perturbation theory first examines the perturbation inside the whole ground-state subspace.

A classical continuous-symmetry model can instead have a manifold of minimizing configurations, such as the sphere of orientations of a Néel state in an isotropic Heisenberg antiferromagnet. That geometric description does not by itself count the dimension of a finite quantum ground-state subspace.

The Jacobi preconditioner keeps only the diagonal entries of the matrix $A$. For a symmetric positive-definite matrix, these entries are positive, so its auxiliary solve is $z_i=r_i/a_{ii}$. It rescales coordinates with very different diagonal sizes, but cannot remove arbitrary off-diagonal coupling.^[ref]

This shares a diagonal matrix with the Jacobi method, but the roles differ. The Jacobi method specifies a whole stationary iterative method for a linear system; a Jacobi preconditioner supplies an auxiliary solve to another iterative method, including the Preconditioned conjugate gradient method.

A matrix decomposition expresses a matrix as a matrix product of factors with simpler structure. An LU decomposition uses triangular factors, a QR decomposition uses an orthogonal matrix and a triangular factor, and a singular value decomposition separates input directions, stretching factors and output directions. The factorization is an identity; the chosen structure determines what it helps compute.

For an original two-by-two example,The outer factors are a lower triangular matrix and an upper triangular matrix, each with unit diagonal as in a unit triangular matrix; the middle factor is a diagonal matrix. Their matrix product is $M$, and the final diagonal entry $5/2$ is the Schur complement of the first entry $2$. This is the scalar version of the factorization under block elimination.

To solve $Mz=(3,4{)}^T$, solve with the three factors in their displayed order from left to right: first the lower triangular factor, then the diagonal factor, then the upper triangular factor. The intermediate vectors are $(3,5/2{)}^T$ and $(3/2,1{)}^T$, giving the concise answer $z=(1,1{)}^T$. A new right-hand side uses the same factors. This is why Reusing LU factors for multiple right-hand sides is useful: factorization records work that otherwise would be repeated, while forward substitution and back substitution apply it cheaply.^[ref]

Different decompositions reveal different features. A Cholesky decomposition uses positive definiteness of a symmetric matrix, while a singular value decomposition also applies to rectangular matrices and exposes matrix rank. The existence and numerical behaviour of the factors depend on their own hypotheses; for example, an unpivoted LU decomposition can fail when a chosen pivot element is zero even though the matrix is an invertible matrix.

A matrix preconditioner is an auxiliary invertible matrix whose matrix inverse is cheap to apply and whose use improves the eigenvalues of a linear system. In symmetric positive-definite matrix problems, a symmetric positive-definite matrix $M$ preserves that structure under the symmetric transformation $M^{-1/2}A{M}^{-1/2}$.

Applying $M^{-1}$ means solving $Mz=r$ for a supplied vector $r$. This notation does not require forming a general matrix inverse. A diagonal matrix gives componentwise divisions; a factored matrix gives auxiliary solves such as forward substitution and back substitution.^[ref]

Choosing $M=I$ makes the solve trivial but changes nothing; choosing $M=A$ gives the exact correction but asks us to solve the original linear system.^[ref]

Two square matrices $A,B$ over the same field are similar if $B=S^{-1}AS$ for an invertible matrix $S$. They can describe one linear operator using two different ordered bases, with the same basis used for input and output in each description. Fix the convention $x=Sy$: $x$ is the old coordinate vector, $y$ the new coordinate vector, and the columns of $S$ are the new basis vectors in old coordinates. This is the convention of the change-of-basis matrix.^[ref]

For a concrete linear operator on the physical plane, let $T$ double horizontal components and multiply vertical components by five. In the old standard basis $\mathcal{E}=(e_1,e_2)$, its matrix of a linear map is $A=\mathrm{diag}(2,5)$. Choose the new ordered basis $\mathcal{F}=(b_1,b_2)$ with $b_1=e_1$, $b_2=e_1+e_2$. ThenThe second column of $B$ says $T(b_2)=-3b_1+5b_2=2e_1+5e_2$. Its off-diagonal entry records how the chosen basis vectors describe the output; the physical linear operator still stretches along the old horizontal and vertical directions.

Take the physical vector $p=2e_1+e_2=b_1+b_2$. Its two coordinate vectors, and the two descriptions of $T(p)$, areReconstruction gives $T(p)=4e_1+5e_2=-b_1+5b_2$. The output coordinate vectors differ, but the endpoint in the physical plane is the same.

This is why matrix similarity is useful: a convenient basis can simplify calculations without changing the linear operator. A diagonalizable matrix can be replaced by a diagonal matrix, making repeated application a calculation with powers of its eigenvalues. In the example, $A$ makes the two independent stretch factors visible, while $B$ expresses the same action in directions chosen for another purpose. Matrix similarity invariants separate properties of the linear operator from features of its address system.

A splitting $A=M-N$ with invertible $M$ produces

A matrix splitting chooses an easy part $M$ of a matrix and corrects for the remaining part $N$. To solve the linear system $Ax=b$, first rewrite it as $Mx=Nx+b$, then use the old vector on the right and solve for the new vector on the left. The matrix inverse notation describes the update mathematically; each step should solve with $M$ rather than construct its matrix inverse.^[ref]

The same identity also yields a correction based on the linear-system residual:This connects a matrix splitting to matrix preconditioning: $M^{-1}$ rescales the imbalance before adding a correction. Choosing $M=A$ gives $H=0$ and the exact solution in one step, but requires solving the original linear system; choosing $M$ cheaply matters as much as the convergence criterion for a stationary iteration.

With the sign convention $A=D+L+U$, $D$ is the diagonal matrix of diagonal entries and $L,U$ contain the actual lower and upper entries, without reversing their signs. The Jacobi method uses $M=D$, $N=-(L+U)$; the Gauss-Seidel method uses $M=D+L$, $N=-U$. The first requires only division by diagonal entries; the second uses a lower triangular matrix and forward substitution. These choices explain their different information flows.

The Moore-Penrose inverse extends the useful part of a matrix inverse to rectangular and singular matrices. A linear system $Ax=b$ may have no exact solution, or may have many. The pseudoinverse resolves both ambiguities: $x_{\ast }=A^{+}b$ is the minimum-norm least-squares solution. First minimize the Euclidean norm $\Vert Ax-b{\Vert }_2$; among all vectors attaining that minimum, choose the one with smallest Euclidean norm $\Vert x{\Vert }_2$. It does not pretend that inconsistent equations can all be satisfied.

For $A\in {\mathbb{C}}^{m\times n}$, its Moore-Penrose inverse is the unique $n\times m$ matrix satisfying the four Penrose equations. The singular value decomposition gives a direct construction:Here $U,V$ are unitary matrices, the star is the conjugate transpose, and ${\Sigma }^{+}$ is the $n\times m$ rectangular diagonal matrix obtained by replacing each nonzero singular value ${\sigma }_j$ by $1/{\sigma }_j$ and leaving zeros at zero. Over the real numbers, use orthogonal matrices and transposes. The pseudoinverse reverses the stretches that can be reversed; it discards output components the original linear map cannot produce and does not invent an input component in its kernel. ^[ref]

The two products have especially clear meanings:They are orthogonal projection matrices: $A{A}^{+}$ selects the producible part of an output, while $A^{+}A$ selects the detectable part of an input. They are generally not identity matrices. For an invertible matrix, both are identities and $A^{+}=A^{-1}$.

This gives a useful order of operations for least-squares solutions: project the requested output onto the column space, then invert the restriction of $A$ from $(\ker A{)}^{\perp }$ to its column space. Adding a kernel vector leaves the residual unchanged but increases the squared Euclidean norm by the Pythagorean theorem in an inner-product space, because $A^{+}b$ is orthogonal to that kernel. The pseudoinverse therefore provides one canonical answer without hiding either lost information or irreconcilable data.

The name records a history with several contributors. E. H. Moore gave an earlier generalization in 1920; M. H. Pearl's 1966 paper records that work, Arne Bjerhammar's 1951 rediscovery for maximal-rank rectangular matrices, and Roger Penrose's independent 1955 treatment. ^[ref] Penrose's original paper proves existence and uniqueness from the four Penrose equations for arbitrary complex matrices, including rectangular ones. ^[ref]