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If $Ax=\lambda x$, then $M=A-\lambda I$ is a normal matrix by part (ii), and $Mx=0$. Part (i), applied to $M$, yields $\Vert M^{†}x\Vert =\Vert Mx\Vert =0$. ThusThe nonzero vector $x$ is consequently an eigenvector of the conjugate transpose with eigenvalue $\overline{\lambda }$. This adjoint eigenvector identity for a normal matrix will also determine the special spectral restrictions below.

Set $M=A-\lambda I$. Its conjugate transpose is $M^{†}=A^{†}-\overline{\lambda }I$, soSince $A$ is a normal matrix, the two products coincide. Therefore $A-\lambda I$ is normal for every complex $\lambda$.

For the standard complex inner product, $\Vert y{\Vert }^2=y^{†}y$. Using the conjugate transpose and the defining equality for a normal matrix,Both norms are nonnegative, so $\Vert Ax\Vert =\Vert A^{†}x\Vert$ for every $x$.

The Moore--Penrose inverse is characterized by

Here $A^{†}=A^{+}$ denotes the Moore-Penrose inverse, while the star denotes the conjugate transpose. The first two identities express consistency when applying and reversing the linear map; the last two make the resulting orthogonal projection matrices Hermitian. They must hold together. For example, $A=(12)$ and $B=(10{)}^T$ satisfy $ABA=A$ and $BAB=B$, butis not Hermitian. This $B$ is a right inverse, yet $Bb=(3,0{)}^T$ is not the minimum-norm least-squares solution for $b=3$. The last two Penrose equations impose the orthogonality that selects the pseudoinverse.

The strict positivity of a quadratic form on every nonzero vector. A real symmetric matrix has positive definiteness exactly when its associated quadratic form does, making it a positive-definite matrix. A Hermitian form is positive definite when its value on $(v,v)$ is real and strictly positive for every nonzero vector.

Think of a positive-definite quadratic form as an energy cost for a displacement: every direction costs something, and zero cost identifies the origin. For example, the real symmetric matrixis a positive-definite matrix because the two squares can vanish together only at $(0,0)$. The mixed term couples the coordinates, but cannot overwhelm their positive contributions: $q(1,1)=8$ and $q(1,-1)=4$. The graph is a bowl, and its positive level sets are ellipses with axes along the eigenvectors, rather than along the original coordinates.

The real and complex settings require different symmetry assumptions. For a real symmetric matrix use $q(x)=x^{T}Ax$ with $x\in {\mathbb{R}}^n$; for a complex Hermitian matrix use $q(z)=z^{\ast }Az$ with $z\in {\mathbb{C}}^n$, where $z^{\ast }$ is the conjugate transpose. The latter expression is real, so strict positivity is meaningful. A complex quadratic form such as $z_1^2+z_2^2$ has no such positivity: it vanishes at $(1,i)$ and takes negative real values at $(i,0)$. A Hermitian form uses complex conjugation, which changes this example to $|z_1{|}^2+|z_2{|}^2$.

For a real matrix $M$ without a symmetry hypothesis, its quadratic form depends only on its symmetric part of a matrix:Positive eigenvalues of $M$ alone do not establish positive definiteness of this quadratic form. For example, $M=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has both eigenvalues equal to $1$, yet $q(1,-1)=1-4+1=-2$. The relevant test is positivity of the symmetric part of a matrix, not the spectrum of an arbitrary matrix.^[ref]

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]

The Rayleigh quotient measures the quadratic form of a self-adjoint operator per unit squared norm. For a nonzero vector $x$ and a finite-dimensional Hermitian matrix $A$,For a real symmetric matrix, replace the conjugate transpose $x^{\ast }$ by $x^T$. More generally, for a self-adjoint operator $H$ on a Hilbert space, use $R_H(\psi )=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$ for nonzero $\psi$ in its operator domain. On an eigenvector it returns the exact eigenvalue. Multiplying the vector by any nonzero scalar leaves the quotient unchanged, so it measures a direction rather than a chosen length.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ from the finite-dimensional spectral theorem example,The first value is not an eigenvalue: it is an average of the two directional responses. To see every direction, write the unit vector $q(\theta )=\cos \theta e_{+}+\sin \theta e_{-}$ in the orthonormal eigenbasis. ThenThe largest value is $3$ along $e_{+}$; the smallest is $1$ along $e_{-}$. A direction halfway between them has quotient $2$. In the left panel, the color on the unit circle represents the quotient; in the right panel, the same values are plotted against the angle measured from $e_{+}$.

The Hermitian Rayleigh quotient range explains why this is useful: evaluating one trial vector gives an upper bound on the smallest eigenvalue and a lower bound on the largest one, without solving a characteristic polynomial. At an extreme eigenvector, a small angular error changes the quotient only quadratically. Near $e_{+}$ in this example, $3-R_A(q(\theta ))=2{\sin }^2\theta =2{\theta }^2+O({\theta }^4)$. This is the geometric reason for quadratic Rayleigh-quotient improvement for the power method and motivates the Rayleigh-Ritz method.

There is a concrete historical link with normal modes. In The Theory of Sound (1877), Volume I, Sections 88–89, Lord Rayleigh studies the period of a constrained trial vibration. He shows that the period has a stationary point at a normal mode and that a trial type bounds the longest natural period. These energy ratios are a mechanical setting for the modern Rayleigh quotient and generalized Rayleigh quotient; the book describes the result through vibration periods, rather than today's matrix notation.^[ref]

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.

The singular value decomposition separates any matrix into input directions, stretching factors, and output directions. Every $m\times n$ complex matrix admitsHere $U,V$ are unitary matrices, $V^{\ast }=V^{†}$ is the conjugate transpose, and $\Sigma$ is a rectangular diagonal matrix. Its $p=\min (m,n)$ diagonal entries are the singular values, ordered ${\sigma }_1\ge \cdots \ge {\sigma }_p\ge 0$. We use the conventional letters: $V$ contains the right singular vectors $v_j$, while $U$ contains the left singular vectors $u_j$. Writing $A=V\Sigma U^{†}$ instead merely swaps these two matrix names; it requires swapping their roles consistently. For real matrices, choose orthogonal matrices and replace the conjugate transpose with the transpose.^[ref]

Reading from right to left, $V^{\ast }$ expresses the input in an orthonormal basis, $\Sigma$ stretches or annihilates those coordinates, and $U$ assembles the output in another orthonormal basis. Unlike matrix similarity, the input and output bases can differ, and even have different sizes. A $3\times 2$ matrix has full factors of sizes $3\times 3$, $3\times 2$, and $2\times 2$. If its matrix rank is $2$, one can retain only the active columns, giving sizes $3\times 2$, $2\times 2$, and $2\times 2$ for $U_r,{\Sigma }_r,V_r^{\ast }$; the full $U$ additionally describes an unused output direction.

For a fully computed nonsymmetric example, takeMultiplication gives $U\Sigma V^T=A$. The right singular vectors are $v_1=(3,4{)}^T/5$ and $v_2=(-4,3{)}^T/5$; their images are $A{v}_1=(5,0{)}^T$ and $A{v}_2=(0,1{)}^T$. Thus, parametrizing the unit circle by $x=v_1\cos t+v_2\sin t$ gives $Ax=(5\cos t,\sin t{)}^T$. Its image is the ellipse $y_1^2/25+y_2^2=1$. The two singular values are its semiaxis lengths, and the left singular vectors point along its axes. The singular value decomposition finds precisely the perpendicular input directions whose images stay perpendicular.

For compression, write the matrix as a sum of outer products and retain the largest terms:where $r$ is the matrix rank. Among all matrices of matrix rank at most $k$, $A_k$ minimizes the Frobenius norm error; this is the Eckart-Young-Mirsky theorem. The operator norm error is ${\sigma }_{k+1}$ for $k<r$. These statements concern specified norms, rather than guaranteeing that discarded details are visually or scientifically unimportant.^[ref]

Consider an $8\times 10$ brightness matrix $B=p{q}^T+a{b}^T/2$, withand $a_i=(-1{)}^{i-1}$, $b_j=(-1{)}^{j-1}$. The broad bright patch $p{q}^T$ and alternating checkerboard $a{b}^T/2$ have orthogonal row and column directions: $p^{T}a=q^{T}b=0$. Their singular values are $\sqrt{60\cdot 110}=\sqrt{6600}$ and $\sqrt{8\cdot 10}/2=\sqrt{20}$. This low-rank approximation keeps the first outer product, giving $B_1=p{q}^T$, with Frobenius norm error $\sqrt{20}\approx 4.472$ and relative error $\sqrt{20/6620}\approx 5.50\%$. For example, $B_{11}=1.5$ becomes $1$, and $B_{12}=1.5$ becomes $2$: every entry changes by $0.5$ in magnitude. The discarded pattern is small but entirely lost.

For this illustrative image compression, storing one normalized left singular vector, one normalized right singular vector, and one singular value requires $8+10+1=19$ scalars instead of $80$. This counts scalars at equal precision, before storage overhead. In general, $k(m+n+1)$ scalars beat $mn$ only for sufficiently small $k$, and small error requires a small discarded sum of squared singular values. The usefulness comes from measuring the tradeoff before throwing information away.^[ref]

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.