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For a finite-dimensional Hermitian operator $A$ with decreasing eigenvalues ${\lambda }_j$ and integer $0\le r\le d$, maximizing $\mathrm{Tr}(AB)$ over positive contractions $B$ with trace $r$ gives the sum of its leading $r$ eigenvalues. In an orthonormal eigenbasis, the diagonal entries of $B$ lie in $[0,1]$ and sum to $r$. Moving their weight to the largest eigenvalues can only increase the objective. An orthogonal projection onto their eigenvectors attains the maximum.

A bounded operator $T:H\to H$ on a real or complex Hilbert space is Hermitian, or self-adjoint, when $T=T^{\ast }$. Equivalently,Its directional values $q_T(x)=⟨Tx,x⟩$ are real, since this identity makes $q_T(x)$ equal its complex conjugate. Reality allows either sign; a positive operator additionally requires nonnegative values in every direction.

The diagram compares Hermitian quadratic forms along paths of unit vectors. The Hermitian matrix $D=\mathrm{diag}(1,-1)$ takes the values $1$ and $-1$ on the standard basis vectors; $P=\mathrm{diag}(1,0)$ takes the values $1$ and $0$; andsatisfies $H{e}_1=e_1$ and $H{e}_3=3e_3$. These eigenvectors form an orthonormal basis, and every $x=a{e}_1+b{e}_3$ has $q_H(x)=|a{|}^2+3|b{|}^2$. Thus $H$ is a positive operator, with eigenvalues $1,3$ and operator norm $3$.

Three conditions describe different geometry. A Hermitian operator satisfies $T=T^{\ast }$; a normal operator satisfies $T^{\ast }T=T{T}^{\ast }$; a unitary operator satisfies $T^{\ast }T=T{T}^{\ast }=I$, where $I$ is the identity operator. Every Hermitian operator and every unitary operator is a normal operator. Concrete matrices separate the conditions:

The spectrum of a bounded Hermitian operator is real. In finite dimension, the spectral theorem supplies an orthonormal basis of eigenvectors, turning its action into independent real scalings. This makes the geometry easy to interpret: each eigenvalue records a signed response along an orthogonal direction. An arbitrary bounded Hermitian operator in infinite dimension need not have any eigenvectors; the general spectral theorem uses a projection-valued measure. MIT's lecture 21 develops the real spectrum and the orthogonality of eigenspaces.^[ref]

A historical connection is the study of forms with conjugate variables. In his 1901 account of Charles Hermite's work, Camille Jordan describes these forms and emphasizes their role in reducing forms with real or complex coefficients. In today's notation, $x^{\ast }Hx$ is the diagonal value of a Hermitian form; the conjugate variables make a real-valued expression possible even with complex coefficients.^[ref]

For a densely defined operator outside this bounded setting, the displayed pairing identity alone defines a symmetric operator. A self-adjoint operator must also have the same operator domain as its adjoint operator.

A bounded Hermitian operator determines a Hermitian form $b_T(x,y)=⟨Tx,y⟩$ and its real-valued diagonal expression $q_T(x)=b_T(x,x)$. On a complex Hilbert space,Thus it is a quadratic form on the underlying real vector space; it is generally not a quadratic form over the complex field, whose scaling rule would involve ${\alpha }^2$. This distinction explains why replacing $x^{\ast }$ by $x^{\mathrm{T}}$ loses the geometry of a complex inner product.

For a finite-dimensional Hermitian operator with an orthonormal basis of eigenvectors $e_j$ and corresponding eigenvalues ${\lambda }_j$, the spectral theorem givesFor a unit vector, this is an average of the eigenvalues with nonnegative weights summing to one. It is also the Rayleigh quotient, so its minimum and maximum are the smallest and largest eigenvalues.

In the Hermitian operator diagram, take $x(\theta )=(\cos \theta ,\sin \theta {)}^{\mathrm{T}}$ for $D,P$, and $x_H(\theta )=\cos \theta e_1+\sin \theta e_3$ for $H$. The computed curves areThese paths illustrate the extremal values; the spectral theorem supplies the bounds for every complex unit vector, including those outside the plotted paths.

For a finite-dimensional Hermitian matrix $A$, the Rayleigh quotient has precisely the range $[{\lambda }_{\min },{\lambda }_{\max }]$, where the endpoints are its extreme eigenvalues. The finite-dimensional spectral theorem writes a nonzero vector as $x={\sum }_j{c}_j{e}_j$ in an orthonormal eigenbasis, givingThus the quotient is a convex combination of eigenvalues. The bounds follow immediately, and equality at either endpoint means that $x$ lies in the corresponding extreme eigenspace, including any repeated eigenvalue. Every intermediate value occurs: for orthonormal extreme eigenvectors, $x=\sqrt{1-t}{e}_{\min }+\sqrt{t}{e}_{\max }$ gives $R_A(x)=(1-t){\lambda }_{\min }+t{\lambda }_{\max }$ for $0\le t\le 1$. If all eigenvalues are equal, the quotient is constant.

The Hermitian hypothesis matters. The real matrix $B=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$ has only the eigenvalue $1$, but $(1,1)B(1,1{)}^T/2=3$. Its quotient is governed by the symmetric part of a matrix $(B+B^T)/2$, not by a real orthonormal eigenbasis of $B$. A nonsymmetric matrix therefore need not obey the displayed eigenvalue bounds.

The Hessian matrix describes how the slope of a function changes in different directions. For a real-valued function $f$ with continuous second partial derivatives on an open subset of ${\mathbb{R}}^n$, it isThe symmetry of second derivatives makes $H_f$ a symmetric matrix under these hypotheses. It is also the Jacobian matrix of the gradient: near $x$, a small displacement $h$ changes $\nabla f$ by $H_f(x)h+o(\Vert h\Vert )$. Thus the gradient measures the current slope, while the Hessian measures its local variation. ^[ref]

For example, take the quadratic formIts gradient and Hessian matrix areThe only critical point is the origin. Completing the square gives $q=(x+y{)}^2+2y^2$, so the origin is a strict global minimum. The eigenvalues $4-2\sqrt{2}$ and $4+2\sqrt{2}$ are both positive, recovering the same conclusion from the positive definiteness of the Hessian. Its eigenvectors identify the rotated directions in which the quadratic form has its smallest and largest bending.

For a fixed vector $v$, restriction to a straight line gives the directional interpretationThis is the quadratic form associated with the Hessian matrix. For a unit eigenvector $v$, it equals the corresponding eigenvalue. Opposite signs therefore mean that some directions bend upwards and others downwards. The second-derivative test turns this information into a classification when the gradient vanishes.

The same object is useful for computation. Newton's method in optimization chooses a displacement $h$ by solving $H_f(x)h=-\nabla f(x)$ when the Hessian matrix is an invertible matrix; this makes the gradient of the local quadratic model vanish. For the displayed $q$, the equation gives $h=-(x,y{)}^T$, so one exact step reaches the minimum from any starting point. For a general function, a Hessian with opposite-sign eigenvalues or a large condition number need not give a useful minimizing step, and the local minimum need not be a global minimum.

The name honors Otto Hesse. ^[ref] An older geometric use is the Hessian criterion for flexes of a plane cubic, connecting a determinant of second partial derivatives with inflection points of an algebraic curve. Hesse's 1844 paper Über die Wendepuncte der Curven dritter Ordnung is an early source on that geometric problem. ^[ref]

On a positive eigenvalue label $\lambda$ and a clean quantum ancilla, the HHL algorithm applies a quantum variable rotation with angle ${\theta }_{\lambda }=\arcsin (c/\lambda )$, where $0<c\le {\lambda }_{\min }$:Uncomputation of the eigenvalue register followed by conditioning on flag one multiplies each input eigenvector amplitude by $c/\lambda$. If $|b⟩$ is normalized, the success probability is $c^2\Vert A^{-1}|b⟩{\Vert }^2$. Choosing $c={\lambda }_{\max }/\kappa$ from a valid condition number bound gives success probability at least $1/{\kappa }^2$.

For a one-dimensional first-order system $v_t+A(v,x,t)v_x=s$, hyperbolicity means that the coefficient matrix has real eigenvalues and a complete set of eigenvectors. Distinct real eigenvalues give strict hyperbolicity and separate characteristic curves. Along a curve of speed $\lambda$, a left eigenvector $\ell A=\lambda \ell$ gives the compatibility equation $\ell dv/dt=\ell s$.

The identity operator maps every vector to itself. Every nonzero vector is an eigenvector with eigenvalue one.

Applying the product of the other eigenvalue factors to a linear relation isolates each coefficient. Distinct eigenvalues therefore give linearly independent eigenvectors. A complete eigenvector basis diagonalizes the associated linear system of ordinary differential equations.

A square nonnegative matrix $A$ is irreducible if for each pair $(i,j)$ there is an integer $k\ge 0$ with $(A^k{)}_{ij}>0$. This says that every index can reach every other through positive entries. The Perron–Frobenius theorem then gives a positive leading eigenvector and an algebraically simple eigenvalue. Irreducibility alone permits other eigenvalues of the same modulus: the cyclic permutation matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has eigenvalues $1,-1$.

An irreducible group representation is nonzero and has no nonzero proper invariant subspace. These are the pieces that cannot be decomposed further by taking a smaller symmetry-preserved vector space.

The real two-dimensional rotation through $120$ degrees illustrates the role of the coefficient field. A one-dimensional real invariant subspace would supply a real eigenvector, but the characteristic polynomial ${\lambda }^2+\lambda +1$ has no real root. The representation is therefore irreducible over the real numbers. Over the complex numbers, that polynomial has distinct roots of unity $e^{2\pi i/3}$ and $e^{-2\pi i/3}$, giving two invariant eigenlines. Irreducibility over the real numbers does not imply absolute irreducibility.

For commuting diagonalizable operators $A_1,\dots ,A_r$ on a finite-dimensional vector space, the joint spectrum is the set of tuples $(a_1,\dots ,a_r)$ for which there is a common nonzero eigenvector $v$ with $A_{i}v=a_{i}v$. Commuting self-adjoint operators on a finite-dimensional inner product space admit a simultaneous orthonormal eigenbasis.

In a Hermitian neutral-kaon mass matrix, independently choosing square-root signs of $z$ and $z^{\ast }$ can interchange the eigenvalue labels. Choose $z=r{e}^{i\phi }$, $a=\sqrt{r}{e}^{i\phi /2}$ and $b=\sqrt{r}{e}^{-i\phi /2}$, so $ab=r>0$, continuously from the desired CP-conserving limit. Then $(a,-b)$ is the lower-mass eigenvector. At a degenerate off-diagonal zero the mass eigenbasis is not fixed by diagonalization alone.

The kernel of the Graph Laplacian consists exactly of vectors constant on each connected component of a graph. Indeed, $Lx=0$ implies $x^{T}Lx=0$, so the Graph Laplacian quadratic form forces equality at the endpoints of every edge; equality then propagates along graph paths. Conversely, these equalities make every entry of $Lx$ zero. Thus the dimension of the kernel, equivalently the multiplicity of the zero eigenvalue, counts the connected components of a graph.

For the disconnected graph with just two edges, $AB$ and $CD$, in order $A,B,C,D$,The two displayed vectors form a basis of the kernel. The other two eigenvectors $(1,-1,0,0{)}^T$ and $(0,0,1,-1{)}^T$ each have eigenvalue two, so the full list of eigenvalues, including multiplicities, is $0,0,2,2$. Separate connected components of a graph allow independent constant levels; a single global constant is only one of these possibilities.

The Kronecker sum of square matrices combines their Kronecker products with identity matrices. If $Ax=\lambda x$ and $By=\mu y$, then $x\otimes y$ is an eigenvector with eigenvalue $\lambda +\mu$. Tensor-product spatial grids therefore turn separated coordinate Laplacians into a Kronecker sum. Two Hermitian matrices give a Hermitian Kronecker sum, allowing an energy method without computing every eigenvector.

For a real symmetric matrix $X$ and $1\le k\le n$,Indeed $Y=Q{Q}^T$ is a rank-$k$ orthogonal projection matrix in the fantope, and projecting onto the eigenvectors of the largest eigenvalues attains the support function maximum. Equivalently, the largest matrix trace of a compression to a $k$-dimensional vector subspace is the sum of the largest eigenvalues.