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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.

An eigenstate of an observable is a quantum state represented by an eigenvector of that observable.

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]

The eigenvalue equation $Av=\lambda v$ asks for a nonzero eigenvector $v$ and its eigenvalue $\lambda$.

An eigenvalue problem asks for the eigenvalues and corresponding eigenvectors or eigenfunctions of a linear operator. It is commonly written as the eigenvalue equation $Av=\lambda v$.

A nonzero vector $v$ is an eigenvector of a linear operator $A$ when $Av=\lambda v$ for some eigenvalue $\lambda$. For a real vector space and $\lambda \ne 0$, the line through $v$ is preserved: the image can point forward or backward, but cannot turn away from that line. When $\lambda =0$, the image is zero and has no direction. The zero vector is excluded because it satisfies the equation for every scalar and identifies no special direction.

Every nonzero scalar multiple of an eigenvector is an eigenvector for the same eigenvalue, since $A(cv)=cAv=\lambda (cv)$. Thus there is usually no preferred length or sign; choosing a unit vector is a convenient normalization. Eigenvectors for a fixed eigenvalue, together with zero, form its eigenspace.

For a fully computed example, consider the real symmetric matrixIts characteristic polynomial isFor $\lambda =3$, $(A-3I)v=0$ gives $-x+y=0$, so $v_3=(1,1{)}^T$ is an eigenvector. For $\lambda =1$, $(A-I)v=0$ gives $x+y=0$, so $v_1=(1,-1{)}^T$ is an eigenvector. Direct multiplication checks both:The map stretches the line $y=x$ threefold and fixes the line $y=-x$ pointwise. These eigenvectors are orthogonal because $v_3^T{v}_1=0$; dividing each by $\sqrt{2}$ gives an orthonormal basis. A general vector turns because its components along these two lines receive different scalings. For example, $(1,0{)}^T$ maps to $(2,1{)}^T$ and is not an eigenvector.

A contrasting example is the planar rotation through $90$ degrees:Every nonzero real vector turns to a perpendicular direction, so there are no real eigenvectors. Its characteristic polynomial is ${\lambda }^2+1$, which has no real roots. Over the complex numbers, however, $R(1,-i{)}^T=i(1,-i{)}^T$ and $R(1,i{)}^T=-i(1,i{)}^T$. The same matrix therefore has an eigenbasis in ${\mathbb{C}}^2$ even though it has none in ${\mathbb{R}}^2$: the choice of field matters.

Under $B=S^{-1}AS$, an eigenvector in old coordinates becomes $S^{-1}v$ in new coordinates. Indeed,Since $S^{-1}$ is an invertible matrix, the new coordinate vector is nonzero. The inverse conversion $S$ proves that the two eigenspaces have the same dimension, hence the same geometric multiplicity.

For $A=\mathrm{diag}(2,5)$, the eigenvector $e_2=(0,1{)}^T$ has eigenvalue $5$. In the oblique basis of the opening example, its coordinate vector isThe tuple changes from $(0,1)$ to $(-1,1)$, but reconstruction gives $-b_1+b_2=e_2$: the physical vertical direction is unchanged. Calling the eigenvectors unchanged is accurate only for the underlying vectors, not for their coordinate vectors.