Eigenvector ($v$)

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.

A zero mode is a nonzero eigenvector of an operator with eigenvalue zero. For a differential fluctuation operator it must also satisfy the specified boundary conditions. A zero mode prevents ordinary inversion of the operator and makes its unmodified functional determinant vanish. Gauge directions can produce zero modes of a kinetic operator, while stationary-path focusing can produce them in a semiclassical boundary-value problem. Removing, separately integrating or otherwise treating these directions is necessary before using a Gaussian determinant formula.

For unstressed central Hookean springs, the quadratic elastic energy is $\frac{1}{2}{\sum }_{ij}{k}_{ij}[{\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j){]}^2$, with unit edge direction ${\mathbf{n}}_{ij}$. A common translation gives zero displacement difference. An infinitesimal rotation ${\mathbf{u}}_i=\mathbf{\Omega }\times {\mathbf{r}}_i$ also gives zero extension because a vector is perpendicular to its cross product with $\mathbf{\Omega }$. Thus a noncollinear three-dimensional free framework has three translational and three rotational zero modes. Extra mechanisms can give additional zero modes; six is a lower bound until rigidity is established.

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.

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

A right eigenvector satisfies $Av=\lambda v$.

A left eigenvector satisfies $u^{T}A=\lambda u^T$, equivalently $A^{T}u=\lambda u$ over the real numbers.