= Eigenvector
{title2=$v$}
{wiki}
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 $\lambda\ne0$, 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 matrix>
$$
A=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2x+y\\x+2y\end{pmatrix}.
$$
Its <characteristic polynomial> is
$$
p_A(\lambda)=\det\begin{pmatrix}\lambda-2&-1\\-1&\lambda-2\end{pmatrix}=(\lambda-2)^2-1=(\lambda-3)(\lambda-1).
$$
For $\lambda=3$, $(A-3I)v=0$ 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:
$$
Av_3=\begin{pmatrix}3\\3\end{pmatrix}=3v_3,\qquad Av_1=\begin{pmatrix}1\\-1\end{pmatrix}=v_1.
$$
The map stretches the line $y=x$ threefold and fixes the line $y=-x$ pointwise. These <eigenvectors> are <orthogonal> because $v_3^Tv_1=0$; dividing each by $\sqrt2$ 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>.
\Image[/eigenvector-geometry.png]
{title=Eigenvector geometry: symmetric stretch, quarter-turn rotation, and shear}
A contrasting example is the <planar rotation> through $90$ degrees:
$$
R=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad R\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-y\\x\end{pmatrix}.
$$
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.
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