= Eigenspace
{title2=$E_\lambda=\ker(A-\lambda I)$}
{wiki}
For an <eigenvalue> $\lambda$ of a <linear operator> $A:V\to V$, its eigenspace is
$$
E_\lambda=\ker(A-\lambda I)=\{v\in V:Av=\lambda v\}.
$$
It 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> $A=\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$ from <eigenvector>,
$$
E_3=\operatorname{span}\{(1,1)^T\},\qquad E_1=\operatorname{span}\{(1,-1)^T\}.
$$
Each <eigenspace> is a line. An <eigenspace> can also be a plane or the whole <vector space>: for $B=\operatorname{diag}(2,2,5)$, $E_2=\{(x,y,0)^T:x,y\in\mathbb R\}$ 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 as
$$
\begin{pmatrix}x\\y\end{pmatrix}=\frac{x+y}{2}\begin{pmatrix}1\\1\end{pmatrix}+\frac{x-y}{2}\begin{pmatrix}1\\-1\end{pmatrix}.
$$
This decomposition makes the independent scalings visible without multiplying out the original <matrix> again.
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