For one eigenvalue there is usually not just one eigenvector but a whole subspace of them, a line or a plane that the map stretches by the same factor. How large that subspace is, compared with how often the eigenvalue repeats as a root, decides whether the map can be described simply.

A number repeated in a list can stand for one thing written twice or for two different things. An eigenvalue repeated as a root may come with one line of eigenvectors or with a whole plane of them.

The eigenspace of an eigenvalue $\mu$ is $E_\mu=\ker(A-\mu I)$: its eigenvectors together with $\mathbf{0}$. It is a subspace, and a basis of it comes from the free variables of $(A-\mu I)\mathbf{v}=\mathbf{0}$. The geometric multiplicity of $\mu$ is $\dim E_\mu$. Its algebraic multiplicity, written $a_\mu$ here, is the number of times $\mu$ is a root of the characteristic polynomial, that is, the power of $(\lambda-\mu)$ in it.

• $\dim E_\mu=n-\operatorname{rank}(A-\mu I)$, by rank–nullity.
• $1\le\dim E_\mu\le a_\mu$ for every eigenvalue: the theorem below.
• $\begin{pmatrix}2&1\\0&2\end{pmatrix}$ has $2$ as a double root, , but is only the line spanned by : geometric multiplicity .
• $2I$ has the same characteristic polynomial, $(\lambda-2)^2$, and $E_2$ is the whole plane: geometric multiplicity . The characteristic polynomial alone cannot tell the two apart.
• An eigenvalue that is a simple root has an eigenspace that is a line.

Take a basis of $E_\mu$, $k$ vectors, and extend it to a basis $B$ of $\mathbb{R}^n$. In that basis the first $k$ columns of the matrix of $A$ hold only $\mu$ on the diagonal, because those basis vectors are eigenvectors. Similar matrices have the same characteristic polynomial, and expanding along those $k$ columns pulls out a factor $(\mu-\lambda)^k$. So $\mu$ is a root at least $k$ times.

$k=\dim E_\mu$, and $k\ge1$ because $\mu$ has an eigenvector.

Extend it to a basis of $\mathbb{R}^n$.

$A\mathbf{v}_j=\mu\mathbf{v}_j$, so column $j\le k$ holds $\mu$ in position $j$ and zeros elsewhere.

The two matrices are similar, so they have the same characteristic polynomial; expanding along each of the first $k$ columns takes out a factor $\mu-\lambda$.

$(\lambda-\mu)^k$ divides the characteristic polynomial, so $\mu$ is a root at least $k$ times.

Row reduce $A-\mu I$ and solve $(A-\mu I)\mathbf{v}=\mathbf{0}$. The free variables give a basis of $E_\mu$, one vector each.

So $2x+y=0$, and $E_2$ is the line spanned by $(1,-2)$.

The algebraic multiplicity counts how often $\mu$ is a root of $p_A$. The geometric multiplicity is $\dim E_\mu$. They can differ, and the second is never the larger.

• $E_\mu=\ker(A-\mu I)$ is a subspace; a basis comes from the free variables.
• Geometric multiplicity $\dim E_\mu=n-\operatorname{rank}(A-\mu I)$.
• Algebraic multiplicity: how often $\mu$ is a root of $p_A$.
• $1\le\dim E_\mu\le a_\mu$, and the two can differ.

Great work! You now know how to:

• find a basis of an eigenspace by row reduction;
• compute the geometric and the algebraic multiplicity;
• use $1\le\dim E_\mu\le a_\mu$;
• recognise when a repeated eigenvalue comes with too few eigenvectors.