There is a geometric interpretation that I find helpful.
Consider for example the matrix $\begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}$, and think of it as a linear transformation on the plane $\mathbb{R}^2$, namely a linear function $T :\mathbb{R}^2 \to \mathbb{R}^2$ given by the formula
$$T(x,y) = (x,y) \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix} = (x+5y,y)
$$
The eigenvalue $\lambda=1$ has algebraic multiplicity 2 but there is only 1 linearly independent eigenvector, namely $(1,0)$ (or anything parallel to it). You can visualize the linear transformation $T$ as follows:
- The transformation $T$ fixes each point on the $x$-axis, $T(x,0)=(x,0)$ (this, geometrically, is what it means to say that the $x$-axis is the eigenspace generated by the eigenvector $(1,0)$ with eigenvalue $1$).
- For each horizontal line $L$, given by an equation $y=b$, the transformation $T$ takes each point on that line to another point on that line, with the effect of translating the entire line by a horizontal amount equal to $5b$:
$$T(x,b) = (x+5b,b) = (x,b) + (5b,0)
$$
This kind of behavior is sometimes described by saying that $T$ is a "shearing transformation" or a "shear mapping".
Shearing maps can be used to give a general description of the behavior when the algebraic multiplicity of an eigenvalue is higher than its geometric multiplicity: the matrix can be decomposed as a product of a "pure shearing transformation" (like the example I just gave) composed with a diagonalizable matrix.
