What does a single eigenvector and eigenvalue for a $2 \times 2$ matrix represent geometrically?

Question

I know that eigenvectors for a matrix $A$ represent lines in the plane that are invariant under the the transformation given by $A$, and the corresponding eigenvalues are the scaling that gets applied along these lines. But what does it mean if we only have one eigenvalue and eigenvector for a $2\times 2$ matrix?

For example,

$$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$

has a single eigenvalue $\lambda = 1$ and the its corresponding eigenvector $x = (1, 0)$ (or any multiple of it).

So what does this mean geometrically...what kind of scaling and rotation does $A$ provide?

Answer 1

The $2 \times 2$ case is quite simple: such an $A$ causes a shear transformation: one vector is fixed, the one perpendicular to it maps to one with the same perpendicular component, but the parallel component is changed (think of the effect on the basis consisting of $(1,0)$ and $(0,1)$: $(1,0) \mapsto (1,0)$ and so is unchanged, whereas $(0,1) \mapsto (1,1)$.

The general case of only having one eigenvector is similar: in the two-dimensional case, the possible Jordan Normal Forms are $$ \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}, \quad \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}, \quad \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, $$ where $\lambda \neq \mu$, and it is a significant theorem that any $2 \times 2$ matrix is related to one of these by a change-of-basis.

Now, the characteristic polynomial of any matrix of the first type is $(t-\lambda)(t-\mu)$, and its coefficients are real if the original matrix is real, so they are either both real or a complex conjugate pair. The complex case can be shown to be a rotation matrix, which has no real eigenvectors in two dimensions. The real case is a pair of unequal scalings.

The second case is a simple scaling, and it is easy to show that the only matrices of this type are of the form $\lambda I$ for real $\lambda$.

The last case is what your $A$ falls into, and it is easy to see that these are all shears, in the same way as $A$ is, but with additional scaling (since it factors into $\lambda I \begin{pmatrix} 1 & 1/\lambda \\ 0 & 1 \end{pmatrix}$).