I'm hoping the #mathstodon community can help me feel better about some basic linear algebra.

I'm thinking about a \(2\times 2\) matrix \(M\) and its invariant lines.

Let \(M\) have two distinct eigenvalues \(e_1\neq1\) and \(e_2\neq1\), with associated eigenvectors \(\bf{v_1}\) and \(\bf{v_2}\), respectively. Then there are exactly two invariant lines: \(\lambda \bf{v_1}\) and \(\lambda \bf{v_2}\), and they both pass through the origin.

If we want to have an invariant line which does not pass through the origin, we must have 1 as one of our eigenvalues. Let \(e_1=1\), and \(e_2\neq 1\). Of course \(\lambda\bf{v_1}\) is one invariant line, but consider the line \(\lambda \bf{v_2} + \bf{r}\) where \(\bf{r}\) is a general vector.

Since we have distinct eigenvalues, the eigenvectors span the plane, and so \(\bf{r}=\alpha \bf{v_1} + \beta \bf{ v_2}\). Now
\[ M ( \lambda \bf{v_2} + \bf{r}) = ((\lambda+\beta) e_2 - \beta)\bf{v_2} + \bf{r}\]
which means the line is invariant.

So far, so good. But I get uncomfortable in the case \(e_1=e_2=1\). Then there is one eigenvector \(\bf{v}\). I find it weird that not only \(\lambda\bf{v}\), but \(\lambda\bf{v}+\bf{r}\), for any \(\bf{r}\), is also invariant, even though there is no second eigenvector to "generate" this family of lines which don't pass through the origin. The implication is that \((M-\mathbb{I})\bf{r}\propto\bf{v}\) for all vectors \(\bf{r}\).

I'd like to construct an \(M\) where I can start with \(e_1=1,e_2\neq1\) and then smoothly deform it to the \(e_1=e_2=1\) case, to somehow see that the infinite family of invariant lines persists even after the eigenvector that generated them has become equal to the other eigenvector.

Can you help?