Does this subset of $GL_2(\mathbb R)$ under Vandermonde-like parametrization have precisely two path-connected components?

Question

Let us fix $2$ distinct nonzero real numbers $\lambda_1 < \lambda_2\in \mathbb R$. Let $A \in M_2(\mathbb R)$ be a $2\times 2$ matrix given by \begin{align} \label{eq:q} \tag{$\star$} \begin{pmatrix} a_1 & a_1 \lambda_1 \\ a_2 & a_2 \lambda_2 \end{pmatrix}. \end{align}

Let us define a set \begin{align*} \mathcal E = \{A \in GL_2(\mathbb R): A \text{ has form \eqref{eq:q}}\}. \end{align*}

I am interested on the number of path-connected components in $\mathcal E$. Is it four or two? It should be at least $2$ since $(1,1)$ would give positive determinant and $(-1, 1)$ would give negative determinant. It is also clear if $(a_1, a_2)$ and $(b_1, b_2)$ have same sign in each coordinate, i.e., $\text{sign}(a_i) = \text{sign}(b_i)$, we could obviously use a straight line to connect them. But does this mean it has $4$ connected components determined by all the combinations of signs?

Answer 1

Yes, there are precisely four components.

Hint The determinant of the matrix is $$\det A = a_1 a_2 (\lambda_2 - \lambda_1) ,$$ which vanishes iff $a_1 = 0$ or $a_2 = 0$.