# Is this subset of $GL_5(\mathbb R)$ under Vandermonde-like parametrization of square matrices path-connected?

Let us fix $$5$$ distinct nonzero real numbers $$\lambda_1, \dots, \lambda_5\in \mathbb R$$. Let $$A \in M_5(\mathbb R)$$ be a $$5\times 5$$ matrix given by \begin{align} \label{eq:q} \tag{\star} \begin{pmatrix} a_1 & a_1 \lambda_1 & b_1 & b_1 \lambda_1 & b_1 \lambda_1^2 \\ a_2 & a_2 \lambda_2 & b_2 & b_2 \lambda_2 & b_2 \lambda_2^2 \\ a_3 & a_3 \lambda_3 & b_3 & b_3 \lambda_3 & b_3 \lambda_3^2 \\ a_4 & a_4 \lambda_4 & b_4 & b_4 \lambda_4 & b_4 \lambda_4^2 \\ a_5 & a_5 \lambda_5 & b_5 & b_5 \lambda_5 & b_5 \lambda_5^2 \end{pmatrix}, \end{align} where where the vectors $$a=(a_1, \dots, a_5)^T$$ and $$b=(b_1, \dots, b_5)^T$$ are linearly independent.

Let us define a set \begin{align*} \mathcal E = \{A \in GL_5(\mathbb R): A \text{ has form \eqref{eq:q}}\}. \end{align*} Note: By saying $$A$$ has form \eqref{eq:q} I mean that a matrix in $$\mathcal E$$ is generated by taking a linearly independent pair $$(a, b)$$ with the formula given by \eqref{eq:q} and meanwhile the matrix generated is nonsingular. For a linearly independent pair $$a, b$$, I will write $$A(a,b)$$ as the matrix generated by the formula in \eqref{eq:q}.

My question is: suppose we are given two linearly independent pairs $$(a, b)$$ and $$(\hat{a}, \hat{b})$$ (here I mean $$a, b$$ are linearly independent, $$\hat{a}, \hat{b}$$ are linearly independent and no assumption on the linearly independence of $$a, b, \hat{a}, \hat{b}$$). We assume $$A(a, b), A(\hat{a}, \hat{b})$$ are nonsingular. I am interested in determining whether we can connect $$A(a,b)$$ and $$A(\hat{a}, \hat{b})$$ in $$\mathcal{E}$$. Clearly we can connect the pairs $$(a,b)$$ and $$(\hat{a}, \hat{b})$$ with a path $$\gamma$$ such that the image of $$\gamma$$ is always a linearly independent pair (by identifying the pair by a matrix with rank $$2$$ and rank $$2$$ matrices are connected). But I am not sure how to guarantee we stay in $$\mathcal E$$.

No, for all choices $$(\lambda_i)$$ of parameters, $$\mathcal E$$ has at least two components.
The determinant has the form $$\sum \pm (\lambda_k - \lambda_l) (\lambda_n - \lambda_p) (\lambda_p - \lambda_m) (\lambda_m - \lambda_n) a_k a_l b_m b_n b_p$$ for appropriate signs $$\pm$$, where the sum is taken over the partitions of $$\{1, 2, 3, 4, 5\}$$ into a pair $$\{k, l\}$$ and a triple $$\{m, n, p\}$$. Since the $$\lambda_i$$ are distinct, all of the coefficients are nonzero. For the two pairs $$(a, b_{\pm})$$, $$a = (1, 1, 0, 0, 0), \quad b_{\pm} = (0, 0, \pm 1, 1, 1) ,$$ the $$\det A(a, b_{\pm}) = \pm C$$ for some nonzero constant. Thus, these $$\mathcal E$$ contains elements in both components of $$GL_5(\Bbb R)$$ and so cannot be connected.
• Thanks for answering my question. But why the path will be in $\mathcal E$ by construction? I understand we can connect the two frames but how do we know the matrix generated by the two frames will in $\mathcal E$ by the prescribed formula in $\star$? – user1101010 Oct 25 '18 at 20:11
• By definition $\mathcal{E}$ contains $A(a, b) \in \mathcal E$ for all $2$-frames $(a, b)$ of $\Bbb R^5$. In particular it contains all points $(a(t), b(t))$ on the path. – Travis Willse Oct 25 '18 at 20:34
• I might not have presented my question in a clear way. I defined $\mathcal E$ to be a subset of general linear maps and also the elements satisfy the formula prescribed by $\star$ relation. – user1101010 Oct 25 '18 at 20:36
• Oh, I see, when you define $\mathcal E$ you imposing the condition that $A(a, b)$ be nondegenerate. – Travis Willse Oct 25 '18 at 20:42