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 $\mathcal E$.


1 Answer 1


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.

  • $\begingroup$ 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 $\star$? $\endgroup$ Commented Oct 25, 2018 at 20:11
  • $\begingroup$ 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. $\endgroup$ Commented Oct 25, 2018 at 20:34
  • $\begingroup$ 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. $\endgroup$ Commented Oct 25, 2018 at 20:36
  • $\begingroup$ Oh, I see, when you define $\mathcal E$ you imposing the condition that $A(a, b)$ be nondegenerate. $\endgroup$ Commented Oct 25, 2018 at 20:42
  • $\begingroup$ Yes. Is your answer applicable with some change? $\endgroup$ Commented Oct 25, 2018 at 20:44

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