[proofplan] We use the clutching construction directly. Over the interval $[0,1]$ the bundle is a product bundle, so an orientation is represented by choosing one of the two orientations of the model fibre $\mathbb{R}^k$ continuously along the interval. Since $[0,1]$ is connected, such a choice cannot switch sign. The only remaining compatibility condition is at the glued endpoint, where the clutching matrix $A$ must preserve the chosen orientation; this is exactly the condition $\det A > 0$. [/proofplan] [step:Model the bundle by a quotient of the trivial bundle over the interval] Let \begin{align*} q: [0,1]\times \mathbb{R}^k &\to E_A \end{align*} denote the quotient map in the clutching construction, defined by the [equivalence relation](/page/Equivalence%20Relation) $(1,v)\sim (0,Av)$ for every $v\in\mathbb{R}^k$. Let \begin{align*} \pi_A:E_A&\to S^1 \end{align*} be the induced real rank-$k$ vector bundle projection, where $S^1$ is identified with the quotient space $[0,1]/(0\sim 1)$. For each $t\in[0,1]$, let $\iota_t:\mathbb{R}^k\to \{t\}\times\mathbb{R}^k$ denote the standard fibre parametrization given by $v\mapsto (t,v)$ before the endpoint identification. Let $\omega_{\mathrm{std}}$ denote the standard orientation of $\mathbb{R}^k$, namely the orientation for which the ordered basis $(e_1,\dots,e_k)$ is positive. A linear isomorphism $T:\mathbb{R}^k\to\mathbb{R}^k$ preserves $\omega_{\mathrm{std}}$ exactly when $\det T>0$, and reverses it exactly when $\det T<0$. [/step] [step:Construct a global orientation when the clutching matrix preserves orientation] Assume $\det A>0$. On the product bundle $[0,1]\times\mathbb{R}^k\to[0,1]$, declare the ordered frame $t\mapsto \big((t,e_1),\dots,(t,e_k)\big)$ to be positive for every $t\in[0,1]$. In the sense of orientability of real vector bundles, this gives a continuous orientation of every fibre before the endpoint identification. We must check that this orientation descends through the endpoint identification. The endpoint identification on model fibres is the [linear map](/page/Linear%20Map) $A:\mathbb{R}^k\to\mathbb{R}^k$ given by $v\mapsto Av$. Since $\det A>0$, the ordered basis $(Ae_1,\dots,Ae_k)$ is positive with respect to $\omega_{\mathrm{std}}$. Therefore the positive orientation on the fibre over $t=1$ is identified with the positive orientation on the fibre over $t=0$. Hence the orientation on the trivial bundle over $[0,1]$ passes to a well-defined continuous orientation of $E_A\to S^1$. [guided] Assume $\det A>0$. Before the endpoint identification, the bundle is the product $[0,1]\times\mathbb{R}^k\to[0,1]$, so there is an immediate candidate for an orientation: use the standard orientation of the fibre $\mathbb{R}^k$ at every parameter value $t\in[0,1]$. More explicitly, declare the frame $t\mapsto \big((t,e_1),\dots,(t,e_k)\big)$ to be positive in the fibre $\{t\}\times\mathbb{R}^k$ for every $t\in[0,1]$. The only possible obstruction is compatibility with the endpoint identification. The quotient identifies the vector $(1,v)$ in the fibre over $1$ with the vector $(0,Av)$ in the fibre over $0$. Thus the endpoint compatibility condition says that the linear map $A:\mathbb{R}^k\to\mathbb{R}^k$, given by $v\mapsto Av$, must send a positive ordered basis at $t=1$ to a positive ordered basis at $t=0$. Since $\det A>0$, the image basis $(Ae_1,\dots,Ae_k)$ has the same orientation as $(e_1,\dots,e_k)$. Therefore the chosen orientation is unchanged by the gluing relation. Consequently the standard fibre orientation on $[0,1]\times\mathbb{R}^k$ is constant along the interval and compatible at the glued endpoint. It descends through the quotient map $q:[0,1]\times\mathbb{R}^k\to E_A$, giving a well-defined continuous orientation of the vector bundle $E_A\to S^1$. [/guided] [/step] [step:Show that any global orientation forces the clutching matrix to preserve orientation] Assume that $E_A\to S^1$ is orientable. For each $t\in[0,1]$, restrict the quotient map to the model fibre to obtain the linear isomorphism $q_t:\mathbb{R}^k\to (E_A)_{[t]}$ induced by $v\mapsto q(t,v)$, where $[t]\in S^1=[0,1]/(0\sim1)$. Transport the given orientation of $(E_A)_{[t]}$ back through $q_t$; this gives a continuous choice of orientation $\omega_t$ on the model fibre $\mathbb{R}^k$ for each $t\in[0,1]$. The orientation double cover of the product interval bundle $[0,1]\times\mathbb{R}^k\to[0,1]$ is explicitly the product $[0,1]\times\{\omega_{\mathrm{std}},-\omega_{\mathrm{std}}\}$, because every fibre is identified with the same model [vector space](/page/Vector%20Space) $\mathbb{R}^k$. Under this identification, the pulled-back orientation section has the form $t\mapsto (t,\omega_t)$. Since $\{\omega_{\mathrm{std}},-\omega_{\mathrm{std}}\}$ has the discrete topology and the section is continuous, the sign map $t\mapsto\omega_t$ is locally constant. Since $[0,1]$ is connected, every locally constant map from $[0,1]$ to a discrete two-point set is constant. Hence $\omega_0=\omega_1$. The endpoint gluing relation identifies the fibre at $t=1$ with the fibre at $t=0$ by the linear map $A:\mathbb{R}^k\to\mathbb{R}^k$. Because the orientation on $E_A$ is well-defined after quotienting, this identification must carry the orientation $\omega_1$ to the orientation $\omega_0$. Since $\omega_1=\omega_0$, the map $A$ preserves an orientation of $\mathbb{R}^k$. Therefore $\det A>0$. [guided] Assume that $E_A\to S^1$ is orientable. By the definition of orientability of real vector bundles, there is a continuous choice of orientation on every fibre of $E_A$. For each $t\in[0,1]$, the quotient map restricts to a linear isomorphism $q_t:\mathbb{R}^k\to (E_A)_{[t]}$ given by $v\mapsto q(t,v)$. Pull the chosen orientation of $(E_A)_{[t]}$ back along $q_t$; this defines an orientation $\omega_t$ of the fixed model vector space $\mathbb{R}^k$. Why must the family $t\mapsto\omega_t$ be constant? Over the product interval bundle, the orientation double cover is not abstract: it is the product $[0,1]\times\{\omega_{\mathrm{std}},-\omega_{\mathrm{std}}\}$. The second factor records which of the two orientations of $\mathbb{R}^k$ has been chosen. The pulled-back global orientation gives a continuous section of this two-sheeted cover, so it has the form $t\mapsto(t,\omega_t)$. Since the set $\{\omega_{\mathrm{std}},-\omega_{\mathrm{std}}\}$ is discrete, continuity implies that the sign choice cannot change on a small neighbourhood of any $t$. Thus $t\mapsto\omega_t$ is locally constant. Because $[0,1]$ is connected, a locally constant map from $[0,1]$ to a discrete two-point set is constant, and therefore $\omega_0=\omega_1$. Now use the gluing. The quotient relation identifies $(1,v)$ with $(0,Av)$, so the endpoint fibre identification is the linear map $A:\mathbb{R}^k\to\mathbb{R}^k$. Since the orientation on $E_A$ is a well-defined orientation after quotienting, this endpoint identification must carry the orientation at $t=1$ to the orientation at $t=0$. In symbols, $A$ sends $\omega_1$ to $\omega_0$. But $\omega_1=\omega_0$, so $A$ preserves an orientation of $\mathbb{R}^k$. A real linear isomorphism preserves an orientation exactly when its determinant is positive. Hence $\det A>0$. [/guided] [/step] [step:Exclude orientability when the clutching determinant is negative] If $\det A<0$, then $A$ reverses every orientation of $\mathbb{R}^k$. If $E_A$ were orientable, the preceding step would imply $\det A>0$, contradicting $\det A<0$. Therefore $E_A$ is not orientable when $\det A<0$. Combining the construction for $\det A>0$ with the obstruction for $\det A<0$ proves that $E_A$ is orientable exactly when the clutching matrix has positive determinant. [guided] Assume $\det A<0$. A real linear isomorphism with negative determinant reverses both possible orientations of $\mathbb{R}^k$: if an ordered basis is positive for one orientation, its image under $A$ is negative for that same orientation. Suppose, for contradiction, that $E_A\to S^1$ were orientable. The preceding step applies to any global orientation of $E_A$: pulling it back to the interval forces the two endpoint orientations on the model fibre to agree, and compatibility with the quotient then forces the clutching map $A$ to preserve that orientation. Therefore the preceding step gives $\det A>0$. This contradicts the standing assumption $\det A<0$. Hence $E_A$ is not orientable when $\det A<0$. Together with the explicit construction in the positive-determinant case, this proves exactly the asserted dichotomy: positive determinant gives orientability, while negative determinant obstructs orientability. [/guided] [/step]