I am looking at some old lecture notes on Linear Algebra, in which the following exercise is given:
My first thought was to simply define $x_{i}(t) = (1-t)u_{i} + tv_{i}$, but this would fail if $(1-t)u_{i} = -tv_{i}$ for all $i$. Given the phrasing of the question, I feel like I am missing a trick involving the structure of the complex numbers. Would anyone care to elaborate on a potential solution, or what I am missing here?
There will be such continuous functions if and only if for each $t\in [0,1]$ we get a change of basis matrix $A(t)$ that sends $\{u_k\}^n_{k=|}$ to $\{u_k(t)\}^n_{k=1}$, that is if there is a continuous path $\varphi:t\mapsto \text{GL}(n,\mathbb{C})$ such that $\varphi(0)=A(0)=I;\ \varphi(1)=A(1),\ $ the change of basis matrix from $\{u_k\}^n_{k=1}$ to $\{v_k\}^n_{k=1}$.
So it is enough to find a path fron $A(1)$ to $I$ or equivalently, a path from $z_0A(1)$ to $I$ for some $z_0\neq 0$ (because there is always a path from $A(1)$ to $z_0A(1)$ that avoids $0$ and we obtain the result joining the two paths).
To do this, choose $z_0\in S^1$ such that the line segment from the origin to $z_0$ does not intersect any eigenvalue of $A(1)$. Now define $\varphi(t)=tI+(1-t)z_0A(1).$ If $t\neq 0,\ \det \varphi(t)=0$ if and only if $z_0\left(\frac{1-t}{t}\right)$ is an eigenvalue of $A(1)$ and by construction, this never happens for $t\in (0,1]$. On the other hand, $\varphi(0)=A(1)$ is invertible and so has non-zero determinant. We conclude that $\varphi$ is a path in $\text{GL}(n,\mathbb{C})$.
Now, if $A(1)\in \text{GL}(n,\mathbb{R})$ and $\det A(1)<0$, any path $\varphi$ from $I$ to $A(1)$ has $\det \varphi(0)>0$ and $\det \varphi(1)<0$ and by the intermediate value theorem applied to $\det$, we get a $0<t<1$ such that $\det \varphi(t)=0$ and since $\varphi(t)\neq \text{GL}(n,\mathbb{R})$, we have a contradiction. Hence, there is no such path.