Eigenvector of a $C^n$ class matrix Let $A$ be the following matrix function:
$\Bbb{R} \to \Bbb{R}^{a \times (a+1)}$
$t \mapsto A(t)$
Let us suppose that $A$ is $C^{\infty}$, meaning that all of $A$ coefficients are $C^{\infty}$. Let us suppose that we also know that $\forall t, rg(A(t))=a$.
Then, at any time $t$, the kernel of $A(t)$ is a line. Let us suppose that we get the base of the kernel by taking the only vector which has a norm of 1 and that has its first non null component positive. We call this vector $n(t)$.
Is there any result making it possible to show that $t \mapsto n(t)$ is $C^{\infty}$ ?
Any help appreciated.
Thank you !
 A: We cannot always have a smooth $n\colon \mathbb{R}\to \mathbb{R}^{a+1}$ such that $n(t)$ spans $\ker A(t)$ for all $t$ and $\lVert n(t)\rVert = 1$ if we demand that the first non-zero component is positive. For example, let $A(t) = \begin{pmatrix} 1 & \sin^2 t\end{pmatrix}$, then for $t\notin \pi\mathbb{Z}$, we have $n(t) = \frac{1}{\sqrt{1+\sin^4 t}} \begin{pmatrix} \sin^2 t\\ -1\end{pmatrix}$ as the unit vector spanning $\ker A(t)$ with positive first component, but then $\lim\limits_{t\to k\pi} n(t) = \begin{pmatrix} 0\\ -1\end{pmatrix}$ has negative first non-zero component.
If we drop the positivity requirement, there is however a smooth function $n \colon \mathbb{R} \to \mathbb{R}^{a+1}$ such that $n(t)$ is a unit vector spanning $\ker A(t)$, and the only other such function is $-n$.
The uniqueness up to multiplication with $-1$ follows since $\ker A(t)$ is always one-dimensional, hence if $n,m$ are two such functions, the inner product $t \mapsto \langle n(t), m(t)\rangle$ is a smooth function whose range is contained in the two-element set $\{-1,1\}$, thus constant.
For the existence, note that the exterior product of the rows of $A(t)$ yields a vector orthogonal to all rows - the case $a+1 = 3$ is the well-known cross product. If we denote the $a\times a$ matrix obtained from $A$ by deleting the $k^{\text{th}}$ column of $A$ by $A_k$, the exterior product of the rows of $A$ has the components
$$b_k(t) = (-1)^{k+1} \det A_k(t).$$
The components are polynomials in the components of $A$, hence $b$ is smooth. If we augment the matrix $A$ by adding $b$ as a first row, Laplace expansion of the determinant by the first row shows
$$\det (A(t)\oplus b(t)) = \sum_{k=1}^{a+1} (-1)^{k+1} b_k(t)\det A_k(t) = \sum_{k=1}^{a+1} (\det A_k(t))^2 > 0,$$
since at least one of the $a\times a$ minors must have full rank. And if we augment the matrix by adding the $m^{\text{th}}$ row of $A$ as a first row, we have two identical rows, therefore
$$0 = \det (A(t) \oplus a_m(t)) = \sum_{k=1}^{a+1} (-1)^{k+1} a_{mk}(t)\det A_k(t) = \sum_{k=1}^{a+1} a_{mk}(t)b_k(t),$$
which shows that $b(t)\in \ker A(t)$. So $b$ is a smooth function such that $b(t)$ spans $\ker A(t)$ for all $t$. Then
$$n(t) = \frac{1}{\lVert b(t)\rVert}\cdot b(t)$$
is also smooth [I presume we use the Euclidean norm, but we could take any other norm that is smooth on $\mathbb{R}^{a+1}\setminus \{0\}$, not, however, norms like the sum-norm or the maximum norm, then $\lVert b(t)\rVert$ would not necessarily be smooth].
