Extending linear bijection $T$ of a subspace, preserving the property $Tx\neq x$ for all $x$. My problem is the one described in the title.

Let $V$ be a finite dimensional vector space over a field $k$, $W$ a subspace, and $T:W\to V$ an injective linear map such that $Tx\neq x$ for all $x\in W\setminus 0$. Does there exist a linear bijection $S:V\to V$ extending $T$ with $Sx\neq x$ for all $x\in V\setminus 0$?

This is something that appeared in my work. I can do that for $k=\mathbb{R}$, which is actually enough for my work, but I think this might hold in general. However my  solution for $k=\mathbb{R}$ uses some basic calculus, so the same arguments don't work in general.
One possible generalization would be the following:

Suppose $V$ a finite dimensional $k$-space, $W$ a subspace and $T:W\to V$ linear. Does there exist $S:V\to V$ a linear extension of $T$ with same spectrum as $T$?

Here, the spectrum of $T:W\to V$ is the set of ''eigenvalues'', that is, $sp(T)=\left\{\lambda\in k:\exists x\neq 0\text{ in }W, Tx=x\right\}$. The spectrum of $S$ is the usual set of eigenvalues.
There might be some analogue questions in infinite dimension, say for ($\mathbb{R}$-)Banach spaces, but I can't formulate one precisely. Any comments on that would be interesting, however.

The proof of the first problem for $k=\mathbb{R}$ (this also works for $k=\mathbb{Q}$): First extend $T$ arbitrarily to a linear bijection $\tilde{S}:V\to V$. Given $n=\dim V$ and $m=\dim W$, take a basis $v_1,\ldots,v_n$ of $V$ such that $v_1,\ldots,v_m$ is a basis of $W$. Let's identify linear maps with their matrices on this basis.
Let's define new maps $S_1,\ldots,S_n$ such that the first $i$ columns of $S_i-1$ are LI, and $S_i$ extends $T$. The determinant function is continuous, so there exists a number $r$ such that if the entries of a matrix $A$ are at distance at most $r$ of the respective entries of $\tilde{S}$, then $A$ is also invertible. Fix such $r$. Also, let $e_{j,i}$ be the matrix unit, with $1$ in the $(i,j)$ entry and zeroes everywhere else.
Let $S_1=S_2=\ldots=S_m=\tilde{S}$. Given $S_i$, we define $S_{i+1}$ as follows: if the $i+1$ first columns of $S_i-1$ are LI, let $S_{i+1}=S_i$. If not, then the $i+1$-th column, let's call it $c$, of $S_i-1$ is a linear combination of the $i$ first. But then there exists $j$ such that the $c+rv_j$ is LI to the first $i$ columns of $S_i-1$. Let $S_{i=1}=S_i+re_{j,i+1}$ (i.e., we change the columns $c$ by $c+rv_j$).
Let $S=S_n$. Then $S$ extends $T$, the columns of $S-1$ are all $LI$, so $Sx\neq x$ for all $x$, and moreover, the entries of $S$ are at distance at most $r$ from the respective entries of $\tilde{S}$, so $S$ is also bijective.
 A: Here's a proof for $\operatorname{char}(k)\neq 2$.
Let's proceed by induction on the codimension of $W$. Let's think how to do it.
A subspace of $V$ which contains $W$ and has codimension $\operatorname{cod}(W)-1$ is of the form $W'=W+kx$, where $x\not\in W$. We simply need to define $Tx$, let's call $y\neq 0$ the desired value. The conditions on $T$ are: for all $w\in W$ and $\lambda\in k$,
$$Tw+\lambda y=0 \Rightarrow w=0\text{ and }\lambda=0$$
$$(Tw-w)+\lambda(y-x)=0\Rightarrow w=0\text{ and }\lambda=0$$
So, the easiest ways to obtain such $x$ and $y$ are to look for $y\not\in TW$ and $x-y\not\in(T-1_W)W$.
Putting all this together, let's try to find $x,y\in V$ satisfying:


*

*$x\not\in W$;

*$y\not\in TW$;

*$y-x\not\in (T-1_W)W$.


Claim There exist $x_1,x_2\not\in W$ with $x_1-x_2\not\in (T-1_W)W$.
$\triangleright$Suppose not. Let $z\in V\setminus W$ and $x\in W$, we have $z-x,z\not\in W$, so $z-(z-x)=x\in(T-1_W)W$. But also $2z,z\not\in W$ (here is where we use $\operatorname{char}(k)\neq 2$), so $2z-z=z\in (T-1_W)W$. Thus $(T-1_W)W=V$, which contradicts the fact that $(T-1_W)W$ is a proper subspace of $V$.$\triangleleft$
Take $x_1,x_2$ as above. Choose any $y\not\in TW$. The cosets $(T-1_W)W+x_1$ and $(T-1_W)W+x_2$ are distinct, so $y$ belongs to at most one of them. Let $x$ be the $x_i$ with $y\not\in (T-1_W)W+x$, i.e., $y-x\not\in(T-1_W)W$.
This way, we obtain conditions 1. to 3. above. Extend $T$ to $W+kx$ by setting $Tx=y$. We are now in codimension $\operatorname{cod}(W)-1$, and the result follows by induction.

In fact the case of characteristic $2$ does not hold in general: Let $k=\mathbb{F}_2$, $V=k^3$, $W=k^2\oplus 0$, and
$$T=\begin{bmatrix}1&1&0\\1&0&0\\0&0&0\end{bmatrix}$$
(here we identify linear maps with their matrices on the canonical basis), which is injective without nonzero fixed points. The extensions of $T$ to linear bijections have full rank, and thus must have the form.
$$S_i=\begin{bmatrix}1&1&a\\1&0&b\\0&0&1\end{bmatrix}$$
But then
$$S_i-1=\begin{bmatrix}0&1&a\\1&1&b\\0&0&0\end{bmatrix}$$
does not have full rank, and thus has nonzero fixed points.
A: I refer to the second question. Is there an extension with the same spectrum? The answer is yes if $T$ has an eigenvalue $\lambda$. Just complete a basis of $W$ to a basis of $V$. Let, e.g., $v_1,\ldots,v_k$ be the basis vectors you added and put $Sv_j = \lambda v_j$ for each $j$. Then you have your extension having the same spectrum as $T$. The answer is no (meaning there are counterexamples) if the spectrum of $T$ is empty. Just consider $k = \mathbb R$ and $\dim W$ being even, but $\dim V$ being odd. Take any $T$ on $W$ with empty spectrum. But $S$ on $V$ MUST have an eigenvalue.
The first question can always be answered in the affirmative. Again, complete a basis of $W$ to a basis $\{v_1,\ldots,v_k\}$ of $V$ and put $Sv_j = 2v_j$ for each $j$. Then $Sx\neq x$ for any $x\neq 0$ (why?).
