Given this limit involving T can I imply that T is injective? Recently, trying to solve a problem, I got stuck in the following fact: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$ such that $$\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=0.$$ Is it true that $T \equiv 0$? I think I was able to prove it by supposing there is some $w \in \mathbb{R}^m \setminus \{0\}$ such that $T \cdot w \neq 0$. Then the sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ given by $v_k=\frac{1}{k} \cdot w$ is such that $\lim\limits_{k \to \infty} v_k=0$ and$$0=\lim\limits_{v \to 0} T \cdot \frac{v}{|v|}=\lim\limits_{k \to \infty}T \cdot \frac{v_k}{|v_k|}=\lim\limits_{k \to \infty} T \cdot \frac{w}{|w|}=T \cdot \frac{w}{|w|}\neq 0.$$ Hence $T \cdot w =0 $ for all $w \in \mathbb{R}^m$ and $T$ must be the zero transformation.
In weakening the hypothesis, I tried to find an answer to the following problem: Given a linear transformation $T:\mathbb{R}^m \to \mathbb{R}^n$, a sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ such that $\lim\limits_{k \to \infty}v_k=0$ and $$\lim\limits_{k \to \infty} T \cdot \frac{v_k}{|v_k|}=0.$$ Is it true that $Ker(T) \neq 0$?
First I tried to find a counterexample, but I couldn't. And since It is easy to prove the converse (i.e. if $Ker(T) \neq 0$ then there is sequence $(v_k)_{k \geq 1} \subset \mathbb{R}^m \setminus \{0\}$ satisfying the above conditions) I thought this might be true but have no idea how to prove it. Any help with a proof or counterexample is appreciated.
 A: If $\operatorname{Ker}(T) = 0$, then the square matrix $T^\intercal T$ is invertible, as $T^\intercal Tx = 0$ implies $|Tx|^2 = x^\intercal T^\intercal Tx = 0$ and hence $Tx = 0$.
We also have $x^\intercal T^\intercal T x = |Tx|^2 \geq 0$ for any $x\in \Bbb R^m$. Thus $T^\intercal T$ is a positive definite symmetric matrix.
We may write $T^\intercal T = U^\intercal D^2 U$, where $U \in O(m)$ is orthogonal and $D = \operatorname{diag}(\lambda_1, \dots, \lambda_m)$ is diagonal, with all $\lambda_i$ strictly positive. This can be rewritten as $T^\intercal T = (DU)^\intercal DU$.
Letting $\lambda = \min_{1 \leq i \leq m} \lambda_i$, we have $$|Tx|^2 = x^\intercal T^\intercal T x = x^\intercal (DU)^\intercal DU x = |DUx|^2 \geq \lambda^2 |Ux|^2 = \lambda^2 |x|^2$$ for all $x\in \Bbb R^m$, i.e. $|Tx| \geq \lambda |x|$.
This means that for any $x \in \Bbb R^m \backslash \{0\}$, we have $\left|T\cdot \frac x{|x|}\right|\geq \lambda$. Therefore there cannot exist such a sequence $(v_k)_k$ as in your question.
A: Let $\ S\ $ be the set of cluster points of $\ \big\{\frac{v_k}{|v_k|}\big\}_{k=1}^\infty\ $.  The kernel of $\ T\ $ must contain  $\ S\ $, and $\ S\ $ must contain a non-zero element.  Therefore $\ Ker(T)\ne\{0\}\ $.
A: Great question! I think it is true. A somewhat analytical way to show the contrapositive would roughly be "$|T(v)|$ must attain a minimum on the unit sphere in $\mathbb R^m$ because $T$ is continuous and the unit sphere is compact. This minimum can't be $0$ by injectivity, so if $v_k$ is a nonzero sequence then $T(v_k / |v_k|) \not \to 0$, as this second sequence is bounded below in magnitude.
A somewhat different (but secretly the same) angle: Effectively what you're asking is "if $T(u_i) \to 0$ for some sequence of unit vectors, can $T$ be injective?" Again by compactness $u_i$ has a convergent subsequence, and by continuity its limit is in the kernel of $T$ (and of course the limit can't be $0$).
A: Well, that's a high mathematics!  But ...can we make it simpler, maybe?
The expression $\frac v{|v|}$ is a unit vector, for any $v$. All possible values of the fraction make a unit sphere centered at $0$.
And $T$ transforms every point of the sphere to $0$. (*)
Pairs of antipodes among them.
If a linear transformation transforms some $q$ and $-q$ to $0$, then it transforms the whole line to $0$. And due to (*) it does so to all lines, hence it is zero everywhere.
Am I right?
