= Trivial kernel
A <linear map> $T:V\to W$ has a trivial <kernel> when $\ker T=\{0_V\}$: the <zero vector> is the only input producing the zero output. The <kernel> is a <singleton set>, not the <empty set>, because every <linear map> sends zero to zero. Equivalently, its <nullity> is zero and $T$ is <injective>. This holds even without a finite-dimensional hypothesis; see <Linear Algebra Done Right by Sheldon Axler (4th edition)>.
The reason this zero-output test controls all outputs is <linearity>:
$$
T(v)=T(w)\iff T(v-w)=0\iff v-w\in\ker T.
$$
With a <trivial kernel>, equal outputs force equal inputs. Therefore every equation $T(x)=b$ has at most one solution. Existence is a separate question: $b$ must belong to the <image of a linear map>. In the <fiber> description under <nullity>, $x_0+\{0\}$ is just the single point $x_0$.
For example, the <linear map> $J:\mathbb R^2\to\mathbb R^3$, $J(x,y)=(x,y,0)$, has <trivial kernel> and <full column rank>:
$$
J=\begin{pmatrix}1&0\\0&1\\0&0\end{pmatrix},\qquad
\ker J=\{0\},\qquad\operatorname{rank}J=2.
$$
The output $(3,4,0)$ comes from exactly $(3,4)$, while $(3,4,1)$ is unreachable. Conversely, $P:\mathbb R^3\to\mathbb R^2$, $P(x,y,z)=(x,y)$, is <surjective> and has <full row rank>, but its <kernel> is $\operatorname{span}\{(0,0,1)\}$: every output $(a,b)$ comes from the solution <straight line> $(a,b,s)$.
For an $m\times n$ <matrix>, a <trivial kernel> is equivalent to <matrix rank> $n$ and <linear independence> of its columns. It can therefore occur only if $n\leq m$. If $m=n$, the <rank-nullity theorem> also forces <surjectivity>, giving an <invertible matrix>; if $m>n$, the example $J$ shows why unique recovery of reachable inputs does not imply that every output is reachable. Finally, a <trivial kernel> guarantees exact uniqueness without guaranteeing numerical stability: the <singular matrix> example $A(t)$ has <trivial kernel> for every $t\ne0$, even as its <condition number> grows without bound when $t$ approaches zero.
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