Preimage of Image of Linear Transformation

Theorem

Let $K$ be a field.

Let $X$ and $Y$ be vector spaces over $K$.

Let $T : X \to Y$ be a linear transformation.

Let $A \subseteq X$ be a non-empty set.

Then $T^{-1} \sqbrk {T \sqbrk A} = \ker T + A$

Let $x \in T^{-1} \sqbrk {T \sqbrk A}$.

Then $T x \in T \sqbrk A$.

Then there exists $y \in A$ such that $T x = T y$.

Hence we have $\map T {x - y} = 0$.

Hence $x - y \in \ker T$.

So $x \in y + \ker T$.

We have $y + \ker T \subseteq A + \ker T$, so we obtain:

if $x \in \in T^{-1} \sqbrk {T \sqbrk A}$ then $x \in \ker T + A$

so that:

$T^{-1} \sqbrk {T \sqbrk A} \subseteq \ker T + A$

Now let $x \in \ker T + A$.

Then there exists $v \in \ker T$ and $y \in A$ such that $x = v + y$.

From linearity, we have $T x = \map T {v + y} = T v + T y$.

Since $v \in \ker T$, we have $T x = T y$.

Since $T y \in T \sqbrk A$, we have $T x \in T \sqbrk A$.

So $x \in T^{-1} \sqbrk {T \sqbrk A}$.

$\blacksquare$