Exercise.
Suppose $T \in \mathcal{L}(V,W)$ and $c \in W$. Prove that $\{x \in V : Tx = c\}$ is either the empty set or is a translate of $\text{null} \ T$.
Source.
Linear Algebra Done Right, Sheldon Axler, 4th edition, Exercises 3E, first part of problem 8.
Notation.
- $\mathcal{L}(V,W)$ is the set of all linear maps from $V$ to $W$.
- $v + \text{null} \ T$ is the set $\{v + n: n \in \text{null} \ T\}$ .
What I've tried.
My strategy is to show
$$
\{x \in V : Tx = c\} \neq \emptyset \iff \{x \in V : Tx = c\} = v + \text{null} \ T
$$
I first prove the forward direction. Suppose
$$\{x \in V : Tx = c\} \neq \emptyset$$
Then $\exists \ v \in \{x \in V : Tx = c\}$ such that $Tv = c$.
Consider any $n \in \text{null} \ T$. We must have $Tn = 0$. Adding $Tn$ to both sides of $Tv = c$:
\begin{align}
Tv + Tn &= c + 0 = c \implies \\
T(v + n) &= c
\end{align}
This implies $$(v + n) \in \{x \in V : Tx = c\}$$
But $(v + n) \in v + \text{null} \ T$ by definition of $v + \text{null} \ T$.
Question: Am I now allowed to conclude that $\{x \in V : Tx = c\} = v + \text{null} \ T$? Or have I only shown $\{x \in V : Tx = c\} \subseteq v + \text{null} \ T$?
Best Answer
You just proved $\{x \in V: Tx = c\} \subseteq v + \mathrm{null}~T$. But you could write instead
\begin{align} \{x\in V: Tx=c\}&=\{x \in V: T(x-v)=0\} \\ &=\{x \in V : x-v \in \mathrm{null}~T \} \\ &= \{x\in V: x \in \mathrm{null}~T +v\}. \\ &= \mathrm{null}~T +v. \end{align}