Worksheet 3

Solution.

Since $V$ is finite-dimensional and $T$ is linear, we may use the rank-nullity theorem, which tells us that

Here, $R\p{T}$ is the range of $T$ and $N\p{T}$ is the null-space (or kernel) of $T$.

1. Since $\dim V < \dim W$ and $0 \leq \dim N\p{T}$, we get

   $$\begin{aligned} \dim R\p{T} &\leq \dim R\p{T} + \dim N\p{T} \\ &= \dim V \\ &< \dim W. \end{aligned}$$

   If $T$ were surjective, then $R\p{T} = W \implies \dim R\p{T} = \dim W$, which is impossible.

2. Since $R\p{T} \subseteq W$, we know $\dim R\p{T} \leq \dim W$. Thus, by rank-nullity,

   $$\begin{aligned} \dim N\p{T} &= \dim V - \dim R\p{T} \\ &\geq \dim V - \dim W \\ &> 0. \end{aligned}$$

   Thus, $N\p{T}$ is non-trivial (i.e., $N\p{T} \neq \set{0}$), so $T$ cannot be injective.