Showing properties of the kernal and range of a linear transformation.

Question

Let $T:\mathbb{R}^3\to V$.

Show directly that the Ker($T$) is a subspace of $\mathbb{R}^3$ and that dim(Ker($T$)) $\leq 3$.

Show that $R(T)$ is a subspace of $V$ and that dim$(R(T)) \leq 3$.

How could on approach this problem? I considered using the dimension theorem, but I am very unfamiliar with this material in particular.

Answer 1

Hint:

1. Let $W$ be a subset of a vector space $V$ (whose underlying field is $\Bbb R$). Then $W$ is a subspace of $V$ iff $cx+y\in W$ for any $x,y\in W$ and $c\in\Bbb R$.

2. If $W$ is a subspace, then $\dim W\leq\dim V$, where the equality holds iff $W=V$.

3. Suppose $f\colon V\to W$ is a linear map. Then $\dim V=\dim\ker f+\dim f(V)$.