= Solution
For a <linear map> $T:V\to W$ with $V$ finite-dimensional, the <rank-nullity theorem> states
$$
\boxed{\dim V=\dim\ker T+\dim\operatorname{im}T.}
$$
To prove it, choose a <basis> $v_1,\ldots,v_k$ of the <kernel of a linear map> and use <basis extension> to obtain $v_1,\ldots,v_n$ as a basis of $V$. The vectors $Tv_{k+1},\ldots,Tv_n$ span the <image of a linear map>: applying $T$ to a basis expansion kills the first $k$ terms. They are also linearly independent. Indeed, if $\sum_{j>k}a_jTv_j=0$, then $\sum_{j>k}a_jv_j\in\ker T$, so it is a combination of $v_1,\ldots,v_k$. Independence of the full basis forces every $a_j=0$. Consequently $\dim\operatorname{im}T=n-k$, proving the theorem.
Here the <image of a linear map> has dimension $r\in\{0,1,2,3\}$, since it lies in $\mathbb R^3$. Thus
$$
\boxed{\dim\ker\alpha\in\{2,3,4,5\}.}
$$
Every possibility occurs: for $0\le r\le3$, map $(x_1,\ldots,x_5)$ to $(x_1,\ldots,x_r,0,\ldots,0)\in\mathbb R^3$. This map has rank $r$ and kernel dimension $5-r$.
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