Let $V$ and $W$ be finite dimensional real vector spaces and $T\colon V\to W$ be linear.
(a)Prove that if $\dim(V) < \dim(W)$, then $T$ cannot be onto.
(b)Prove that if $\dim(V) > \dim(W)$, then $T$ cannot be one to one
I'm having trouble understanding how to even begin this question. Since we know that every linear transformation is a matrix transformation then…
$AV=W$, but by matrix multiplication, $A$ times $V$ will always result in the same number of vectors in $V$. So how is it possible for a transformation to result in more vectors than it began with?
Best Answer
Since $V$ and $W$ are finite dimensional, rank-nullity says $\dim(V)=\dim(\ker T)+\dim(\operatorname{im} T)$. So if $\dim V<\dim W$, you have
$$\dim(\operatorname{im} T)\leq\dim V<\dim W,$$ so $\dim(\operatorname{im} T)<\dim W$. So $T$ cannot be onto.
Suppose now $\dim V>\dim W$. Since $\operatorname{im} T$ is a subspace of $W$, $\dim(\operatorname{im} T)\leq\dim W$. Then by rank-nullity $$ \dim V=\dim(\ker T)+\dim(\operatorname{im} T)>\dim W\geq\dim(\operatorname{im} T) $$ so that $\dim(\ker T)>0$. Now just recall that $T$ is one-to-one iff $\ker T$ is trivial, iff $\dim(\ker T)=0$.