- Theorem $T:\quad$ Let $S = \{v_1, v_2,..., v_p\}$ be a set of vectors in $\mathbb R^m$. If $p>m$, then this set is linearly dependent.
Theorem $T$ can be proved to be true.
- Converse $T_c$ of theorem $T:\quad$ If the set $S$ is linearly dependent then $p > m$.
The converse of theorem $T$ is not a true statement. Does it mean that its negation is true?
- Negation $T_{nc}$ of $T_c:\quad$ The set $S$ is linearly dependent and $p \le m$.
In simple words, what does $T_{nc}$ mean, and how to prove it?
Negating a statement logically flips its truth value; this means that even if you abstractly redefine the phrases (e.g., 'linearly dependent') and non-logical symbols (e.g., the 'less than' symbol) within it, the statement and its negation continues to have opposite truth values. So, the negation of a false statement has to be true.
You have not fully stated the converse $T_c:$ the part about $S$ being an arbitrary set of $p$ vectors in $\mathbb R^m$ is important, because in full: $$\forall p{\in}\mathbb Z_0^+\quad\forall m{\in}\mathbb Z^+\\\forall S{\in}\{\text{sets of $p$ vectors in }\mathbb R^m\}\;(S\text{ is linearly dependent}\implies p>m).\tag1$$
Consequently, your negation is incomplete; compare it with the full negation $$\exists p{\in}\mathbb Z_0^+\quad\exists m{\in}\mathbb Z^+\\\exists S{\in}\{\text{sets of $p$ vectors in }\mathbb R^m\}\;(S\text{ is linearly dependent and } p\le m).\tag{1n}$$
As suggested by Anne Bauval, just exhibit $(p,m,S)=\left\{1,1,\{0\}\right\}.$
Addendum
In my second paragraph above, the point is that the "Let $S = \{v_1, v_2,..., v_p\}$ be a set of vectors in $\mathbb R^m$" part of theorem $T$ is understood to mean that $S$ is an arbitrary set of $p$ vectors in $\mathbb R^m,$ and that this stipulation is equivalent to the sequence of universal quantifiers that I made explicit in statement $(1).$
Another point worth noting is that when taking the converse or contrapositive, the external quantifiers (the quantifiers outside the "if..., then...") are unaltered, like in the above.