Cool riddle involving integer vectors
If any real numbers are allowed then $ T=1 $: choose the vector $(e, e^2,\ldots , e^n) $ and use the fact that $ e $ is transcendental. It is not constructive but you know that different vectors will output different dot products.
Edit: if $(a_1,\ldots, a_n) $ and $(b_1,\ldots, b_n) $ are two vectors that output the same number, then $ e $ is a root of the polynomial with integer coefficients $$\sum (a_i -b_i) X^i$$ therefore all its coefficients are zero.
If $V$ were limited to entries at most $b$, then the solution would be easy: query $$ V \cdot [b^0, b^1, \dots, b^{k-1}]$$ The result, when read in base-$b$, would be the vector $V$. However, this would be foiled for any constant choice of $b$ because the entries of $V$ could be arbitrarily large. (Any scheme to "guess" such a $b$, for example, by repeatedly doubling, could be similarly fooled).
However, it's easy to get a $b$ large enough: $V \cdot [1, 1, \dots, 1]$ is the sum of the entries in $V$. Since all of the entries in $V$ are strictly positive, this dot product is strictly bigger than any entry in $V$ (as it's the sum of the largest element and the rest of the elements, which are at least 1).
Thus, the first query is $b = V \cdot [1, 1, 1, \dots, 1]$.
The second query is $v = V \cdot [b^0, b^1, b^2, \cdots, b^{k-1}]$.
Then $V$ is the result from reading the digits of $v$ in base-$b$, so $t \leq 2$.
(Since any number of vectors have the same dot product, $t \geq 2$ so this does give $t = 2$ exactly)
This relies heavily on the positiveness of the entries of $V$. Can the problem be solved in fewer than $k$ queries when the entries of $V$ are allowed to be negative?
I remember a very similar problem, that tasks you with determining the coefficients of a polynomial $q$ with only non-negative integer as coefficients with as few integer queries as possible. (The method looks about the same as this problem).
A link to a blog post concerning this problem is here.
Here is my answer:
Let $I = [ln(2),ln(3),...,ln(p_k)]$ where $p_k$ is the $k^{th}$ prime.
$V*I=n_1ln(2)+n_2ln(3)+...+n_kln(p_k) = ln(2^{n_1}*3^{n_2}*...*p_k^{n_k})$
Therefore, by taking the prime factorizations of $e^{V*I}$, and looking at the exponents, we can deduce what $V$ is.
So by choosing $I$ and only $I$ to be our vector, we can fully deduce $V$ (and $minT=1$).