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Source. Theorem 2c, printed pp. 150–152 (published PDF).

Statement. For finite matroids $M_i=(E,\mathcal F_i)$ on one finite set, $1\le i\le k$, there are pairwise disjoint bases $B_i$ of the respective matroids if and only if

$$|A|\ge\sum_{i=1}^k\bigl(r_i(E)-r_i(E\setminus A)\bigr) \qquad(A\subseteq E). \tag{1}$$

Proof. If the bases exist, then

$$|B_i\cap A|=r_i(E)-|B_i\setminus A| \ge r_i(E)-r_i(E\setminus A).$$

Disjointness permits summing these lower bounds inside $A$, proving (1).

Conversely, (1) at $A=E$ gives the nonnegative integer $N=|E|-\sum_i r_i(E)$. It is at most $|E|$. Let $M_0$ be the uniform matroid consisting of all subsets of $E$ of size at most $N$; it is a truncation of the free matroid and has rank $r_0(A)=\min(N,|A|)$.

A family of disjoint bases $B_i$ leaves exactly $N$ elements, and so extends to a partition into independent sets of $M_0,M_1,\ldots,M_k$. Conversely, in any such partition,

$$|E|=|I_0|+\sum_i|I_i| \le N+\sum_i r_i(E)=|E|.$$

Equality forces $|I_0|=N$ and $|I_i|=r_i(E)$ for each $i$. Thus the $I_i$ are the required bases.

By Theorem 1c, the partition exists exactly when

$$|A|\le\min(N,|A|)+\sum_i r_i(A)\qquad(A\subseteq E).$$

For $|A|\le N$ this is automatic, as is $|A|\le N+\sum_i r_i(A)$; for $|A|>N$ the two tests are identical. Hence the criterion is equivalent to

$$|A|\le |E|-\sum_i r_i(E)+\sum_i r_i(A).$$

Replacing $A$ by its complement gives exactly (1). The nonnegativity check preceded the construction of $M_0$, so it also covers zero ranks and an empty ground set. $\square$