[proofplan] The necessity follows by counting, for each basis, how many elements must lie outside a fixed subset $A\subset E$. For the converse, we apply the matroid union theorem to $k$ copies of $M$ on the same ground set. The rank formula for the union matroid translates the displayed inequality exactly into the assertion that the union contains an independent set of size $k r(E)$. Such an independent set decomposes into $k$ independent sets, and the rank bound forces each one to have size $r(E)$, hence to be a basis. [/proofplan]

[step:Count the elements outside an arbitrary subset for disjoint bases] Assume that $B_1,\dots,B_k\subset E$ are pairwise disjoint bases of $M$. Fix a subset $A\subset E$. For each index $j\in\{1,\dots,k\}$, the set $B_j\cap A$ is independent in $M$ because $B_j$ is independent and independence is hereditary. Since $B_j\cap A\subset A$, the definition of rank gives \begin{align*} |B_j\cap A|\le r(A). \end{align*} Also $|B_j|=r(E)$ because $B_j$ is a basis. Therefore \begin{align*} |B_j\setminus A|=|B_j|-|B_j\cap A|\ge r(E)-r(A). \end{align*} The sets $B_1\setminus A,\dots,B_k\setminus A$ are pairwise disjoint subsets of $E\setminus A$, since the bases $B_1,\dots,B_k$ are pairwise disjoint. Hence \begin{align*} |E\setminus A|\ge \sum_{j=1}^k |B_j\setminus A|\ge k\bigl(r(E)-r(A)\bigr). \end{align*} Since $A\subset E$ was arbitrary, the displayed inequality holds for every subset $A\subset E$. [/step]

[step:Use matroid union to convert the inequalities into a rank statement]Assume conversely that \begin{align*} |E\setminus A|\ge k\bigl(r(E)-r(A)\bigr) \end{align*} for every subset $A\subset E$. Let $M_1,\dots,M_k$ denote $k$ copies of the matroid $M$ on the same ground set $E$. Let $N$ denote their union matroid. Thus a subset $I\subset E$ is independent in $N$ if and only if there exist subsets $I_1,\dots,I_k\subset E$ such that $I=I_1\cup\dots\cup I_k$ and each $I_j$ is independent in $M$. The hypotheses of the [Matroid Union Theorem](/theorems/5815) are satisfied because $M_1,\dots,M_k$ are matroids on the same finite ground set $E$. Therefore the rank function $r_N:2^E\to \mathbb N\cup\{0\}$ of $N$ satisfies, for every subset $X\subset E$, \begin{align*} r_N(X)=\min_{A\subset X}\bigl(|X\setminus A|+k r(A)\bigr). \end{align*} Applying this formula with $X=E$ gives \begin{align*} r_N(E)=\min_{A\subset E}\bigl(|E\setminus A|+k r(A)\bigr). \end{align*} The assumed inequality is equivalent, for each $A\subset E$, to \begin{align*} |E\setminus A|+k r(A)\ge k r(E). \end{align*} Therefore $r_N(E)\ge k r(E)$. On the other hand, taking $A=E$ in the minimum gives \begin{align*} r_N(E)\le |E\setminus E|+k r(E)=k r(E). \end{align*} Thus \begin{align*} r_N(E)=k r(E). \end{align*}[/step]

[guided]We want to extract $k$ disjoint bases of $M$. The natural way to make several independent sets at once is to use matroid union. Let $M_1,\dots,M_k$ be $k$ copies of $M$ on the same finite ground set $E$, and let $N$ be their union matroid. By definition of the union matroid, a subset $I\subset E$ is independent in $N$ precisely when $I$ can be written as a union \begin{align*} I=I_1\cup\dots\cup I_k, \end{align*} where each $I_j$ is independent in the original matroid $M$. We now apply the [Matroid Union Theorem](/theorems/5815). Its hypotheses are satisfied because $M_1,\dots,M_k$ are matroids on the same finite ground set $E$. Since each copy has the same rank function $r$, the theorem gives \begin{align*} r_N(X)=\min_{A\subset X}\bigl(|X\setminus A|+k r(A)\bigr) \end{align*} for every subset $X\subset E$. In particular, \begin{align*} r_N(E)=\min_{A\subset E}\bigl(|E\setminus A|+k r(A)\bigr). \end{align*} The assumed inequality says exactly that every term inside this minimum is at least $k r(E)$, because \begin{align*} |E\setminus A|\ge k\bigl(r(E)-r(A)\bigr) \end{align*} is equivalent to \begin{align*} |E\setminus A|+k r(A)\ge k r(E). \end{align*} Hence $r_N(E)\ge k r(E)$. The opposite inequality comes from one specific admissible choice in the minimum, namely $A=E$: \begin{align*} |E\setminus E|+k r(E)=k r(E). \end{align*} Therefore the minimum is exactly $k r(E)$, so \begin{align*} r_N(E)=k r(E). \end{align*} This is the key conversion: the subset inequalities are precisely the condition that the union matroid has full possible rank $k r(E)$.[/guided]

[step:Decompose a maximum independent set in the union into bases] Since $r_N(E)=k r(E)$, there exists an independent set $I\subset E$ in the union matroid $N$ such that \begin{align*} |I|=k r(E). \end{align*} By the definition of independence in $N$, there exist independent sets $J_1,\dots,J_k\subset E$ in $M$ such that \begin{align*} I=J_1\cup\dots\cup J_k. \end{align*} Choose, for each element $e\in I$, one index $j(e)\in\{1,\dots,k\}$ with $e\in J_{j(e)}$. Define subsets $B_1,\dots,B_k\subset E$ by \begin{align*} B_j=\{e\in I:j(e)=j\} \end{align*} for each $j\in\{1,\dots,k\}$. Then $B_1,\dots,B_k$ are pairwise disjoint and \begin{align*} I=B_1\cup\dots\cup B_k. \end{align*} Moreover $B_j\subset J_j$, so $B_j$ is independent in $M$ by heredity. For each $j$, every independent subset of $M$ has cardinality at most $r(E)$, so \begin{align*} |B_j|\le r(E). \end{align*} Summing over $j$ and using the disjoint union gives \begin{align*} k r(E)=|I|=\sum_{j=1}^k |B_j|\le \sum_{j=1}^k r(E)=k r(E). \end{align*} Thus equality holds throughout, and in particular $|B_j|=r(E)$ for every $j\in\{1,\dots,k\}$. Each $B_j$ is therefore an independent subset of $E$ of size $r(E)$, hence a basis of $M$. The sets $B_1,\dots,B_k$ are pairwise disjoint, so $M$ has $k$ pairwise disjoint bases. [/step]