Prove that there exists five matrices $B\_i \in \mathbb{F}\_2^{5\times 10}$, $i\in \{1,2,3,4,5\}$, such that any two $B\_i$'s form an invertible matrix in $\mathbb{F}\_2^{10\times 10}$.


I am interested in a proof of existence that could be generalized to matrices with other dimensions.


Follow-up question when there are constraints on the $B\_i$'s: [Existence of matrices in $\mathbb{F}\_2$ with some invertibility properties](https://mathoverflow.net/questions/405296/existence-of-matrices-in-mathbbf-2-with-some-invertibility-properties)



Existence of matrices with some invertibility properties
If $B$, $B'$ are $5\times 10$ matrices of rank $5$ over $\mathbb{F}\_2$, the condition that the matrix formed by stacking $B$ on top of $B'$ is invertible is equivalent to the condition that the row span of $B$ and the row span of $B'$ have trivial intersection. The row spans are $5$-dimensional subspaces of $\mathbb{F}\_2^{10}$, so given five $5$-dimensional subspaces of $\mathbb{F}\_2^{10}$ with pairwise trivial intersection, we can choose bases for them to construct matrices $B\_1,\ldots,B\_5$.


Note that $\mathbb{F}\_2^{10}$ is isomorphic to $\mathbb{F}\_{2^5}^2$ as an $\mathbb{F}\_2$ vector space. A $1$-dimensional $\mathbb{F}\_{2^5}$-subspace of $\mathbb{F}\_{2^5}^2$ can be identified with a $5$-dimensional $\mathbb{F}\_2$ subspace of $\mathbb{F}\_2^{10}$, and every pair of $1$-dimensional $\mathbb{F}\_{2^5}$-subpsaces of $\mathbb{F}\_{2^5}^2$ have trivial intersection. There are 33 such subspaces, so we can find even find $B\_1,\ldots,B\_{33}\in\mathbb{F}\_2^{5\times 10}$ with the desired property.


