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What does it mean for a collection of bases $\{B_i\}_i$ for a corresponding set of vector subspaces $\{W_i\}_i$ to be pairwise disjoint? Thank you.

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Disjoint means "having empty intersection", so pairwise disjointness means $B_i \cap B_j = \emptyset$ for $i \ne j$.

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Ah, so there is no implication that the elements in the $B_i$'s are linearly independent? – johann Sep 12 '11 at 21:10
@johann: If each $B_i$ is a basis (implied by saying it is a "collection of bases"), then of course each $B_i$ is linearly independent, so the elements of each $B_i$ form a linearly independent set. – Arturo Magidin Sep 12 '11 at 21:12
@Arturo: Thanks! Ah, I see it's an English slip... I meant that the elements from the different $B_i$'s. – johann Sep 12 '11 at 21:14
I would imagine that if there is a vector in the intersection of different bases $B_i, B_j$ , that the change-of-basis matrix $T$ taking the matrix for $B_i$ (with vectors as columns) has an eigenvalue =$1$, with a vector in common as eigenvector associated with that eigenvalue. – gary Sep 12 '11 at 21:28
@johann: No, there is no assumption on linear independent. For example, in $\mathbb{R}^3$, $B_1=\{(1,0,0), (0,1,0), (0,0,1)\}$ and $B_2=\{(1,1,0),(1,0,1), (0,1,1)\}$ are disjoint bases, but lots of subsets of 3 elements taken from the two bases are linearly dependent. – Arturo Magidin Sep 12 '11 at 21:29

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