Question 10.7 of schaum's outlines linear algebra regarding Invariant Direct-Sum Decompositions

Question

The following theorem is proposed: let W, W', W",... be subspaces of V with B, B', B'',... bases. V is the direct sum of all W's if and only if the union of all the B's bases is a basis of V. I wonder what is wrong with the following counter example: let W be the subspace of R3 formed by the basis B={(1,0,0),(0,1,0)} and let W' be the subspace of R3 formed by the basis B'={(0,1,0),(0,0,1)}. The union of B and B' forms the canonical base for R3, yet R3 is not the direct sum of W and W', for the intersection between W and W' is not empty.enter image description here

Answer 1

I think their proof falls apart when they say "because $B$ is a basis of $V$, $a_{ij} = b_{ij}$ for each $i$ and each $j$". This is not correct because if, say $w_{11} = w_{12}$, then we can only conclude that $a_{11} + a_{12} = b_{11} + b_{12}$, so uniqueness can fail.