Prove or give a counterexample: If $U_1,U_2,W$ are subspaces of $V$ such that

then $U_1 = U_2$.

Let $V = \mathbf R^2$, $W = \{(x,0) : x \in \mathbf R\}$, $U_1 = \{(0,y) : y \in \mathbf R\}$ and $U_2 = \{(y,y) : y \in \mathbf R\}$.

We have $V = U_1 \oplus W = U_2 \oplus W$ but $U_1 \ne U_2$, thus this is a counterexample.

This was counterintuitive to me at first, until I realized the only requirement for a direct sum (of two subspaces) being $U_1 \cap W = \{0\}$ means any subspace $U_1$ is direct if it doesn't intersect $W$ (this seems obvious in hindsight, but thinking about it this way makes the non-uniqueness obvious)

Consequences:

• The direct sum in 2.34 is not unique
• Extending a basis of $W$ to a basis of $V$ can be done in multiple ways, such that the list we add has a different span. (my intuition before was that it would be the same, which is wrong)

Doing the previous exercises 1c/21 and 1c/23 made coming up with a counterexample relatively easy.