orthogonal direct sum - counter example

Question

Let V be inner product space and $U_1, U_2, U_3$ be its sub-spaces.

If $U_1 \oplus U_2 \oplus U_3=V$ then $U_1^\perp \oplus U_2^\perp \oplus U_3^\perp=V$

It seems to be not true but i cant find any counter example.

Any hints?

Answer 1

To summarise what was said in the chat, we show that the case $\dim V = 3$ leads to a contradiction.

Suppose that $U_{1} + U_{2} + U_{3} = V$ where $\dim V = 3$. One possibility for the subspaces $U_{i}$ is that $$ \dim U_{1} = \dim U_{2} = \dim U_{3} = 1. $$ For convenience, you may view $U_{1}, U_{2}, U_{3}$ as the coordinate axes (though there are plenty of other subspaces that could have this property).

By the direct sum $\oplus$ conditions, we must have that $A \cap B = \{0\}$ whenever we see $A \oplus B$. This is true for $U_{1}, U_{2}, U_{3}$.

But now consider the orthogonal spaces $U_{1}^{\perp}, U_{2}^{\perp}, U_{3}^{\perp}$. We know that $$ \dim U_{i}^{\perp} = \dim V - \dim U_{i} $$ so that each orthogonal subspace has dimension $$ \dim U_{1}^{\perp} = \dim U_{2}^{\perp} = \dim U_{3}^{\perp} = 2. $$ But in a three-dimensional space, these must have a pairwise one-dimensional intersection. In the case that we view $U_{1}, U_{2}, U_{3}$ as the coordinate axes, then $$ U_{1} = \text{$x$-axis} \implies U_{1}^{\perp} = \text{$yz$-plane} $$ and $$ U_{2} = \text{$y$-axis} \implies U_{2}^{\perp} = \text{$xz$-plane} $$ so that the intersection of $U_{1}^{\perp}$ and $U_{2}^{\perp}$ is the $z$-axis (and analogously for the other two pairs). This is not trivial, so each pair fails the condition to be 'directly summed' together.

You can also see this from the dimensions. If $$ U_{1}^{\perp} \oplus U_{2}^{\perp} \oplus U_{3}^{\perp} = V $$ where $$ \dim U_{1}^{\perp} = \dim U_{2}^{\perp} = \dim U_{3}^{\perp} = 2, $$ then we must have $$ \dim V = \dim U_{1}^{\perp} + \dim U_{2}^{\perp} + \dim U_{3}^{\perp} = 6. $$ This is clearly false. Hence a contradiction to the statement.