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As the title says, I would like to know if that equality is true if we consider $\mathbb{R^4}$.

I would write down my attempt, but I have no idea how to prove it, or how to choose a counterexample.

$(U^{\perp}\cap V^{\perp})$ are the vectors which are orthogonal to both $U$ and $V$, while $(U+V)^{\perp}$ are the vectors which are orthogonal to the linearly independent vectors of $U$ and $V$

Lorenzo B.
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1 Answers1

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This is a very basic property and can be shown like a set theoretic result:

Let $x \in (U^{\perp} \cap V^{\perp})$. Then $(x,u) = (x,v) = 0$ for all $u \in U$ and all $v \in V$, where $(\cdot,\cdot)$ denotes the inner product.

Now take $y \in U + V$. By definition of this sum, there exist $u_y \in U$ and $v_y \in V$ such that $y = u_y + v_y$. Now we have $$(x,y) = (x,u_y+v_y) = (x,u_y) + (x,v_y) = 0+0=0$$ and as we started with $y$ arbitrary, we get that $x \in (U+V)^{\perp}$.

For the other direction let $x \in (U+V)^{\perp}$. Then...

I'm sure you can fill in the rest now.

Dirk
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  • Thank you, now I'll write my attempt at the other direction:

    Let $x \in (U+V)^{\perp}$. Then $(x,u_x+v_x)=0$ where $u_x+v_x \in (U+V)$. Then by linearity of the inner product, $(x,u_x)+(x,v_x)=0=0+0 (*)$

    Then $x \in U^{\perp}$ and $x \in V^{\perp}$, so $x \in (U^{\perp} \cap V^{\perp}) $. Am I right? I'm not convinced by $(*)$

     – Lorenzo B. Sep 12 '17 at 09:44