1

Assume that $X_1\subseteq V$ and $X_2 \subseteq V$

How can we prove that:

1) If $X_1 \subseteq X_2$ then $X_2^\perp\subseteq X_1 ^\perp$

2) a) $(X_1+X_2)^\perp=X_1^\perp\cap X_2^\perp$ and b) $(X_1\cap X_2)^\perp=X_1^\perp+X_2^\perp$

I found these 2 posts which are related to at least one of these problems, I could not comprehend the proof in the first post and the second post was unanswered. I was also hinted that I should use the definition of the orthogonal supplement to prove at least the first one but I couldn't do it despite my efforts.

orthogonal complement of a sum

Two proof problems about orthogonal complement

Edit: This is the proof I came up for (2a) after the help I got from the answers:

$\begin {align}X_1^\perp\cap X_2^\perp=\{y:\langle y,x\rangle=0\ \ \forall x_1\in X_1,\ \ \forall x_2\in X_2\}=\{y\in V:\langle x_1,y\rangle=0\ \ \& \ \ \langle x_2,y\rangle=0\ \ \forall x_1\in X_1,\ \ \forall x_2\in X_2\}=\{y\in V:\langle x_1,y\rangle + \langle x_2,y\rangle=0 \ \ \forall x_1\in X_1, \ \ \forall x_2\in X_2 \}=(X_1+X_2)^\perp \end{align}$

Is this correct?

As for (2b), I think it's $X_1^\perp + X_2^\perp = \{y_1 + y_2 : y_1 \in X_1^\perp \text{ and }y_2 \in X_2^\perp\}\\$ we got to work with, but its definition seems to differ a bit compared to the other ones on the list. I'm not sure how I'm supposed to proceed with it to reach the desired result, which would be:

$(X_1 \cap X_2)^\perp = \{y : \langle y,x \rangle = 0 \text{ for all } x \in X_1 \cap X_2\}$

Desperate Soul
  • 13
  • 3
  • 1
    Have you tried anything on your own? Do you understand the definition of the orthogonal complement? – Ben Grossmann May 12 '20 at 21:12
  • Please note the changes that I have made. In the future, try to put the entirety of each mathematical expression into one $...$. For instance, $A + B \geq C$ is better than A + B $\geq$ C. – Ben Grossmann May 12 '20 at 21:23
  • Regarding the proof you added, your first line $X_1^\perp\cap X_2^\perp={y:\langle y,x\rangle=0\ \ \forall x_1\in X_1,\ \ \forall x_2\in X_2}$ doesn't make sense. – Ben Grossmann May 13 '20 at 21:01

1 Answers1

1

Here's an answer to the first part.

We want to show that if $y \in X_2^\perp$, then $y \in X_1^\perp$. Consider any $y \in X_2^\perp$. By definition, this means that we have $\langle y,x \rangle = 0$ for every $x \in X_2$.

Now, consider any $x \in X_1$. It is also true that $x \in X_2$. So from what we just said, it follows that $\langle y, x \rangle = 0$.

Because $\langle y,x \rangle = 0$ for every $x \in X_1$, $y$ is an element of $X_1^\perp$, which is what we wanted.

Answering the second part requires that you similarly "unpack" the definition of the orthogonal complement.

In response to your comments: here is a way to write out the definition of all relevant sets: $$ \begin{align} X_1 + X_2 &= \{x: x = x_1 + x_2 \text{ with } x_1 \in X_1,x_2 \in X_2\} \\ & = \{x_1 + x_2 : x_1 \in X_1, x_2 \in X_2\}\\ (X_1 + X_2)^\perp &= \{y: \langle y,x \rangle = 0 \text{ for all } x \in X_1 + X_2\}\\ &= \{y: \langle y,x_1 + x_2 \rangle = 0 \text{ for all } x_1 \in X_1, x_2 \in X_2\}\\ X_1^\perp + X_2^\perp &= \{y_1 + y_2 : y_1 \in X_1^\perp \text{ and }y_2 \in X_2^\perp\}\\ X_1 \cap X_2 &= \{x: x \in X_1 \text{ and }x \in X_2\}\\ X_1^\perp \cap X_2^\perp &= \{y: y \in X_1^\perp \text{ and } y \in X_2^\perp\} \\ &= \{y : \langle y,x \rangle = 0 \text{ for all } x \text{ such that } x \in X_1 \text{ and all } x \text{ such that } x \in X_2\}\\ (X_1 \cap X_2)^\perp &= \{y : \langle y,x \rangle = 0 \text{ for all } x \in X_1 \cap X_2\} \end{align} $$

Ben Grossmann
  • 225,327
  • Thanks, that really hellped a lot. If my understanding is correct, then, regarding the second part it'd be $(X_1 + X_2)^\perp = {y \in V: \langle x_1,y\rangle=0 \ \ for \ \ all \ \ x_1\in X_1 \ \ and \ \ for \ \ all \ \ x_2\in X_2 }$

    Is this right? I also don't quite get how $X_1^\perp+X_2^\perp$ as well as $(X_1\cap X_2)^\perp$ and $X_1^\perp \cap X_2^\perp$ would be.

     – Desperate Soul May 13 '20 at 07:26
  • @DesperateSoul That's not the literal definition, but your statement is technicially correct. Do you understand what the subspace $X_1 + X_2$ is? – Ben Grossmann May 13 '20 at 15:18
  • @DesperateSoul In response to that last statement, I have edited my answer to include the definitions of some of the sets that we need to think about. If you are still confused, I recommend that you try thinking of examples. For instance, what if $X_1$ is the $xy$-plane and $X_2$ is the $xz$-plane with $V = \Bbb R^3$? – Ben Grossmann May 13 '20 at 15:31
  • Using the definitions you've included:

    $X_1^\perp\cap X_2^\perp={y:\langle y,x\rangle=0\ \ \forall x_1\in X_1,\ \ \forall x_2\in X_2}\Rightarrow{y\in V:\langle x_1,y\rangle=0\ \ & \ \ \langle x_2,y\rangle=0\ \ \forall x_1\in X_1,\ \ \forall x_2\in X_2}\Rightarrow{y\in V:\langle x_1,y\rangle + \langle x_2,y\rangle=0 \ \ \forall x_1\in X_1, \ \ \forall x_2\in X_2 }=(X_1+X_2)^\perp$

    Is this correct? As for (b), it is $X_1^\perp+X_2^\perp$ that we gotta work with I think, but I'm cluelesss as to how I'm supposed to approach it, it's very different from the others

     – Desperate Soul May 13 '20 at 18:09
  • @DesperateSoul I can't read your math properly, since you put it all into one line of a comment. Could you instead add this to your post after your question? – Ben Grossmann May 13 '20 at 19:50
  • Sorry, I'm not very familiar with LaTeX and formatting in general. I've updated the main post as you suggested. – Desperate Soul May 13 '20 at 20:23