0
$\begingroup$

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:

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

$\endgroup$
3
  • 1
    $\begingroup$ Have you tried anything on your own? Do you understand the definition of the orthogonal complement? $\endgroup$ – Ben Grossmann May 12 '20 at 21:12
  • $\begingroup$ 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. $\endgroup$ – Ben Grossmann May 12 '20 at 21:23
  • $\begingroup$ 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. $\endgroup$ – Ben Grossmann May 13 '20 at 21:01
0
$\begingroup$

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: $$ \begin{align} X_1 + X_2 &= \{x: x = x_1 + x_2 \text{ with } x_1 \in X_1,x_2 \in X_2\} \\ & = \{x_1 + x_2 : x_1 \in X_1, x_2 \in X_2\}\\ (X_1 + X_2)^\perp &= \{y: \langle y,x \rangle = 0 \text{ for all } x \in X_1 + X_2\}\\ &= \{y: \langle y,x_1 + x_2 \rangle = 0 \text{ for all } x_1 \in X_1, x_2 \in X_2\}\\ X_1^\perp + X_2^\perp &= \{y_1 + y_2 : y_1 \in X_1^\perp \text{ and }y_2 \in X_2^\perp\}\\ X_1 \cap X_2 &= \{x: x \in X_1 \text{ and }x \in X_2\}\\ X_1^\perp \cap X_2^\perp &= \{y: y \in X_1^\perp \text{ and } y \in X_2^\perp\} \\ &= \{y : \langle y,x \rangle = 0 \text{ for all } x \text{ such that } x \in X_1 \text{ and all } x \text{ such that } x \in X_2\}\\ (X_1 \cap X_2)^\perp &= \{y : \langle y,x \rangle = 0 \text{ for all } x \in X_1 \cap X_2\} \end{align} $$

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ – Desperate Soul May 13 '20 at 7:26
  • $\begingroup$ @DesperateSoul That's not the literal definition, but your statement is technicially correct. Do you understand what the subspace $X_1 + X_2$ is? $\endgroup$ – Ben Grossmann May 13 '20 at 15:18
  • $\begingroup$ @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 $V = \Bbb R^3$? $\endgroup$ – Ben Grossmann May 13 '20 at 15:31
  • $\begingroup$ 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 $\endgroup$ – Desperate Soul May 13 '20 at 18:09
  • $\begingroup$ @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? $\endgroup$ – Ben Grossmann May 13 '20 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.