To prove this, we need to prove $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$ and $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$.
To prove $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$:
We know since each vector in $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$ is a linear combination of $\{\vec{v}_1, \vec{v}_2 \}$, any linear combination of elements from the set $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$ can also be viewed as a linear combination of elements from the set $\{\vec{v}_1, \vec{v}_2 \}$. Therefore, $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$.
Now going the other way, we need to prove $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
Starting with the set $\{\vec{v}_1, \vec{v}_2 \}$, linear combinations of $\vec{v}_1$ and and $\vec{v}_2$ can be made. Two of these combinations include $(1)\vec{v}_1 + (1)\vec{v}_2$ and $(1)\vec{v}_1 + (-1)\vec{v}_2$. These combinations can be simplified to form $\vec{v}_1 + \vec{v}_2$ and $\vec{v}_1 - \vec{v}_2$. Since both $\vec{v}_1 + \vec{v}_2$ and $\vec{v}_1 - \vec{v}_2$ are within the set $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$, the set $\{\vec{v}_1, \vec{v}_2 \}$ is a subset of $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$. Therefore, $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
We have now proven $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$ and $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$. Thus we can conclude, $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle = \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
I am not completely certain that this the right way to prove this problem, especially the second part of the proof.
I could not figure out how to get the curly brackets to appear between the span symbols. The brackets are supposed to be there, but for some reason they will not appear. Please just assume that they are there.
To get the curly bracket to show, try /langle\\{\vec{v}_1\\}\langle
You have to use \\{ on the forums as opposed to the standard\{. I don't know why.
here is the above text compiled:
$\langle{ \{\vec{v}_1\} }\rangle$
@Student23 First off, please note that I made a few small edits to your post. The words that I added in place of yours are in bold. I also added the set braces like so: \\{..\\}.
Generally, I think you have the right idea here. I certainly like where you say:
any linear combination of elements from the set $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$ can also be viewed as a linear combination of elements from the set $\{\vec{v}_1, \vec{v}_2 \}$.
Now, to justify that symbolically, I guess you could start with a linear combination
$$
a (\vec{v}_1 + \vec{v}_2) + b (\vec{v}_1 - \vec{v}_2)
$$
from the first set and try to express it as a linear combination of elements from the second, say
$$
c\vec{v}_1 + d\vec{v}_2.
$$
Of course, $c$ and $d$ would be expressed in terms of $a$ and $b$.
Comments
To prove this, we need to prove $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$ and $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$.
To prove $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$:
We know since each vector in $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$ is a linear combination of $\{\vec{v}_1, \vec{v}_2 \}$, any linear combination of elements from the set $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$ can also be viewed as a linear combination of elements from the set $\{\vec{v}_1, \vec{v}_2 \}$. Therefore, $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$.
Now going the other way, we need to prove $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
Starting with the set $\{\vec{v}_1, \vec{v}_2 \}$, linear combinations of $\vec{v}_1$ and and $\vec{v}_2$ can be made. Two of these combinations include $(1)\vec{v}_1 + (1)\vec{v}_2$ and $(1)\vec{v}_1 + (-1)\vec{v}_2$. These combinations can be simplified to form $\vec{v}_1 + \vec{v}_2$ and $\vec{v}_1 - \vec{v}_2$. Since both $\vec{v}_1 + \vec{v}_2$ and $\vec{v}_1 - \vec{v}_2$ are within the set $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$, the set $\{\vec{v}_1, \vec{v}_2 \}$ is a subset of $\{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \}$. Therefore, $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
We have now proven $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$ and $\langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle \subseteq \langle \{\vec{v}_1, \vec{v}_2 \} \rangle$. Thus we can conclude, $\langle \{\vec{v}_1, \vec{v}_2 \} \rangle = \langle \{\vec{v}_1 + \vec{v}_2, \vec{v}_1 - \vec{v}_2 \} \rangle$.
I am not completely certain that this the right way to prove this problem, especially the second part of the proof.
I could not figure out how to get the curly brackets to appear between the span symbols. The brackets are supposed to be there, but for some reason they will not appear. Please just assume that they are there.
To get the curly bracket to show, try /langle\\{\vec{v}_1\\}\langle
You have to use \\{ on the forums as opposed to the standard\{. I don't know why.
here is the above text compiled:
$\langle{ \{\vec{v}_1\} }\rangle$
@Student23 First off, please note that I made a few small edits to your post. The words that I added in place of yours are in bold. I also added the set braces like so:
\\{..\\}.Generally, I think you have the right idea here. I certainly like where you say:
Now, to justify that symbolically, I guess you could start with a linear combination
$$
a (\vec{v}_1 + \vec{v}_2) + b (\vec{v}_1 - \vec{v}_2)
$$
from the first set and try to express it as a linear combination of elements from the second, say
$$
c\vec{v}_1 + d\vec{v}_2.
$$
Of course, $c$ and $d$ would be expressed in terms of $a$ and $b$.