Can someone plz help with last two?
#Linear Algebra
tranquil pewter
torn gyroBOT
Can someone plz help with last two?
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close
- Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
viscid oxide
Can someone plz help with last two?
what's the definition of a 0 vector?
tranquil pewter
Wouldnt it just be 0?
viscid oxide
is $u\boxplus 0=u$ for all $u$?
frail bladeBOT
Omegabet_
viscid oxide
again, what is the definition of a 0 vector? (ie, what does the axiom of there being a 0 vector say?)
tranquil pewter
all components equal to 0?
viscid oxide
No
cause the objects in V need not 'have components'
the 0 vector is an element of V first and foremost, so whatever type of object is in V is the type of object the 0 vector will be. Here V is R, so the 0 vector is a real number.
anyway, since ig you're not going to look at your notes, $0_V\in V$ is the 0 vector if $u\boxplus 0_V=0_V\boxplus u=u$ for all $u\in V$.
frail bladeBOT
Omegabet_
viscid oxide
therefore it's clear $u+0_V+1=u$, so $0_V+1=0$
frail bladeBOT
Omegabet_
viscid oxide
hence $0_V=-1$.
frail bladeBOT
Omegabet_
tranquil pewter
ohh i got it now
viscid oxide
alternatively, $0_V=0\boxdot u=0u+0-1=-1$
frail bladeBOT
Omegabet_
viscid oxide
what's the definition of an additive inverse?
viscid oxide
what is the axiomatic statement of what an additive inverse is?
tranquil pewter
An additive inverse of a number is defined as the value, which on adding with the original number results in zero value.
viscid oxide
that's for fields but sure
For $x\in V$, the additive inverse of $x$ is the vector $y\in V$ such that $x\boxplus y=0_V$
frail bladeBOT
Omegabet_
tranquil pewter
so its like x+y+1=-1
viscid oxide
yes
tranquil pewter
so does it mean the answer is just simply y
viscid oxide
clearly not?
y is a 'function' of x
for each x in V I get a y in V
y depends on x explicitly.
and if you sit there and say the inverse is 'y' with no actual definition of what y is concretely, you've not said anything remotely useful
tranquil pewter
is it y= -x-2
viscid oxide
yes.
tranquil pewter
ye i think i got it rn
viscid oxide
which, as you should know from the axioms, is also $-1\boxdot x$
frail bladeBOT
Omegabet_
