Determine if the subset f(x)=c , the set of all constant functions is a subspace of all continous functions
sure seems that way to me c+d=e another constant a*c=ac another constant so I think it is closed under addition and multiplication
or a*c=d however you want to write it...
okkie dokkie
I have one more lol
wait I totally misunderstood that question :/
ohh lol
I thought constant functions lol
I mean, wait... no that is what they mean, right? if constants are a subset of continuous functions?
i think so that is what i thought
ok then I'm right after all sorry, other question?
umm when f(0)=1
umm that means that c=1
no?
talking about the same problem?
yes like instead of it being f(x)=c it is f(x)=1
whoops f(0)=1
f(x)=c it doesn't matter what x is, c is whatever we choose it to be if you mean a different problem, like asking about if f(0)=1 is a subspace of c, the answer is no
noo ok let me rephrase the question
Determine if the subset f(0)=1 is a subspace of all continous functions
now that i am looking at it i think the answer wld be yes
that is what I meant by if f(0)=1 is a subspace of c ask yourself this: what would be f(0)+g(0) in this vector space ?
c+d which is a constant which is a straight line. so it wld still be continous
but\[f(0)+g(0)=1+1=2\neq1\]
the question is not whether the result of our addition is in c, but whether it is in our subspace W
ohhh ok just digesting that
if the space is not closed by addition and scalar multiplication it is not a subspace, not matter if the result is in the original space V
like if it was f(0)=0 then it wld be a subspace right?
Right
kkk got it Thanks turing, U r a lifesavor
that, by the way, is called the null space it is a subspace of every vector space
oh cool
sorry it's called 'zero space'*
null space is something else
every vector space has at least two subspaces 1)the zero space 2)itself and it must be closed in /itself/ hope that helps
yes i am sord of getting the hang of it
glad, good luck :D
thanks
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