Thread: determine weather the given formula defines an inner product

Hi this was an exercise left for the reader, so i did part (a) and (c) but i really wanted to know (b) and (d) and i really tried but can't seem to figure it out

the exercise is as follows


here are the axioms i believe its referring to


and here is an example


I am really curious to see the solution, since i spent so much time on it.
Thank you in advance

Originally Posted by ohYeah

I am really curious to see the solution, since i spent so much time on it.
Thank you in advance

Why don't you tell us how far you've gotten? Then we can focus on where you need the help.

-Dan

ok this is what I did for (a) and (c)


and this is what i was thinking for (b)
just take f ∈ C[0,1] to be:

f(x) = 0, x∈ [0,1/2]

f(x) = x − 1/2, x∈ [1/2,1]

Then <f,f> = 0, but f ≠ 0.

<f,f> ≥ 0 is not enough for an inner product; it must satisfy the stronger condition:

<f,f> = 0 ⇔ f = 0

but i believe the approach i am taking for ( b ) is not correct and i don't know what to do for (d)

is anyone able to do this

what you did for (b) is just fine. it's not positive-definite (although it is positive), so its not an inner product.

let's just try to verify the axioms for (d):

<u,v> = (Au).(Av) = (Av).(Au) (since the standard dot product is symmetric)

= <v,u> well, that was easy.

<u,v+w> = (Au).(A(v+w)) = (Au).(Av+Aw) (since matrix multiplication is linear)

= (Au).(Av) + (Au).(Aw) (since the standard dot product is linear in the second variable)

= <u,v> + <u,w>.

i'm pretty sure you can handle the 3rd axiom yourself.

so it all boils down to the 4th axiom:

<u,u> > 0 when u ≠ 0 (it's pretty clear that <0,0> = (A0).(A0) = 0.0 = 0).

suppose u ≠ 0.

we need to know that Au ≠ 0 that is, that: null(A) = {0}. recall that n = dim(null(A)) + rank(A). under what conditions on rank(A) can we be sure that dim(null(A)) = 0?

so the rank(A) would have to be zero
so A would have to consist of zero vectors
so it does pass axiom 4
so its an inner product
is my logic near correct

no. it's very exactly totally wrong.

let me ask you, does 0 + 0 = n (if n > 0)?

oh wow yeah that makes absolutely no sense
so then the rank(A) = n
and since n = dim(null(A)) + rank(A) so if rank(A) = n then dim(null(A)) would have to be 0

yes, that's the right way to deduce it. sometimes this is phrased as: a matrix of full rank gives a 1-1 function.