Proof of inner spaces

[bonappler]

I have tried going through all the inner product space proofs for this question and the condition still holds, am I missing something?

 

[pka]

Let \(\displaystyle u = \left\langle {1,0,0} \right\rangle \) then \(\displaystyle \left\langle u,u \right\rangle=~? \).

What is wrong with that???

 

[stapel]

I have tried going through all the inner product space proofs for this question and the condition still holds, am I missing something?

We can't tell if you're missing something until you post your work and reasoning. So please be complete. Thank you!

 

[bonappler]

Let \(\displaystyle u = \left\langle {1,0,0} \right\rangle \) then \(\displaystyle \left\langle u,u \right\rangle=~? \).

What is wrong with that???

Uh, wouldn't that equal 0? Doesn't the condition still hold?

 

[bonappler]

also can someone help with this?

 

[HallsofIvy]

Uh, wouldn't that equal 0? Doesn't the condition still hold?

What "condition" are you talking about? This one: For any vector, v, \(\displaystyle <v, v>\ge 0\) and is equal to 0 only if v= 0?

 

[bonappler]

What "condition" are you talking about? This one: For any vector, v, \(\displaystyle <v, v>\ge 0\) and is equal to 0 only if v= 0?

Ohhhhh, I thought it had to be greater than or equal to 0 regardless of the value of v, it makes sense now thank you so much guys. I apologise if I seemed like a noob (I kind of am anyway).

 

[pka]

Uh, wouldn't that equal 0? Doesn't the condition still hold?

In any inner product space it must be true that \(\displaystyle \left\langle u,u \right\rangle=0 \) if and only if \(\displaystyle u = 0\).

 

[bonappler]

In any inner product space it must be true that \(\displaystyle \left\langle u,u \right\rangle=0 \) if and only if \(\displaystyle u = 0\).

yep, thanks I get it now

 

[HallsofIvy]

also can someone help with this?View attachment 4117

You are not showing any attempt to do these yourself. Surely whoever gave you this problem expects you to know:
To find the matrix representation of a linear transformation, apply the transformation to each of the basis vectors in turn, writing the result as a linear combination of the basis vectors. The coefficients give the columns of the matrix.