I'm just stuck on this question. How can I represent the additive inverse of all continuous functions?
The additive inverse: For every $\overrightarrow{u}$ in V, there is a vector V denoted by $\overrightarrow{-u}$ such that $\overrightarrow{u}$ + ($\overrightarrow{-u}$) = $\overrightarrow{0}$.
Any help is appreciated.
This is a solution I found to a similiar problem earlier:
Describe the additive inverse of the vector space $P_3$ where $P_3$ is the set of all polynomials of degree 3 or below.
Solution:
$-(a_0 + a_1x + a_2x^2 + a_3x^3) = -a_0 – a_1x – a_2x^2 – a_3x^3$
Best Answer
I'm not sure if this is what you're after, but the additive identity is always given by $0\cdot\mathbf{v}$ for any $\mathbf{v}$ in the vector space.
Proof: $$\mathbf{v} = 1\cdot\mathbf{v}=(1+0)\mathbf{v} = 1\cdot\mathbf{v} + 0\cdot\mathbf{v} = \mathbf{v} + 0\cdot\mathbf{v}$$ Applying $(\mathbf{-v})$ to both sides then yields $\mathbf{0} = 0\cdot\mathbf{v}$
If the additive inverse is what you want instead, then a similar result will show that $(-1)\mathbf{v}$ is the additive inverse for $\mathbf{v}$
Proof: $$\mathbf{v} + (-1)\mathbf{v} = 1\cdot\mathbf{v} + (-1)\mathbf{v} = (1+(-1))\mathbf{v} = 0\cdot\mathbf{v} = \mathbf{0}$$ Again applying $(\mathbf{-v})$ yields $(\mathbf{-v}) = (-1)\mathbf{v}$