Inverse of an element of a linear vector space For a linear vector space $\mathbb{V}$, the inverse $|\phi\rangle$ associated with an element $|\psi\rangle$ is defined as $$|\phi\rangle+|\psi\rangle=|0\rangle$$ where $|0\rangle$ is the additive identity and is not same as number $0$. In that case, is it justified to write $|\phi\rangle=-|\psi\rangle$? All I can see from the definition is that $|\phi\rangle=|0\rangle-|\psi\rangle$. But how is $|0\rangle-|\psi\rangle=-|\psi\rangle$? I used Dirac notation which physicists use all the time.
 A: Yes, $-\left|\psi\right\rangle$ is the additive inverse of $\left|\psi\right\rangle$.
The symbol $-$, when used as unitary prefix operator ($-x$), is universally understood as the function mapping an element $x$ to its additive inverse. It doesn't matter if $x$ is $1$, $i$, $2^{100}$, $\vec v$ or $\left|\psi\right\rangle$.
Note that the use as binary infix operator, $x-y$, is actually equivalent to addition of the inverse, $x+(-y)$.
A short note on your use of $\left|0\right\rangle$ to denote the zero vector: While this is not directly wrong, it is highly unusual in physics. When physicists use $\left|0\right\rangle$, this usually denotes a nonzero (indeed, normalized) vector corresponding to some specific quantum state of the system under consideration; most often the ground state (state of lowest energy), in quantum field theory the vacuum.
A: The definition of a (linear) vector space starts with the assumptions that you can add vectors and multiply them by scalars and that the usual rules of arithmetic apply (when they make sense). This is independent of Dirac notation, which I will forego here.
Further, the definition requires that there be a vector which when added to any other vector leaves that vector unchanged. It's usually called $0$. It's not hard to see that $\lambda 0 = 0$ for any scalar $\lambda$ and that $1v = v$ for any vector $v$.
Now let $v$ be any vector. The arithmetic of vectors then says
$$
(-1)v +  v = (-1 + 1)v = 0v =0.
$$
That says $(-1)v$ is a (in fact the) vector which you can add to $v$ to get $0$. It's natural to call that vector $-v$.
This rather longwinded discussion also explains a puzzle schoolchildren sometimes have distinguishing (when necessary) between the $-$ sign as an operator and as an adjective, as in
$$
3 - 2 = 3 + (-2).
$$
Just get used to it.
