[proofplan] We first prove that the zero set $N_p$ is a linear subspace by using the two seminorm axioms: subadditivity and absolute homogeneity. Once the quotient [vector space](/page/Vector%20Space) $V/N_p$ is available, the main point is well-definedness: representatives of the same coset differ by an element on which $p$ vanishes, so subadditivity forces them to have the same seminorm. The norm axioms on $V/N_p$ then follow directly from the corresponding seminorm axioms on $V$, with definiteness supplied precisely by quotienting out $N_p$. [/proofplan] [step:Show that the null set of the seminorm is a linear subspace] Let $0_V\in V$ denote the zero vector. Since $p$ is absolutely homogeneous, \begin{align*} p(0_V)=p(0\cdot 0_V)=|0|p(0_V)=0. \end{align*} Thus $0_V\in N_p$, so $N_p$ is nonempty. Let $x,y\in N_p$. By subadditivity of $p$, \begin{align*} 0\leq p(x+y)\leq p(x)+p(y)=0. \end{align*} Hence $p(x+y)=0$, and therefore $x+y\in N_p$. Let $\lambda\in\mathbb{F}$ and let $x\in N_p$. By absolute homogeneity, \begin{align*} p(\lambda x)=|\lambda|p(x)=0. \end{align*} Thus $\lambda x\in N_p$. Therefore $N_p$ is a linear subspace of $V$. [guided] We need to verify the linear subspace conditions for $N_p=\{x\in V:p(x)=0\}$. The first condition is nonemptiness, and the natural candidate is the zero vector $0_V$. Because $p$ is a seminorm, it is absolutely homogeneous, meaning $p(\lambda x)=|\lambda|p(x)$ for every $\lambda\in\mathbb{F}$ and every $x\in V$. Applying this with $\lambda=0$ and $x=0_V$ gives \begin{align*} p(0_V)=p(0\cdot 0_V)=|0|p(0_V)=0. \end{align*} So $0_V\in N_p$. Next we prove closure under addition. Take $x,y\in N_p$, so $p(x)=0$ and $p(y)=0$. The subadditivity axiom for the seminorm gives \begin{align*} p(x+y)\leq p(x)+p(y)=0. \end{align*} Since the codomain of $p$ is $[0,\infty)$, we also have $0\leq p(x+y)$. Combining both inequalities gives \begin{align*} p(x+y)=0. \end{align*} Therefore $x+y\in N_p$. Finally we prove closure under scalar multiplication. Let $\lambda\in\mathbb{F}$ and let $x\in N_p$. Absolute homogeneity gives \begin{align*} p(\lambda x)=|\lambda|p(x)=|\lambda|\cdot 0=0. \end{align*} Thus $\lambda x\in N_p$. We have shown that $N_p$ is nonempty, closed under addition, and closed under scalar multiplication, so $N_p$ is a linear subspace of $V$. [/guided] [/step] [step:Prove that the quotient formula is independent of the representative] Define a map candidate \begin{align*} q:V/N_p\to [0,\infty),\qquad x+N_p\mapsto p(x). \end{align*} We prove that $q$ is well-defined. Let $x,y\in V$ and suppose $x+N_p=y+N_p$. By the definition of equality of cosets in the quotient by a subspace, $x-y\in N_p$, so $p(x-y)=0$. Using subadditivity and absolute homogeneity, \begin{align*} p(x)=p(y+(x-y))\leq p(y)+p(x-y)=p(y). \end{align*} Also $y=x+(y-x)$ and $y-x=-(x-y)$, so \begin{align*} p(y)=p(x+(y-x))\leq p(x)+p(y-x)=p(x)+p(-(x-y))=p(x)+|-1|p(x-y)=p(x). \end{align*} Thus $p(x)=p(y)$. Hence the value assigned to the coset $x+N_p$ does not depend on the representative $x$, and $q$ is well-defined. [/step] [step:Verify positivity and definiteness on the quotient] For every coset $x+N_p\in V/N_p$, the definition of $q$ gives \begin{align*} q(x+N_p)=p(x)\geq 0. \end{align*} Thus $q$ is nonnegative. Now suppose $q(x+N_p)=0$. Then $p(x)=0$, so $x\in N_p$. Therefore $x+N_p=N_p$, which is the zero vector of the quotient vector space $V/N_p$. Conversely, \begin{align*} q(N_p)=q(0_V+N_p)=p(0_V)=0. \end{align*} Hence $q(x+N_p)=0$ if and only if $x+N_p=N_p$. [/step] [step:Verify absolute homogeneity on quotient cosets] Let $\lambda\in\mathbb{F}$ and let $x+N_p\in V/N_p$. The scalar multiplication in $V/N_p$ is defined by \begin{align*} \lambda(x+N_p)=\lambda x+N_p. \end{align*} Using the definition of $q$ and absolute homogeneity of $p$, \begin{align*} q(\lambda(x+N_p))=q(\lambda x+N_p)=p(\lambda x)=|\lambda|p(x)=|\lambda|q(x+N_p). \end{align*} Thus $q$ is absolutely homogeneous. [/step] [step:Verify the triangle inequality on quotient cosets] Let $x+N_p,y+N_p\in V/N_p$. Addition in $V/N_p$ is defined by \begin{align*} (x+N_p)+(y+N_p)=(x+y)+N_p. \end{align*} Using the definition of $q$ and subadditivity of $p$, \begin{align*} q((x+N_p)+(y+N_p))=q((x+y)+N_p)=p(x+y)\leq p(x)+p(y)=q(x+N_p)+q(y+N_p). \end{align*} Thus $q$ satisfies the triangle inequality. [/step] [step:Conclude that the induced quotient function is a norm] The preceding steps show that $q:V/N_p\to [0,\infty)$ is well-defined, nonnegative, definite, absolutely homogeneous, and subadditive. These are exactly the norm axioms on the quotient vector space $V/N_p$. Therefore \begin{align*} \|x+N_p\|_{V/N_p}:=p(x) \end{align*} defines a norm on $V/N_p$. [/step]