Invariance of boundary Related : How do I show that for linearly independent set in dual is a dual of a linearly independent set?
Reference: Margalef - Differential Topology p.8

Definition
Let $V$ be a real normed space and $A:=\{\lambda_1,...,\lambda_n\}$ be linearly indepedent continuous linear functionals on $V$. Define $V^+_A:=\{x\in V: \forall 1\leq i \leq n, \lambda_i(x)\geq 0\}$. Then, $V^+_A$ is called the $A$-quadrant of $V$.

My question is the following:

Let $V$ be a real normed space and $A:=\{\lambda_1,...,\lambda_n\}$ and $B:=\{\mu_1,...,\mu_n\}$ be linearly independent continuous functionals on $V$. If $V^+_A=V^+_B$, then how do I prove that there exists a permutation $\sigma\in S_n$ and $r_1,...,r_n>0$ such that $\lambda_i=r_i \mu_{\sigma(i)}$?

Assuming $V^+_A=V^+_B$, I have proven that $\bigcap_{i=1}^n \ker(\lambda_i)=\bigcap_{i=1}^n \ker(\mu_i)$. Let's denote this subspace as $W$. Moreover, I have shown that $V=V^+_A - V^+_A$.
Since the map $V\rightarrow \mathbb{R}^n:x\mapsto (\lambda_1(x),...,\lambda_n(x))$ is surjective, there exists $x_1,...,x_n\in V$ such that $\lambda_i(x_j)=\delta_{ij}$ and $V= W \oplus span(\{x_1,...,x_n\})$. Analogously, there exists $y_1,...,y_n\in V$ such that $\mu_i(y_j)=\delta_{ij}$ and $V=W\oplus span(\{y_1,...,y_n\})$.
Fix $1\leq k\leq n$. Since $x_k\in V^+_A=V^+_B$, For all $1\leq i\leq n$, $\mu_i(x_k)\geq 0$. Write $x_k=\alpha +\sum_{i=1}^n a_i y_i$ where $\alpha\in W$ and $a_i\in \mathbb{R}$. Since $a_i=\mu_i(x_k)$, $a_i\geq 0$ for all $i$. Hence, we can conclude that for given $a_1,...,a_n\geq 0$, there exist $b_1,...,b_n\geq 0$ and $\alpha\in W$ such that $\sum_{i=1}^n a_i x_i= \alpha+ \sum_{i=1}^n b_i y_i$.
So it would suffice to show that $\lambda_i$ is equal to some $\mu_j$ on the above elements since $V=V^+_A-V^+_A$. However, I'm stuck here. How do I prove this? Thank you in advance.
Edited:
Moreover, I have now proven that there exists a linear homeomorphism $T:V\rightarrow V$ such that $T(x)=x$ on $W$ and $T(x_i)=y_i$ so that $T(V^+_A)=V^+_A$ and $\mu_i\circ T = \lambda_i$. However, I don't know if this helps.
 A: Note that $y_i=\sum_j \lambda_j(y_i)x_j \mod W$ and $x_i=\sum_j \mu_j(x_i)y_j \mod W$.
Hence, $x_i=\sum_j\sum_k \mu_j(x_i)\lambda_k(y_j)x_k \mod W$. Thus, $\delta_{ik}=\sum_j \mu_j(x_i)\lambda_k(y_j)$.
Let $A$ be the $n\times n$ matrix with entries $A_{ij}= \lambda_i(y_j)$ and $B$ be the $n\times n$ matrix with entries $B_{ij}=\mu_i(x_j)$. Then, $AB=I_n$.
Since $(AB)_{ii}=1$, $\sum_j A_{ij} B_{ji} = 1$. Since entries of $A,B$ are nonnegative by the assumption, there must exist $j$ such that $A_{ij} B_{ji} >0$, hence $A_{ij}>0$ and $B_{ji}>0$ .Suppose that there exists $k\neq j$ such that $B_{ki}>0$. If $l\neq i$, $(AB)_{il}=0$, and since entries are nonnegative, $B_{jl}$ and $B_{kl}$ must be $0$. This implies that $det(B)=0$, which is a contradiction. Hence for all $k\neq j$, $B_{ki}=0$. This shows that for each column has only one nonzero row. Together with the fact that $\det(B)\neq 0$, we conclude that $B$ is a permutation matrix with multipled entries. Thus, there exists a permutation $\sigma\in S_n$ such that $\mu_i(x_{\sigma(j)})=r_i \delta_{ij} (r_i>0)$.  Since $V=W\oplus span(\{x_1,...,x_n\})=W\oplus span(\{y_1,...,y_n\})$, we have that $\mu_i= r_i \lambda_{\sigma(i)}$. Q.E.D.
