Can I define a linear map on the set $\{X\in \mathcal{S}^{n\times n}: X\succeq 0, \operatorname{X}=1 \}$?

Question

The definition of a linear map $\phi$: $\mathcal{V} \mapsto\mathcal{W}$ is:

$\forall u, v \in \mathcal{V}$ and $\forall a,b \in \mathbb{F}$ such that $\phi(au+bv) = a\phi(u) + b\phi(v)$.

Can I define a linear map between the set $\mathcal{V}=\{X\in \mathcal{S}^{n\times n}: X\succeq 0, \operatorname{tr} (X)=1 \}$ and other group such as $SO(n)$?

(I know $\mathcal{V}$ is not a group since there is no identity element for addition and multiplication. $\mathbf{0}\notin \mathcal{V}$ and $I_{n\times n}\notin \mathcal{V}$)

What I am confused now is that $X, Y\in \mathcal{V}$ does not imply $aX+bY\in \mathcal{V}$ so addition and scalar multiplication are not defined on the set $\mathcal{V}$.

However, $\mathcal{V}$ is a convex set, which is $\operatorname{conv}\{X\in \mathcal{S}^{n\times n}: X\succeq 0, \operatorname{tr} (X)=1, \operatorname{rank}(X)=1\}$. Therefore, it seems that we can define $\phi$ such that $\phi(\lambda X+(1-\lambda)Y) = \lambda \phi(X) + (1-\lambda)\phi(Y), \ \ \lambda \geq 0$. And we know $\lambda X+(1-\lambda)Y\in \mathcal{V}$ if $X,Y\in \mathcal{V}$

So could we define a linear map on $\mathcal{V}$? or actually we could define a homomorphism, which is more general than a linear map, on $\mathcal{V}$?

Hope for a detailed explanation.

Answer 1

As you've mentioned positive semidefiniteness, I suppose that $\mathbb F$ is either $\mathbb R$ or $\mathbb C$ and the symbol $\mathcal S^{n\times n}$ means the set of all Hermitian matrices in $M_n(\mathbb F)$. I also assume that $n\ge2$. Then:

1. There exists a linear map on $M_n(\mathbb F)$ or $\mathcal S^{n\times n}$ that maps $\mathcal V$ into $SO(n)$. One obvious choice is $X\mapsto\operatorname{trace}(X)\,I_n$. Any such linear map cannot be multiplicative, however, because there are elements in $\mathcal V$ whose product is zero.
2. There does not exist any linear map on $M_n(\mathbb F)$ or $\mathcal S^{n\times n}$ that maps $SO(n)$ into $\mathcal V$, because matrices in $SO(n)$ can sum to zero. For instance, when $n=2$, we have $I+(-I)=0\notin\mathcal V$.