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There is any proof about $B(H)\cong\mathbb{M}_n(\mathbb{C})$ for some $n\in\mathbb{N}$? Where $H$ is a Hilbert space finite dimensional and $B(H)$ is bounded operator space of $H$.

Kempa
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    Well, $H\cong \mathbb{C}^n$ for some $n$ (if it is a complex Hilbert space) if $H$ is finite-dimensional. Furthermore, in finite-dimensions, every linear map is bounded. – Severin Schraven May 02 '24 at 21:09

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Note that in general we have that $B(H)$ is a subset of the set of all linear maps $H\rightarrow H$. If $H$ is finite-dimensional, all linear maps are bounded (see here Every linear operator $T:X \to Y$ on a finite-dimensional normed space is bounded). Thus, $B(H)$ is equal to the set of all linear maps $H\rightarrow H$. After picking a basis, we can identify $B(H)\cong M_n(\mathbb{C})$ where $n=\text{dim}_\mathbb{C}(H)$ (see here Proving isomorphism between linear maps and matrices).

Severin Schraven
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