About $B(H)$ containing a embedded copy of $M_2(\mathbb{C})$

Question

Let $H$ be a Hilbert space. Let $B(H)$ be the space of all bounded linear operators on $H$. I came across a statement that states that: If dim(H)>1, then $B(H)$ contains an embedded copy of $M_2(\mathbb{C})$. Can anyone tell how? or give a reference.

Answer 1

Since $\dim(H)>1$, that is, $\dim(H) \geq 2$, there are at least two linearly independent (or even orthonormal) vectors $e_1,e_2$.

Now, consider the subspace $X$ generated by $\{e_1, e_2\}$ and look at the operators $A:H\rightarrow H$ such that $AX \subset X$ and $A|_{X^\bot} = 0_{X^\bot}$ (here, $X^\bot$ is the orthogonal complement to $X$, and $0_{X^\bot}$ is the zero operator on it). Such operators act inside the subspace $X$, and leave the rest of the space untouched.

Then

• All such operators are indeed bounded
• The algebra of this operators is isomorphic to $M_2(\mathbb C)$

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Boundedness follows from the fact that such operators are "essentially finite-dimensional" (technically, they have finite-dimensional image), and all finite-dimensional operators are automatically bounded.

To prove boundedness rigorously, it is enough to consider $x \in H$ such that $\|x\|=1$. Decompose $x$ as $x = x_1e_1 + x_2e_2 + y$, where $x_1, x_2 \in \mathbb C$ and $y \in X^\bot$. Note that $\|x\|^2 = |x_1|^2 + |x_2|^2+\|y\|^2$ (we are using that $e_1, e_2$ are orthonormal), so, since $\|x\|=1$, $|x_1|$, $|x_2|$ and $\|y\|$ are all $\leq 1$. Furthemore, $Ay=0$, since $y \in X^\bot$.

Then

$$\|Ax\| = \|A(x_1e_1 + x_2e_2 + y)\| = \|x_1Ae_1 + x_2Ae_2 + Ay\| = \\ = |x_1|\cdot\|Ae_1\| + |x_2|\cdot\|Ae_2\| + \|Ay\| = |x_1|\cdot\|Ae_1\| + |x_2|\cdot\|Ae_2\| + \|0\| \leq \\ \leq \|Ae_1\| + \|Ae_2\| $$

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This set indeed forms a subalgebra of $B(H)$, since

• $\forall \lambda \in \mathbb C : \lambda A X \subset X$ and $(\lambda A) y = \lambda (A y) = \lambda 0 = 0$ for $y \in X^\bot$
• $(A_1 + A_2) X = A_1 X + A_2 X \subset X$ and $(A_1 + A_2) y = A_1 y + A_2 y = 0 + 0 = 0$ for $y \in X^\bot$
• $A_1 A_2 X \subset A_1 X \subset X$ and $(A_1 A_2) y = A_1 (A_2 y) = A_1 0 = 0$ for $y \in X^\bot$

That is, this set of operators is closed under multiplication by a scalar and addition (thus forms a subspace), and under multiplication (thus forms a subalgebra).

To see the isomorphism with $M_2(\mathbb C)$, observe that this subalgebra is $$\operatorname{End}(X) \cong M_{\dim X}(\mathbb C) = M_2(\mathbb C)$$