Bounded Operators on a finite-dimensional Hilbert space - Linear combination of at most two unitaries and from a partial isometry to a unitary

Question

Good day,

In the lecture of functional analysis the proof of two statements were skipped as a task for us but I'm not sure how I approach these questions.

a) Show that every partial isometry $V \in B(H)$ on a finite-dimensional Hilbert space $H$ can be extended to a unitary on $H$.

b) Show that every bounded operator on a finite-dimensional Hilbert space can be written as a linear combination of at most two unitaries.

Some definitions and results which were proven:

• $B(H)=B(H,H)=\{ V:H \to H ~|~ V ~\text{linear and}~ \exists M< \infty: ||Vx|| \leq M ||x|| ~\forall x \in H\}$ the set of linear bounded operators on $H$
• $V \in B(H)$ is a partial isometry iff $V$ is a isometry wrt. $(ker V)^\perp$ i.e. $||Vx||=||x|| ~\forall x \in (kerV)^\perp$
• $V \in B(H)$ is called unitary iff $V^* V=V V^* = I$ where $V^*$ is the adjoint of $V$ and $I$ is the identity
• The dimension of a Hilbert space is defined by the cardinality of its orthonormal basis
• A bounded operator on a Hilbert space can be decomposed uniquely into its real and imaginary part, i.e. $V=A+iB=Re(V)+iIm(V)$ where $A$ and $B$ are unique and hermitian bounded operatos
• $V$ is a partial isometry iff $U^* U$ is a orthogonal projection on $(ker U)^\perp$ i.e. $(U^* U)^2 = U^* U = U U^*$
• Every bounded operator $V$ can be decomposed with a partial isometry i.e. it exists a partial isometry $P$ s.t. $V=P |V|$ where $| \cdot|$ is the absolute value and is defined by $|V|=\sqrt{V^* V}$ (This is called Polar Decomposition)
• It was shown that every bounded operator is the linear combination of at most 4 unitaries. (Not on a finite-dim. HS)

That are a lot of theorems and definitions that were done. If there is a lack of clarity then just ask.

My problem is that I can't work with the finite-dimensional Hilbert space. What can the finite-dimensional HS do that the infinite-dimensional HS can't? Okay, I have a finite orthonormal basis. But what else? I'm thankful for every help.

Marvin.

Answer 1

a) The first task here is to interpret the question appropriately. I guess it should be read as,

"Show that for every partial isometry $V\in B(H)$ on a finite-dimensional Hilbert space $H$, the restriction $V|_{(\ker V)^{\perp}}$ can be extended to a unitary on $H$."

(The only extension of $V$ to $H$ is $V$ itself, and it is certainly not true that every partial isometry on a finite-dimensional Hilbert space is unitary.)

But let's turn to the question: Choose an orthonormal basis $(e_1,\dots,e_k)$ of $(\ker V)^{\perp}$ and extend it to an orthonormal basis $(e_1,\dots,e_n)$ of $H$. Since $V|_{(\ker V)^\perp}$ is an isometry, $(f_1,\dots,f_k):=(Ve_1,\dots,Ve_k)$ is an orthonormal basis of $\operatorname{ran} V$. Extend it to an orthonormal basis $(f_1,\dots,f_n)$ of $H$. Define $U e_j:=f_j$.

Then $Ue_j=f_j=V e_j$ for $j\in\{1,\dots,k\}$, so $U$ coincides with $V$ on $(\ker V)^\perp$. Since $U$ maps on orthonormal basis to an orthonormal basis, it is unitary.

b)By the polar decomposition, every operator $T\in B(H)$ can be written as $T=V|T|$ with a partial isometry $V$ such that $\ker V=\ker |T|$. Extend $V|_{(\ker V)^\perp}$ to a unitary $U$. Then $T=U|T|$. Now $|T|$ is self-adjoint and can therefore be written as a linear combination of two unitaries (compare the proof that every bounded operator can be written as linear combination of four unitaries). Thus, $T$ can be written as linear combination of two unitaries.