# Lecture 5: Square Root, Ran and Ker, Polar Decomposition ###### tags: `224a 2020` $\renewcommand{\ker}{\mathrm{Ker}}$ $\newcommand{\ran}{\mathrm{Ran}}$ ## Square Root We use the notion of norm convergence to recover a familiar fact from finite dimensional linear algebra: a positive operator has a square root. The proof constructs this square root using an absolutely convergent power series. This is a common technique in the infinite dimensional setting; at a high level, the techniques of analytic function theory are the ones that generalize best beyond finite dimensions. **Theorem.** If $A\ge 0$ then there is a unique $B\in L(H)$ with $B^2=A$, $B\ge 0$, and $AB=BA$. See Reed and Simon Theorem VI.9 for a proof. ## Range and Kernel Let $\ker(A)$ and $\ran(A)$ denote the kernel and range of $A\in L(H)$. These are related as: $$ \ran(A^*)^\perp = \ker(A),\qquad \overline{\ran(A^*)}=\ker(A)^\perp,$$ where the second equality can be seen by taking the orthogonal complement of the first one and noting that $(M^\perp)^\perp =\overline{M}$ for any subspace $M$ (homework). While $\ker(A)$ is always a closed subspace, this is not true of $\ran(A)$, and this distinction is important for instance because the projection theorem, orthogonal decomposition, and existence of orthonormal bases only work for *closed* subspaces. **Example.** To see all of this in action, consider the Volterra operator $$(Vf)(x) = \int_0^x f(y)dy$$ on $L^2(0,1)$ from the last lecture, which we showed was bounded. Its adjoint can be calculated from the definition by observing \begin{align*} (g,Vf) &= \int_0^1 \overline{g(x)} (\int_0^1 \{y\le x\} f(y)dy) dx\\ &= \int_0^1 (\int_0^1 \{y\le x\} \overline{g(x)} dx) f(y)dy\quad\textrm{by Fubini} \\&=(V^*g,f)\end{align*} by definition, so we must have $$ (V^*f)(x)=\int_x^1 f(y)dy.$$ It is now evident that $V$ is not self-adjoint, since for any strictly positive function $f$, $Vf$ is strictly increasing whereas $V^*f$ is strictly decreasing. To compute $\ker(V)$, observe that by the fundamental theorem of calculus (or more precisely its Lebesgue version, https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem), $(Vf)(x)$ is an almost everywhere differentiable function of $f$, with $$\frac{d}{dx}(Vf)(x)=f(x).$$ Thus if $Vf=0$ a.e. we must have $f=0$ a.e., so $\ker(V)=\{0\}$. The above calculation also reveals that $\ran(V)\neq L^2(0,1)$ (since there are continuous functions in $L^2$ which are not differentiable a.e.), yielding an example of a range which is not closed. However, we do know that $$\overline{\ran(V)} = \ker(V^*)^\perp = L^2(0,1)$$ since $\ker(V^*)=0$ as well. **Remark 1.** An operator of type $$(T_Kf)(x):=\int_0^1 K(x,y)f(y)dy$$ is called an *integral kernel* operator, and the Fubini argument above shows that whenever such an operator is bounded we have $$ (T_K)^* = T_{K^*},$$ where $K^*(x,y)=\overline{K(y,x)}$. ## Projection and Unitary Operators A *projection* is an operator which satisfies $P^2=P$, and an *orthogonal projection* further satisfies $P=P^*$. It is easy to check that $\ran(P)$ is closed for a projection, and there is a 1-1 correspondence between orthogonal projections and closed subspaces of $H$ by $H=\ran(P)\oplus \ker(P)$. A *partial isometry* is an operator $U:H\rightarrow H$ satisfying $$ \|Ux\|=\|x\|\quad\forall x\in \ker(A)^\perp,$$ which can be seen to be equivalent to $$ (Ux,Uy)=(x,y)\quad\forall x,y\in\ker(A)^\perp$$ by the polarization identity. Such an operator is called *unitary* if $\ker(U)=\{0\}$, in which case $U$ is a bijection. It is easy to verify that $U^*U=P_{\ker(U)^\perp}$ and $UU^* = P_{\ran(U)}$. The first fact implies that the range of a unitary is always closed since a sequence $Ux_n$ is Cauchy if and only if $x_n$ is Cauchy. Intuitively, we think of unitaries as "rotations" and of partial isometries as "geometry preserving" embeddings of one subspace into another. **Example.** The right shift operator on $\ell^2(\mathbb{N})$ is a partial isometry with $\ker=\{e_1\}$ whereas the right shift operator on $\ell^2(\mathbb{Z})$ is a unitary. ## The Polar Decomposition The polar decomposition allows one to express any bounded operator as a product of a partial isometry and a positive operator. Let $|A|=\sqrt{A^*A}$ denote the absolute value of $A$. **Theorem 1.** If $A\in L(H)$ then there is a unique partial isometry $U$ such that $$A=U|A|$$ with $\ker(U)=\ker(A)$. The proof is given in Reed and Simon Theorem VI.10. Suppose we could prove that every self-adjoint operator (such as $|A|$) can be "diagonalized": $|A|=VDV^*$ for some "diagonal" $D$ (we will prove this soon, though the definition of diagonal is different from the finite case). Then the polar decomposition gives: $$ A = U(VDV^*)=WDV^*$$ for some partial isometries $W,V$, which is a "singular value decomposition" for elements of $L(H)$.