Finite-rank operator

In functional analysis, a branch of mathematics, a finite-rank operator is a bounded linear operator between Banach spaces whose range is finite-dimensional.

Finite-rank operators on a Hilbert space

A canonical form

Finite-rank operators are matrices (of finite size) transplanted to the infinite dimensional setting. As such, these operators may be described via linear algebra techniques.

From linear algebra, we know that a rectangular matrix, with complex entries, $M\in \mathbb {C} ^{n\times m}$ has rank $1$ if and only if $M$ is of the form

$M=\alpha \cdot uv^{*},\quad {\mbox{where}}\quad \|u\|=\|v\|=1\quad {\mbox{and}}\quad \alpha \geq 0.$

Exactly the same argument shows that an operator $T$ on a Hilbert space $H$ is of rank $1$ if and only if

$Th=\alpha \langle h,v\rangle u\quad {\mbox{for all}}\quad h\in H,$

where the conditions on $\alpha ,u,v$ are the same as in the finite dimensional case.

Therefore, by induction, an operator $T$ of finite rank $n$ takes the form

$Th=\sum _{i=1}^{n}\alpha _{i}\langle h,v_{i}\rangle u_{i}\quad {\mbox{for all}}\quad h\in H,$

where $\{u_{i}\}$ and $\{v_{i}\}$ are orthonormal bases. Notice this is essentially a restatement of singular value decomposition. This can be said to be a canonical form of finite-rank operators.

Generalizing slightly, if $n$ is now countably infinite and the sequence of positive numbers $\{\alpha _{i}\}$ accumulate only at $0$, $T$ is then a compact operator, and one has the canonical form for compact operators.

Compact operators are trace class only if the series ${\textstyle \sum _{i}\alpha _{i}}$ is convergent; a property that automatically holds for all finite-rank operators.

Algebraic property

The family of finite-rank operators $F(H)$ on a Hilbert space $H$ form a two-sided *-ideal in $L(H)$, the algebra of bounded operators on $H$. In fact it is the minimal element among such ideals, that is, any two-sided *-ideal $I$ in $L(H)$ must contain the finite-rank operators. This is not hard to prove. Take a non-zero operator $T\in I$, then $Tf=g$ for some $f,g\neq 0$. It suffices to have that for any $h,k\in H$, the rank-1 operator $S_{h,k}$ that maps $h$ to $k$ lies in $I$. Define $S_{h,f}$ to be the rank-1 operator that maps $h$ to $f$, and $S_{g,k}$ analogously. Then

$S_{h,k}=S_{g,k}TS_{h,f},\,$

which means $S_{h,k}$ is in $I$ and this verifies the claim.

Some examples of two-sided *-ideals in $L(H)$ are the trace-class, Hilbert–Schmidt operators, and compact operators. $F(H)$ is dense in all three of these ideals, in their respective norms.

Since any two-sided ideal in $L(H)$ must contain $F(H)$, the algebra $L(H)$ is simple if and only if it is finite dimensional.

Finite-rank operators on a Banach space

A finite-rank operator $T:U\to V$ between Banach spaces is a bounded operator such that its range is finite dimensional. Just as in the Hilbert space case, it can be written in the form

$Th=\sum _{i=1}^{n}\langle u_{i},h\rangle v_{i}\quad {\mbox{for all}}\quad h\in U,$

where now $v_{i}\in V$, and $u_{i}\in U'$ are bounded linear functionals on the space $U$.

A bounded linear functional is a particular case of a finite-rank operator, namely of rank one.