A canonical form edit

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, ${\displaystyle M\in \mathbb {C} ^{n\times m}}$  has rank ${\displaystyle 1}$  if and only if ${\displaystyle M}$  is of the form

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

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

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

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

Therefore, by induction, an operator ${\displaystyle T}$  of finite rank ${\displaystyle n}$  takes the form

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

where ${\displaystyle \{u_{i}\}}$  and ${\displaystyle \{v_{i}\}}$  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 ${\displaystyle n}$  is now countably infinite and the sequence of positive numbers ${\displaystyle \{\alpha _{i}\}}$  accumulate only at ${\displaystyle 0}$ , ${\displaystyle T}$  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.^[2]

Algebraic property edit

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

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

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

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

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

A finite-rank operator ${\displaystyle T:U\to V}$  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

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

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

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