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, ${\displaystyle M\in \mathbb {C} ^{n\times m}}$ 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.

If the series ${\displaystyle \sum _{i}\alpha _{i}}$ is convergent, ${\displaystyle T}$ is a trace class operator.

Algebraic property

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.