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.

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


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

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


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

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


 * $$T h = \sum _{i = 1}  ^n \alpha_i \langle h, v_i\rangle u_i \quad \mbox{for all} \quad h \in H ,$$

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 $ \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


 * $$S_{h,k} = S_{g,k} T S_{h,f}, \,$$

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


 * $$T h = \sum _{i = 1}  ^n \langle u_i, h\rangle v_i \quad \mbox{for all} \quad h \in U ,$$

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.