How to Prove That a Linear Operator Is Surjective

To prove that a linear operator $T$ is surjective, you must show that every element of its codomain is the image of some element of its domain. The operator and the two spaces are not specified here, so there is no particular surjectivity claim to prove. The methods below give the standard approaches and indicate what information a concrete proof needs.

The direct test

If $T:X\to Y$, surjectivity means

$$T(X)=Y.$$

The most direct proof begins with an arbitrary $y\in Y$ and constructs, or proves the existence of, an $x\in X$ such that $Tx=y$. The choice of $x$ may depend on $y$; it need not be unique. Uniqueness would be a statement about injectivity, not surjectivity.

A useful sufficient condition is a right inverse: if you can define a map $S:Y\to X$ satisfying $T(Sy)=y$ for every $y\in Y$, then $T$ is onto. The right inverse does not have to be linear for this implication, though in functional analysis one often seeks a bounded linear right inverse.

Worked example: the backward shift on $\ell^2$

Define $B:\ell^2\to\ell^2$ by

$$B(x_1,x_2,x_3,\ldots)=(x_2,x_3,x_4,\ldots).$$

Take an arbitrary $y=(y_1,y_2,\ldots)\in\ell^2$. Set

$$x=(0,y_1,y_2,\ldots).$$

Because $y$ is square-summable,

$$\sum_{n=1}^{\infty}|x_n|^2 =\sum_{n=1}^{\infty}|y_n|^2<\infty,$$

so $x\in\ell^2$. Applying $B$ gives $Bx=y$. Since this construction works for every $y\in\ell^2$, $B$ is surjective. In fact, the map $S(y)=(0,y_1,y_2,\ldots)$ is a bounded linear right inverse of $B$.

This example also shows why checking only whether an operator preserves norms or is injective is not enough. The right shift $R(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots)$ is an isometry, but it is not onto: every vector in its range has first coordinate zero.

Using the adjoint for Banach spaces

For a bounded linear operator $T:X\to Y$ between Banach spaces, its adjoint is $T^*:Y^*\to X^*$, defined by $T^*y^*=y^*\circ T$. A useful criterion is

$$T\text{ is surjective} \quad\Longleftrightarrow\quad \text{there is }c>0\text{ such that }\|T^*y^*\|\ge c\|y^*\|\text{ for all }y^*\in Y^*.$$

The lower bound is stronger than merely showing that $T^*$ is injective. To see why surjectivity implies the bound, the open mapping theorem gives a constant $C$ such that each $y\in Y$ has a preimage $x$ with $Tx=y$ and $\|x\|\le C\|y\|$. Hence, for $y^*\in Y^*$,

$$|y^*(y)|=|(T^*y^*)(x)|\le C\|T^*y^*\|\|y\|,$$

which yields $\|y^*\|\le C\|T^*y^*\|$.

Conversely, such a lower bound makes $T^*$ injective and its range closed. Injectivity of $T^*$ implies that the range of $T$ is dense, by the Hahn–Banach theorem. The closed-range theorem then implies that the range of $T$ is closed. A dense, closed range is all of $Y$, so $T$ is surjective.

Hilbert-space version

If $T:H_1\to H_2$ is bounded between Hilbert spaces, the Banach-space criterion can be written using the Hilbert-space adjoint. Surjectivity is equivalent to the existence of $c>0$ such that

$$\|T^*y\|\ge c\|y\|\qquad (y\in H_2).$$

Equivalently, there is a constant $a>0$ for which

$$\langle TT^*y,y\rangle=\|T^*y\|^2\ge a\|y\|^2\qquad (y\in H_2).$$

This is often useful when the adjoint or the positive operator $TT^*$ is easier to analyze than the equation $Tx=y$ itself. Be careful: $\ker T^*=\{0\}$ alone guarantees only that the range of $T$ is dense, not necessarily that it is all of $H_2$.

Common mistakes

• Proving injectivity instead. Showing $Tx=0$ only when $x=0$ says that $T$ is one-to-one. Surjectivity requires solving $Tx=y$ for every target $y$.
• Using the wrong codomain. An operator may be onto one space but not onto a larger space. State the codomain explicitly.
• Showing only dense range. Approximation of every target by elements of the range does not prove exact surjectivity unless you also establish that the range is closed.
• Applying a theorem without its hypotheses. The adjoint criterion above assumes a bounded operator between Banach spaces. For unbounded operators, domains and closedness conditions must be addressed separately.

A concrete proof therefore needs the formula for $T$, its domain and codomain, and any relevant assumptions such as boundedness or closedness. With those specified, try first to solve $Tx=y$ for an arbitrary target; use an adjoint criterion when a direct construction is difficult.