Does an Invertible Adjoint Imply an Invertible Operator?

Suppose $T$ is a bounded linear operator on a Hilbert space $H$. The question is whether invertibility of its adjoint $T^*$ forces $T$ itself to be invertible.

Yes. In fact,

$$T \text{ is invertible} \quad\Longleftrightarrow\quad T^* \text{ is invertible},$$

and the inverses satisfy

$$T^{-1} = (T^*)^{-1*} = \bigl((T^*)^{-1}\bigr)^*.$$

Here, “invertible” means that the operator has a bounded two-sided inverse.

Direct proof on a Hilbert space

Assume that $T^*$ is invertible and write

$$B=(T^*)^{-1}.$$

Since $B$ is the inverse of $T^*$, it satisfies both inverse identities

$$T^*B=I \qquad\text{and}\qquad BT^*=I,$$

where $I$ is the identity operator on $H$.

Take adjoints. Using $(AB)^*=B^*A^*$ and $I^*=I$, the first identity gives

$$(T^*B)^*=B^*T=I.$$

The second gives

$$(BT^*)^*=TB^*=I.$$

Thus $B^*$ is both a left and a right inverse of $T$:

$$B^*T=I \qquad\text{and}\qquad TB^*=I.$$

Therefore $T$ is invertible, and

$$T^{-1}=B^*=\bigl((T^*)^{-1}\bigr)^*.$$

This argument works in finite- and infinite-dimensional Hilbert spaces. No finite-dimensional determinant argument is needed.

The converse

The converse follows by exactly the same reasoning. If $T$ is invertible, then

$$T^{-1}T=I \qquad\text{and}\qquad TT^{-1}=I.$$

Taking adjoints gives

$$T^*(T^{-1})^*=I \qquad\text{and}\qquad (T^{-1})^*T^*=I.$$

Hence $T^*$ is invertible and

$$(T^*)^{-1}=(T^{-1})^*.$$

Combining both directions produces the useful equivalence

$$T\text{ invertible}\quad\Longleftrightarrow\quad T^*\text{ invertible}.$$

Worked matrix example

Consider the real matrix

$$A=\begin{pmatrix} 1&2\\ 0&3 \end{pmatrix}.$$

For a real matrix, the adjoint is the transpose, so

$$A^*=A^{\mathsf T}=\begin{pmatrix} 1&0\\ 2&3 \end{pmatrix}.$$

The inverse of $A$ is

$$A^{-1}=\begin{pmatrix} 1&-\frac23\\ 0&\frac13 \end{pmatrix}.$$

The theorem predicts that

$$(A^*)^{-1}=(A^{-1})^*.$$

Indeed,

$$(A^*)^{-1}=\begin{pmatrix} 1&0\\ -\frac23&\frac13 \end{pmatrix} =(A^{-1})^{\mathsf T}.$$

For complex matrices, replace the transpose by the conjugate transpose: $A^*=\overline{A}^{\mathsf T}$.

Banach-space version

The same invertibility equivalence is also true for a bounded operator $T:X\to Y$ between Banach spaces, where

$$T^*:Y^*\to X^*, \qquad (T^*g)(x)=g(Tx).$$

The simple adjoint-taking proof above needs care because $(T^*)^*$ acts between bidual spaces, not necessarily directly between $Y$ and $X$. Nevertheless, the result remains true.

Assume $T^*$ is invertible. Let $C=\|(T^*)^{-1}\|$. For any $x\in X$ and any $f\in X^*$, choose $g\in Y^*$ such that $T^*g=f$. Then

$$\|g\|\le C\|f\|,$$

and therefore

$$|f(x)|=|g(Tx)|\le \|g\|\,\|Tx\|\le C\|f\|\,\|Tx\|.$$

Taking the supremum over all $f$ with $\|f\|\le 1$ gives

$$\|x\|\le C\|Tx\|.$$

Thus $T$ is bounded below. It is consequently injective and has closed range.

Also, because $T^*$ is injective, the range of $T$ is dense in $Y$: if a nonzero functional in $Y^*$ vanished on the range of $T$, it would belong to the kernel of $T^*$. The range is both dense and closed, so it equals $Y$. Hence $T$ is bijective, and the bounded inverse theorem shows that $T^{-1}$ is bounded.

Common mistakes

Confusing an adjoint with an inverse

The adjoint $T^*$ is generally not the inverse of $T$. For example, a matrix can have an invertible transpose without the transpose equaling the matrix inverse. The correct relationship is

$$(T^*)^{-1}=(T^{-1})^*.$$

Only special operators, such as unitary operators, satisfy $T^{-1}=T^*$.

Checking only one inverse identity

For operators on an infinite-dimensional space, having a left inverse or a right inverse alone does not automatically mean that the operator is invertible. The proof above uses both identities for $(T^*)^{-1}$, which yield both identities for $T$.

Ignoring boundedness

The statement is standard for bounded operators on Hilbert or Banach spaces. Unbounded operators require specifying domains carefully; taking adjoints and manipulating inverse identities is then subject to additional domain conditions. The bounded-operator theorem should not be transferred to unbounded operators without those hypotheses.