Why Does an Isometric Linear Operator Satisfy $T^*T=I$?

Let $H$ be a Hilbert space and let $T:H\to H$ be linear and isometric. Here, “isometric” means that $T$ preserves distances—or, equivalently for a linear map, preserves norms: $\|Tx\|=\|x\|$ for every $x\in H$. The goal is to show that applying $T$ and then its adjoint $T^*$ leaves every vector unchanged.

The key link is that a linear isometry preserves not only norms but also inner products. Once that is established, the defining identity for the adjoint gives the operator equality.

From norm preservation to inner-product preservation

A Hilbert-space inner product can be recovered from norms using the polarization identity. In a real Hilbert space, for example,

$$\langle u,v\rangle=\frac{1}{4}\bigl(\|u+v\|^2-\|u-v\|^2\bigr).$$

Since $T$ is linear, $T(x+y)=Tx+Ty$ and $T(x-y)=Tx-Ty$. Since it preserves norms, applying the real polarization identity gives

$$\begin{aligned} \langle Tx,Ty\rangle &=\frac{1}{4}\bigl(\|Tx+Ty\|^2-\|Tx-Ty\|^2\bigr)\\ &=\frac{1}{4}\bigl(\|T(x+y)\|^2-\|T(x-y)\|^2\bigr)\\ &=\frac{1}{4}\bigl(\|x+y\|^2-\|x-y\|^2\bigr)\\ &=\langle x,y\rangle. \end{aligned}$$

The complex polarization identity also recovers the inner product from norms, using the norms of $u+v$, $u-v$, $u+iv$, and $u-iv$. The same argument therefore shows that a complex linear isometry preserves the complex inner product as well.

Use the adjoint to identify the operator

Use the convention that the inner product is linear in its first variable, so the adjoint satisfies $\langle Tx,z\rangle=\langle x,T^*z\rangle$. Set $z=Ty$. Then, for all $x,y\in H$,

$$\langle x,T^*Ty\rangle=\langle Tx,Ty\rangle=\langle x,y\rangle.$$

Subtracting the two sides gives $\langle x,(T^*T-I)y\rangle=0$ for every $x,y\in H$. For any fixed $y$, choose $x=(T^*T-I)y$. Its inner product with itself is zero, so $(T^*T-I)y=0$. This holds for every $y$, and consequently

$$T^*T=I.$$

Equivalently, the inner products of $T^*Ty$ and $y$ with every vector $x$ agree. In a Hilbert space, that determines the vector uniquely.

A direct check on a sequence space

Consider the right shift $S$ on $\ell^2(\mathbb{N})$:

$$S(x_1,x_2,x_3,\ldots)=(0,x_1,x_2,\ldots).$$

It preserves norms because

$$\|Sx\|^2=0+|x_1|^2+|x_2|^2+\cdots=\|x\|^2.$$

Its adjoint is the left shift $S^*(x_1,x_2,x_3,\ldots)=(x_2,x_3,\ldots)$. Applying the right shift and then the left shift gives

$$S^*S(x_1,x_2,x_3,\ldots)=(x_1,x_2,x_3,\ldots),$$

so $S^*S=I$. This example also illustrates why the theorem does not require $T$ to be onto: the right shift never produces a sequence with a nonzero first coordinate, so it is not surjective.

What the result does—and does not—say

The equation $T^*T=I$ says that $T$ is an isometry: it has a left inverse, namely $T^*$. It does not in general say that $T$ is invertible or that $TT^*=I$. For the right shift above, $SS^*$ sends $(x_1,x_2,\ldots)$ to $(0,x_2,x_3,\ldots)$, which is not the identity. If $T$ is also surjective, then it is unitary and both $T^*T=I$ and $TT^*=I$ hold.

Common mistakes

• Assuming surjectivity is needed. It is not needed to prove $T^*T=I$; norm preservation and linearity suffice.
• Claiming both products equal the identity. The isometry condition gives $T^*T=I$. The additional equation $TT^*=I$ requires surjectivity.
• Using norm preservation without linearity. The polarization argument relies on $T(x+y)=Tx+Ty$ (and, in the complex case, complex linearity). A distance-preserving map that is not linear need not satisfy the operator identity in this form.
• Treating the adjoint as an inverse in every case. $T^*$ is a left inverse for an isometry, but it is an inverse on all of $H$ only when $T$ is onto.