Linear Transformation is Isomorphism iff Inverse Equals Adjoint

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Theorem

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.

Let $U : \HH \to \KK$ be a bounded linear transformation.


Then the following are equivalent:

$(1): \quad U$ is an isomorphism
$(2): \quad U$ is invertible and $U^{-1} = U^*$, where $U^*$ denotes the adjoint of $U$.


Proof

$(1)$ implies $(2)$

Suppose that:

$U$ is an isomorphism.

That is:

$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$

for each $g, h \in \HH$.


Let $g, h \in \HH$.

From the definition of the adjoint, we have:

$\innerprod g h_\HH = \innerprod g {U^* U h}_\KK$

So, from Inner Product is Sesquilinear, we have:

$\innerprod g {h - U^* U h}_\HH = 0$

Setting $g = h - U^* U h$, we have:

$\innerprod {h - U^* U h} {h - U^* U h}_\HH = 0$

So from the positiveness of the inner product, we have:

$h - U^* U h = 0$

so:

$h = U^* U h$

for all $h \in \HH$.

So:

$U^* U = I_\HH$

where $I_\HH$ is the identity map on $\HH$.


From Hilbert Space Isomorphism is Bijection, we have that:

$U$ is a bijection.

So:

$U$ is invertible.

Let $U^{-1} : \KK \to \HH$ be the inverse of $U$.

Then, we have:

\(\ds U^{-1}\) \(=\) \(\ds I_\HH U^{-1}\)
\(\ds \) \(=\) \(\ds \paren {U^* U} U^{-1}\)
\(\ds \) \(=\) \(\ds U^* \paren {U U^{-1} }\)
\(\ds \) \(=\) \(\ds U^*\)

So:

$U$ is invertible with $U^{-1} = U^*$.

$\Box$


$(2)$ implies $(1)$

Suppose that:

$U$ is invertible with $U^{-1} = U^*$.

By the definition of an isomorphism, we aim to show that:

$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$

for each $g, h \in \HH$.

Let $g, h \in \HH$.

Then, we have:

\(\ds \innerprod {U g} {U h}_\KK\) \(=\) \(\ds \innerprod g {U^* \paren {U h} }_\HH\) Definition of Adjoint Linear Transformation
\(\ds \) \(=\) \(\ds \innerprod g {U^{-1} \paren {U h} }_\HH\) since $U^* = U^{-1}$
\(\ds \) \(=\) \(\ds \innerprod g {\paren {U^{-1} U} h}_\HH\)
\(\ds \) \(=\) \(\ds \innerprod g h_\HH\)

So:

$U$ is an isomorphism.

$\blacksquare$


Sources

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