# Non-orthogonal extension of an inner product

Let $W$ be a real linear vector space of possibly infinite dimension and let $<\cdot, \cdot>_W$ be an inner product on $W$. Further let $V$ be a superspace of $W$, such that holds $V = W \bigoplus span\{\psi\}$.

I am calling an inner product $<\cdot, \cdot>_V$ an extension of $<\cdot, \cdot>_W$ orthogonal to $\psi$, if it satisfies

1. $<\phi, \rho>_V = <\phi, \rho>_W \quad \forall\,\phi, \rho \in W$
2. $<\phi, \psi>_V = 0 \quad \forall\,\phi \in W$.

Any element $\phi \in V$ has a unique decomposition $\phi = \phi_w + a\cdot\psi$ with $\phi_w \in W$ and $a\in\mathbb{R}$. Hence the mapping $$\alpha_\psi: V \to \mathbb{R}, \; (\phi_w+ a\cdot\psi) \mapsto a$$ is well defined, linear and suffices $\alpha_\psi(\phi)=0$ for all $\phi \in W$.

We can decompose any extension $<\cdot, \cdot>_V$ of $<\cdot, \cdot>_W$ orthogonal to $\psi$ as follows \begin{align} <\phi, \rho>_V &= <\phi_w, \rho_w>_W + \alpha_\psi(\phi)<\rho,\psi>_V+\alpha_\psi(\rho)<\phi,\psi>_V + \alpha_\psi(\phi)\alpha_\psi(\rho)<\psi,\psi>_V\\ &= <\phi_w, \rho_w>_W + \alpha_\psi(\phi)\alpha_\psi(\rho)<\psi, \psi>_V \end{align}. In other words: Any extension $<\cdot, \cdot>_V$ of $<\cdot, \cdot>_W$ orthogonal to $\psi$ is of the form $$<\phi, \rho>_V = <\phi_w, \rho_w>_W + \lambda\cdot \alpha_\psi(\phi)\alpha_\psi(\rho),$$ for some $\lambda > 0$.

My question: Can we say anything about the existence and uniqueness of extensions of the inner product $<\cdot, \cdot>_W$ to $V$ that are not orthogonal to $\psi$ (i.e. suffice 1. but not 2.)?

In finite dimension uniqueness certainly does not hold (but can we say anything about the cardinality of extensions w.r.t. the dimension?) In infinitely dimensions does existence holds? And is uniqueness possible?

Yes, you can, and it is very easy: suppose $A$ is a linear operator on $V$ that is bounded away from $0$ and $\infty$, i. e. there exist constants $0 < c < C$ such that \begin{align*} c \, \mathbb{1} \leq A \leq C \, \mathbb{1} \end{align*} holds. Note that in general $A$ mixes the subspaces $W$ and $\mathrm{span}\{ \psi \}$!
Now define a second scalar product on $V$ which is weighted by $A$, namely \begin{align*} \langle \varphi \, , \, \eta \rangle_A := \bigl \langle \varphi \, , \, A \eta \bigr \rangle_V . \end{align*} You can check that $\langle \, \cdot \, , \, \cdot \, \rangle_A$ satisfies all the axioms of a scalar product, and that the two norms induced by the scalar products $\langle \, \cdot \, , \, \cdot \, \rangle_A$ and $\langle \, \cdot \, , \, \cdot \, \rangle_V$ are equivalent. Thus, you obtain a second Hilbert space $V_A$ that agrees with $V$ as Banach spaces. That means that $A$ is still a bounded operator on $V_A$ that is bounded away from $0$ and $\infty$ (with different constants).
Then unless $A$ is block-diagonal with respect to the direct sum decomposition $V = W \oplus \mathrm{span} \{ \psi \} = V_A$ (the latter equality is the equality as Banach spaces!), the two subspaces $V_A$ (i. e. $V$ equipped with the alternative scalar product) are no longer orthogonal to each other. These arguments also work if you tack on an $N$- or infinite-dimensional subspace rather than a $1$-dimensional one.
• Thanks a lot for your answer! In order to satisfy condition 1. form above, the operator $A$ needs to leave $W$ invariant. So that the question that still remains for me is of the existence of an operator $A$ that is not block diagonal and leaves $W$ invariant... which I now realised is obviously true ... Thanks! – Paul H Aug 28 '18 at 11:41
• Is it really obvious that there exists a non-block diagonal operator on $V$ that leaves $W$ invariant and is bounded away from zero and infinity? – Paul H Aug 28 '18 at 11:49
• The way I read your definition is that you need to satisfy both conditions in order to have an orthogonal extension (you claim that being not orthogonal still means (1) has to be satisfied, but I disagree, mathematically you need both conditions!). Even if you drop only condition (1), the extension is not unique: in that case $A = \mathbb{1} \oplus B$ where $B$ is a linear operator on the subspace you tack on that is bounded away from $0$ and $\infty$. If you add a one-dimensional subspace, $B = \lambda$ is just a positive real number $\lambda > 0$. So your extension is not unique. – Max Lein Aug 29 '18 at 7:52
• You asked whether an extension exists (yes) and is unique (no). This is even the case if you require (1) and (2) to hold, e. g. $A = \mathbb{1} \oplus \lambda$ where $\lambda > 0$. This is still correct if you drop condition (2): condition (1) does not imply that the offdiagonal part of $A$ vanishes. Take the case where $W = \mathbb{C}$ so that $A$ is a $2 \times 2$ matrix. (1) is then satisfied iff $A_{11} = (1,0)^T \cdot A (1,0)^T$ equals $1$. Thus, (1) does not impose any restrictions on the other entries of $A$, including the offdiagonal ones. Also here, the extension is not unique. – Max Lein Aug 30 '18 at 1:09