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Let $\mathcal{H}$ be a complex, $n$-dimensional Hilbert space with two inner products $\langle \cdot, \cdot \rangle_1$, $\langle \cdot, \cdot \rangle_2$.

Show that there exists a basis $ X = x_1, \dots, x_n$ of $\mathcal{H}$ such that for $i \neq j$, $0 = \langle x_i, x_j \rangle_1 = \langle x_i, x_j \rangle_2$. That is, we want to find a basis that is orthogonal with respect to both inner products.

Using Gram Schmidt, I know that we can find an orthonormal basis for each inner product separately. The question is how we then construct a basis that is orthogonal with respect to both $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$ at the same time.

I have thought about trying to use an induction argument on $n$. The $n=1$ case is trivial. For $n > 1$, we can write $\mathcal{H}$ as a direct sum $\mathcal{H} = V \bigoplus \mathbb{C} x $, where $V$ is an $n-1$ dimensional Hilbert space. We can then find a basis $Y$ of $V$ that is orthogonal with respect to both $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$. The question would then be how we make $x$ orthogonal to the rest of the basis elements in $Y$.

Hints or solutions are greatly appreciated.

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  • $\begingroup$ I do not fully understand the question. Since <x_i,x_j>_h=0 (h=0,1)for i≠j and assuming <x_i,x_i>_h=1 then X is an orthonormal set relative to <,>_1 and also to <,>_2. But two orthogonal bases are unitarily related so <x_i,x_j>_2=<Ux_i,Ux_j>_1=<x_i,x_j>_1 since U is unitary. Thus if X is orthonormal the conditions are met. Next you mention a third basis but why? $\endgroup$
    – Urgje
    Apr 28, 2014 at 12:00
  • $\begingroup$ @Urgje - Thank you for the comment. I am looking to find a basis that is orthogonal with respect to both inner products. That is, I need to construct the basis $X$ that I describe above. I do not know a priori that I have such a basis. I know that I can find a basis (say $Y_1$) that is orthonormal with respect to $\langle \cdot, \cdot \rangle_1 $. And I can also find a basis, $Y_2$, that is orthonormal with respect to $\langle \cdot, \cdot \rangle_2$. But how to find one that is orthonormal with respect to both inner products? $\endgroup$
    – JZS
    Apr 28, 2014 at 12:18

2 Answers 2

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Let us write ${\cal H}_i$ to denote ${\cal H}$ equipped with $\langle\cdot,\cdot\rangle_i$, for $i=1,2$. Consider an orthonormal basis $(e_1,\ldots,e_n)$ in ${\cal H}_1$, and consider the operator $$ G:{\cal H}\longrightarrow{\cal H},G(x)=\sum_{i=1}^n \langle e_i,x\rangle_2 e_i $$ we have $$ \langle y, G(x)\rangle_1 =\sum_{i=1}^n \langle e_i,x\rangle_2 \langle y, e_i\rangle_1 =\sum_{i=1}^n \overline{\langle e_i,y\rangle_1}\langle e_i,x\rangle_2 = \left\langle \sum_{i=1}^n\langle e_i,y\rangle_1e_i,x\right\rangle_2 =\left\langle y,x\right\rangle_2 $$ Thus $$ \langle y, G(x)\rangle_1=\left\langle y,x\right\rangle_2 =\overline{\left\langle x,y\right\rangle_2}=\overline{\left\langle x,G(y)\right\rangle_1}= \langle G(x),y\rangle_1 $$ So $G$ is hermitian, and since $\langle G(x),x\rangle_1=||x||_2^2>0$ for nonzero $x$, we conclude that $G$ is definite positive. Thus, there is an orthonormal basis $(f_1,\ldots,f_n)$ of ${\cal H}_1$ that diagonalizes $G$ that is $G(f_k)=\lambda_k f_k$, with $\lambda_k>0$, for $k=1,\ldots,n$. For $k\ne m$, we have by construction,$\langle f_k, f_m\rangle_1=0$ , and $$ \langle f_k, f_m\rangle_2= \langle f_k, G(f_m)\rangle_1=\lambda_m\langle f_k, f_m\rangle_1 =0$$ Thus $(f_1,\ldots,f_n)$ does the job.$\qquad\square$

${\bf Remark.}$ $(f_1,\ldots,f_n)$ is orthonormal in ${\cal H}_1$ and orthogonal in ${\cal H}_2$.

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It is perhaps worth clarifying that this question is equivalent to showing that, if $X$ is a complex inner product space, then a self-adjoint linear map is diagonalizable. (This is a standard result, proved by induction on dimension, relying on the fact that the self-adjoint property of $S$ guarantees that the eigenspaces of $S$ are orthogonal).

The reason is the following: If $\langle -,-\rangle$ is an inner product on $X$, and $S\colon X\to X$ is a self-adjoint linear map, so that $\langle S(x_1),x_2\rangle = \langle x_1,S(x_2)\rangle$, then if we set $$ b_S(x_1,x_2) := \langle S(x_1),x_2\rangle, $$ then clearly $b_S$ is linear in $x_1$ an skew-linear in $x_2$ (assuming this is true of our inner product) and $$\ b_S(x_2,x_1) = \langle S(x_2),x_1\rangle = \langle x_2,S(x_1)\rangle =\overline{\langle S(x_1),x_2\rangle} = \overline{b_S(x_1,x_2)} $$ Thus $b_S$ is a Hermitian symmetric bilinear form on $X$, and the map $S \mapsto b_S$ is a linear map between two vector spaces of the same dimension ($\frac{1}{2}n(n-1)$). Now $b_S(x_1,S(x_1)) = \langle S(x_1).S(x_1)\rangle = \|S(x_1)\|^2$, and thus $b_S(x_1,S(x_1))=0$ implies $S(x_1)=0$. Thus if $b_S$ is identically zero, we must have $S(x_1)=0$ for all $x_1\in X$, that is, $S=0$. Thus $S\mapsto b_S$ is injective, and hence an isomorphism. Now the spectral theorem shows that $S$ has an orthonormal basis of eigenvectors, and this basis will be orthogonal for the form $b_S$.

Note that if we fix an orthonormal basis of $X$ and use it to identify $\text{End}(X)$ with $\text{Mat}_n(\mathbb C)$ then the map $S\mapsto b_S$ just corresponds to taking the matrix $A$ of the linear map $A$ and viewing it as the matrix for the bilinear form given by $(x,y) = x^\intercal A y$.

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