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Let $V=\mathbb{R}[x]$ be the vector space of all polynomials with coefficients in $\mathbb{R}$ with the inner product defined by $\langle f,g\rangle=\int_{0}^{1}f(t)g(t)dt$.

Let $W$ the subspace of $\mathcal{L}(V)$ generated by those linear operators which have no adjoint. What is the dimension of $W$?

I know that the differentiation operator belongs to $W$, which means that $W$ is a nonempty set.

I would appreciate you help.

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How are you considering a space generated by operators on $V$ as a subspace of $V$? –  joriki Feb 20 '12 at 14:42
    
@joriki: I made the corrections. Thanks for point it out. –  spohreis Feb 20 '12 at 14:54
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up vote 7 down vote accepted

In fact $W = \mathcal{L}(V)$, so in particular $W$ is infinite dimensional.

Clearly $W$ contains all operators that have no adjoint. Let $S$ be any operator without an adjoint (you know there is at least one) and $T$ any operator with an adjoint. Then $S+T$ has no adjoint (if it did, we could write $S^* = (S+T)^* - T^*$). So $S+T \in W$. Since $W$ is a subspace, we have $T = (S+T)-S \in W$. Thus $W$ also contains all operators that do have an adjoint.

Edit: The point is that the set $A$ of all operators that do have an adjoint is a subspace of $\mathcal{L}(V)$, and $W = A^c$. The above argument shows that the span of the complement of a proper subspace is always the entire space.

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