Let $V$ be a vector space. Let $A$ be a symmetric bounded multi-linear operator from $V \times V \rightarrow \Bbb{R}$. Suppose that $A(v,v) \neq 0$ for all $v \in V \setminus \{0\}$. This let us produce an inner-product $(\cdot,A \cdot)$ on $V$, and hence we can view $V$ as an inner-product space. Does it follow that the representation of $A$ as an linear transformation from $V$ to $V$ is onto?
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If $V$ is a real Hilbert space, the bounded symmetric bilinear form $A$ does correspond to a bounded self-adjoint linear operator $T$ by $A(x,y) = <x, Ty>$, and your condition $A(v,v) \ne 0$ for $v \in V \backslash \{0\}$ says $T$ is injective. The range of $T$ is dense in $V$, but might not be closed, so it's not necessarily surjective. For example, the operator on $L^2[0,1]$ given by $Tf(x) = x f(x)$ is self-adjoint and injective but not surjective, e.g. it doesn't contain the constant function $1$. In fact, a self-adjoint operator is surjective if and only if its spectrum is bounded away from 0. If $V$ is not a Hilbert space, all you get is a bounded operator from $V$ into its dual $V^*$. |
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I'll assume $V$ is finite-dimensional, as other wise I don't know what to make of the matrix for $A$. If a linear operator on a finite-dimensional vector space is not onto, then it has a non-trivial kernel. Doesn't this answer the question? |
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