# Symmetric Linear Transformations with trivial kernels

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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I don't think you really mean "a symmetric bounded linear operator from $V \times V \to \mathbb R$". An inner product on $V$ (a vector space over the reals) is a bilinear map from $V \times V$ to $\mathbb R$, not a linear one. A linear operator from $V \times V$ to $\mathbb R$ can't have the property $A(v,v) \ne 0$ for all $v \in V \backslash \{0\}$ unless $V$ is one-dimensional. –  Robert Israel Aug 30 '11 at 1:57
Robert: You're correct. I changed the statement of the question to follow this. –  Q-the-curvature Aug 30 '11 at 2:07

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 $V^*$.
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?