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?
Symmetric Linear Transformations with trivial kernels
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0Robert: You're correct. I changed the statement of the question to follow this. – 2011-08-30
2 Answers
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) =
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?
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1Gerr$y$: I meant for V to be infinite-dimensional. It's my mistake that I used the word "matrix" (this is a bad habit I have from working with compact operators on Hilbert spaces). – 2011-08-30