[Math] Inner product and linear transformation

inner-productslinear algebra

Let V be an inner product space over $\mathbb{R}$ with inner product ⟨ , ⟩. Let $L:V\rightarrow\mathbb{R}$ be a linear transformation.

Show that there is a $\vec{u}\in{V}$ such that $L(\vec{x}) = ⟨\vec{x},\vec{u}⟩$ $ \forall\vec{x}\in{V}$

I'm not sure how inner product spaces relate to linear transformations so I'm quite confused with this question.

Best Answer

One way to avoid using a basis is to use a little calculus. Think of $L$ as a scalar field on $V$.

Because $L$ is linear, the gradient $\nabla L$ is a constant vector field.

Then, note that $\nabla (x \cdot u) = u$ for any vector $u$.

Let $M(x) = x \cdot \nabla L$. Then $M$ has the same gradient as $L$, and $M$ is linear. If two linear scalar fields have the same gradient everywhere, they must be equal: they both agree at the origin, after all.

Therefore, any linear scalar field $L$ can be characterized by a vector $u = \nabla L$ and written $L(x) = x \cdot u$.

To use this as a solution, I would consider a few points that must be proven:

  • Prove that a linear scalar field has constant gradient (as an idea, use Taylor expansion).
  • Prove that $\nabla (x \cdot u) = u$ for any constant $u$.
  • Prove that two linear scalar fields are equal if their gradients are equal (again, Taylor expansion is probably your friend here).

The argument may feel a little circular; remember that you're only having to argue that $u$ exists. You don't need to be able to compute it given an arbitrary $L$. With $L$ being a linear scalar field, the existence of such a $u$ is well-founded, so it is not circular to rewrite $L$ in terms of it.

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