📐 Natural Isomorphism to Double Dual

The map $\\phi: V \\to V^{**}$ defined by $\\phi(v)(f) = f(v)$ is a natural isomorphism for finite-dimensional $V$.

Proof: **Linearity:** For $u, v \\in V$ and $f \\in V^*$:

$$\\phi(u + v)(f) = f(u + v) = f(u) + f(v) = \\phi(u)(f) + \\phi(v)(f)$$

**Injectivity:** If $\\phi(v) = 0$, then $f(v) = 0$ for all $f \\in V^*$. Choosing $f$ from the dual basis shows $v = \\mathbf{0}$.

**Dimension:** $\\dim(V) = \\dim(V^*) =...

From: Advanced Linear Algebra

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