If V = R^n is every norm induced by an inner product?
I know that there are several different norms such as:
$$\Vert x \Vert_1 = \sum_{i=1}^{n} \vert x_i \vert$$ or
$$\Vert x \Vert_2 = \sqrt{x\times x} = \sqrt{\sum_{i=1}^{n} \vert x_i \vert ^2}$$
I think the second norm is induced by an inner product, but I'm not sure about the first?
I didn't really understand the question very well. This is all very new to me and I'm not getting the hang of it. Could someone help?
No, not every norm is induced by an inner product.
If $\lVert\cdot\rVert$ is induced by an inner product on $\mathbb{R}^n$, i.e. $\lVert\mathbf{x}\rVert = \sqrt{\langle \mathbf{x},\mathbf{x}\rangle}$, where $\langle\cdot,\,\cdot\rangle$ is the inner product, then this norm must satisfy the parallelogram law: $$2\lVert \mathbf{x}\rVert^2 + 2\lVert \mathbf{y}\rVert^2 = \lVert\mathbf{x}+\mathbf{y}\rVert^2 + \lVert \mathbf{x}-\mathbf{y}\rVert^2.$$ And in fact, the converse holds; that is, any norm that satisfies the parallelogram law is necessarily induced by an inner product; that inner product will be given by $$\langle \mathbf{x},\mathbf{y}\rangle = \frac{1}{4}\left(\lVert \mathbf{x}+\mathbf{y}\rVert^2 - \lVert\mathbf{x}-\mathbf{y}\rVert^2\right),$$ which essentially uses the Polar Identities over $\mathbb{R}$; a similar but more complicated formula works for $\mathbb{C}$.
Your first norm does not satisfy the parallelogram law. For example, with $n=2$, take $\mathbf{x}=(1,0)$ and $\mathbf{y}=(0,1)$. Then $$\begin{align*} 2\lVert\mathbf{x}\rVert^2 + 2\lVert\mathbf{y}\rVert^2 &= 2(|1|+|0|)^2 + 2(|0|+|1|)^2 = 4,\\ \lVert\mathbf{x}+\mathbf{y}\rVert^2 + \lVert\mathbf{x}-\mathbf{y}\rVert^2 &= \lVert(1,1)\rVert^2 + \lVert (1,-1)\rVert^2\\ &= (|1|+|1|)^2 + (|1|+|-1|)^2 \\ &= 4+4=8. \end{align*}$$ Since the norm does not satisfy the parallelogram law, it cannot be induced by an inner product.