Question
Prove from the inner product axioms that, in any inner product space $V,\langle\mathbf{v}, \mathbf{0}\rangle=0$ for all $\mathbf{v}$ in $V$
Step 1
Step 1: We start with the inner product of a vector $\mathbf{v}$ and the zero vector $\mathbf{0}$, which we can write as $\langle\mathbf{v}, \mathbf{0}\rangle$. Show more…
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Prove from the inner product axioms that for all vectors $\mathbf{u}, \mathbf{v},$ and $\mathbf{w}$ in an inner product space $V,$ we have $$\langle\mathbf{u}, \mathbf{v}+\mathbf{w}\rangle=\langle\mathbf{u}, \mathbf{v}\rangle+\langle\mathbf{u}, \mathbf{w}\rangle$$
Inner Product Spaces
Definition of an Inner Product Space
Let $V$ be a real inner product space. Prove that for all $\mathbf{v}, \mathbf{w}$ in $V$ (a) $\|\mathbf{v}+\mathbf{w}\|^{2}-\|\mathbf{v}-\mathbf{w}\|^{2}=4\langle\mathbf{v}, \mathbf{w}\rangle$ (b) $\|\mathbf{v}+\mathbf{w}\|^{2}+\|\mathbf{v}-\mathbf{w}\|^{2}=2\left(\|\mathbf{v}\|^{2}+\|\mathbf{w}\|^{2}\right)$
Let $V$ be a real inner product space. Show that (i) $\|u\|=\|v\|$ if and only if $\langle u+v, u-v\rangle=0$ (ii) $\|u+v\|^{2}=\|u\|^{2}+\|v\|^{2}$ if and only if $\langle u, v\rangle=0$ Show by counterexamples that the above statements are not true for, say, $\mathbf{C}^{2}$.
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