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I am reading a book " Abstract Linear Algebra " - M. L. Curtis

Chapter V is about normed algebras, which is a title. Here normed algebra $X$ is a algebra with norm satisfying the following relation : For $a$, $b\in X$, $|ab|=|a||b|$.

${\bf R}$ is a typical example. So from this we want to consturct ${\bf C}$. So we must define multiplication structure. The book insist that since the norm is comed out from inner product, the multiplication is unique.

From the parallelogram law we can derive inner product from norm. If not I do not understand why the title is normed algebra.

Hurwitz theorem is that the only real normed algebras are ${\bf R}$, ${\bf C}$, ${\bf H}$, and ${\bf O}$.

I will summarize my question : Can we derive inner product from normed algebra ?

Thank you in advance.


My answering : Let $(s,t)=(0,1)^2$ (i.e., $s^2 + t^2=1)$. If $(1,0)$ is a left identity for multiplication, then $2=|(1,1)(0,1)|^2= |(0,1) + (s,t)|^2=2+2t$ Hence $t=0$.

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2 Answers 2

up vote 1 down vote accepted

First, take a look at this MO thread on the same topic. Notice that the accepted answer cites that any normed division algebra is finite-dimensional and therefore isomorphic to the four Hurwitz's theorem describes $\Bbb{R}, \Bbb{C},\Bbb{H}, \text{and} \,\Bbb{O}$. Therefore your result should hold, as all of the norms on those spaces fulfil the parallelogram law.

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Thank you for your referring. –  Hee Kwon Lee Jan 25 '13 at 2:52

Yes, we can derive an inner product from a norm that satisfies the parallelogram law. See this pdf.

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Thank you. I know that inner product iff norm with parallelogram law. My question : Does the norm on normed algebra implies inner product ? –  Hee Kwon Lee Jan 25 '13 at 1:20
    
@user37116 my apologies, I misread your question. See my answer below. –  Sam DeHority Jan 25 '13 at 1:55

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