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Prove that to every $A \in L(R^n,R^1)$ corresponds a unique $y \in R^n$ such that $Ax=\langle x, y\rangle$. Prove that $\|A\|=|y|$

Definition: For $A \in L(R^n,R^m)$, define the norm $\|A\|$ of A to be the sup of all numbers $|Ax|$, where x ranges over all vectors in $R^n$ with $|x| \leq 1$. The inequality $|Ax| \leq \|A\| \cdot |x|$ holds for all $x \in R^n$.

I am thinking about induction on the power of $n$, but not sure how to express each individual $A$, $x$ and $y$.

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For a proof avoiding bases, define a map $T:R^n\to L(R^n, R)$ by sending $y\in R^n$ to $\langle \cdot, y\rangle$. If $y\in \ker(T)$ then, in particular, $\langle y, y\rangle =0$, and hence $y=0$. So $\ker T= 0$. By the rank-nullity theorem, we see that $T$ is a linear isomorphism. Thus for every $A\in L(R^n, R)$ there is $y\in R^n$ with $Ax=\langle x, y\rangle$ for all $x$.

(Of course, the basis-free proof uses bases in disguise in the rank-nullity theorem and in the fact that $\dim L(R^n, R) = n$).

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Hint: What does $A$ do to each standard basis vector $e_k$?

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Hint: Consider a base of $R^n$ like $\{e_i\}$ and apply $A$ to each $e_i$, assuming such $y$ exists. What should be this $Ae_i$?

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