Inner product - Show that there is a unique vector

Question

Let $A$ be a $n\times n$ positive definite real symmetric matrix.

I have shown that $\langle x, y\rangle=y^tAx$ satisfies the properties of an inner product.

I want to show that there is a unique $y_0\in \mathbb{R}^n$ such that for each $x\in \mathbb{R}^n$ with $x^t=(x_1, \ldots , x_n)$ we have $x_1+\ldots +x_n=y_0^tAx$.

How can we show that?

Couls you give me a hint?

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Hint : $A$ is invertible. – Arnaud D. Sep 03 '18 at 16:37

nonuser · Accepted Answer · 2018-09-03T16:32:11.710

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I think that has something to do with Riesz representation theorem.

For linear functional $f:\mathbb{R}^n\to \mathbb{R}$ there is a vector $u_0$ such that $f(v) = \langle v,u_0 \rangle$.

In particual $f(v) = x_1+x_2+...+x_n$ vhere $v=(x_1,...,x_n)$.

edited Sep 03 '18 at 16:32

answered Sep 03 '18 at 16:16

nonuser

90,026

I haven't really understood how the uniqueness implies. Could you explain this further to me? – Mary Star Sep 03 '18 at 16:36
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https://math.stackexchange.com/questions/1738045/riesz-representation-theorem-in-linear-algebra – nonuser Sep 03 '18 at 16:37
Ok! Thank you!! :-) – Mary Star Sep 03 '18 at 17:32

Inner product - Show that there is a unique vector

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