# Inner product - Show that there is a unique vector

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?

• Hint : $A$ is invertible. – Arnaud D. Sep 3 '18 at 16:37

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 $v=(x_1,...,x_n)$.