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$(\cdot | \cdot )_0 $ an inner Product in $\mathbb{R}^n$. Show that there is one and only one symmetric, positive Definite $n \times n$ matrix $A$ such that: $$ (x |y)_0 = (Ax|y) $$ for all vectors in $\mathbb{R}^n$. And $(x|y)$ the standard inner product.

So i have shown that if such A exists then it must be unique, that part is easy. But I have troubles showing the existence. I thought it could be possible, since it is possible to reach any linear combination of the coordinates of x and y with a proper matrix, but I don't know how to write that mathematically and if it is enough...

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  • $\begingroup$ Where is your $x$ on the RHS? Is it a typo [I guess]? $\endgroup$ – xbh Nov 2 '18 at 15:13
  • $\begingroup$ Now it's like it should. Thanks! $\endgroup$ – M-S-R Nov 2 '18 at 15:17
  • $\begingroup$ Could you do it now? I think Gram matrix would help. $\endgroup$ – xbh Nov 2 '18 at 15:19
  • $\begingroup$ btw, why do you use $\mathbb{R}^n$ and euclidean inner product? that is quite restrictive. $\endgroup$ – Enkidu Nov 2 '18 at 15:29
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Now first of all, lets use proper notation!

what we want is a matrix s.t $xAy^*=\langle x , y \rangle$. Since you already have an inner product, you have a basis (i.e. worst case svenario, pick a ONB(orthonormal basis))$\{b_1,...,b_n\}$. Now define $A_{i,j}:= \langle b_i , b_j \rangle$. then this matrix suffices the formula above. I leave the verification (using representation into basis elements) to you.

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  • $\begingroup$ btw, note that that also works for hermitian vector spaces, and we never used that it is $\mathbb{R}^n$ with the standard structure. It is actually a really nice exercise to use this to show: For any euclidean or hermitian vectorspace, there exists an Isomorphism of euclidean or hermitian Vectorspaces to $(\mathbb{R}^n,\langle,\rangle_{\mathbb{R}^n})$ respectively $(\mathbb{C}^n,\langle,\rangle_{\mathbb{C}^n})$ $\endgroup$ – Enkidu Nov 2 '18 at 15:27
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Let $\{e_1,\dots,e_n\}$ be the standard basis for $\Bbb R^n$. If such matrix $A$ exist, put $x=e_i$, $y=e_j$ and the right-hand-side would become $(Ae_i\vert e_j)=a_{ji}$, the $(j,i)$-entry of $A$. The left-hand-side shall be equal to this number, i.e. we have $a_{ji}=(e_i\vert e_j)_0$. This deals with uniqueness of $A$ actually.

You then show that this $A$ satisfies the equation for all $x,y$, because we only know the equality holds when $x,y$ are standard basis vectors. But, you don't quite need to prove this, because there is the following proposition:

Proposition: If two real-valued bilinear functions $B_1,B_2$ on $\Bbb R^n$ satisfy the property that there exists a basis $\{v_1,\dots,v_n\}$ for $\Bbb R^n$ such that $B_1(v_i,v_j)=B_2(v_i,v_j)$ for all $i,j$, then in fact $B_1=B_2$.

If you know how to prove this and how to apply this theorem, you are done.

If not, prove in the old-fashioned way: let $x=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix},y=\begin{bmatrix}y_1\\\vdots\\y_n\end{bmatrix}$, and prove the equality $(x\vert y)_0 = (Ax\vert y)$ with brute force.

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