# Any inner product can be defined through the standard product and a Matrix

$$(\cdot | \cdot )_0$$ 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...

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

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.

• 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 $(\mathbb{C}^n,\langle,\rangle_{\mathbb{C}^n})$ – Enkidu Nov 2 '18 at 15:27

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.