Show that if $x^tAy=x^tBy$ for all $x,y\in \mathbb{R}^n$ then $A=B$.

Question

This problem is from "Notes on Geometry" by Rees.

I am studying the Euclidean Geometry Part in the book.

This problem is right below the definition of the orthogonal and orthonormal, so I assume that $A,B$ in this problem are matrices of $T:\mathbb{R}^n\to \mathbb{R}^n$ with respect to an orthonormal basis.

Moreover, this problem wants to use the fact "$a_{ij}=e_{i}^{t}Ae_j$", which seems like $a_{ij}$ is the element of the matrix $A$ and $e_i,e_j$ are elements of an orthonormal basis. Also, I think I have to use another fact "if $x=\sum_{i}^{n}x_ie_i $ and $y=\sum_{i}^{n}y_ie_i$, then $x\cdot y=\underline{x}^t\underline{y}$." Here, the underline indicates that $x,y$ are column vectors.

The book itself is a little vague for a beginner like me, so I would like to see if my answer is the right solution for this problem.

My answer:

Pick arbitrary $x,y \in \mathbb{R}^n$.

We have

$x^tAy=x^tBy$, so this becomes $x^te_ia_{ij}e_j^ty = x^te_ib_{ij}e_j^ty$ by the first fact listed above.

Then, by the second fact, the equation again becomes $(x\cdot e_i) a_{ij} (e_j\cdot y)=(x\cdot e_i) b_{ij} (e_j\cdot y)$.

Assuming that $x,y$ are non-zero for division, we get $a_{ij}=b_{ij}$.

Hence, $A=B$.

Is this the right approach?

Answer 1

Not quite. It's better to phrase it in the following way.

If $x^T A y = x^T B y$ for all $x,y \in \mathbb{R}^n$, then we may, in particular, set $x = e_i$ and $y = e_j$, so that we get $$a_{ij} = e_i^T A e_j = e_i^T B e_j = b_{ij}$$ This implies $A = B$.

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Note that this differs from your answer in the following two ways:

1. $x$ and $y$ are not arbitrary vectors, but rather chosen such that they "pick off" entries of $A$ and $B$
2. Your attempt tries to replace the matrices $A$ and $B$ by $e_i a_{ij}e_j^T$, but instead you should be replacing $x$ and $y$ with $e_i$ and $e_j$, respectively.