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For $n \in \mathbb{N}$ let $\{v_1, v_2, \ldots v_n \}$ be a orthonormal basis of $\mathbb{R}^{n}$. Further, let $i \in \{ 1, 2, \ldots, n \}$ be arbitrary but fixed. I am trying to prove that there exists a symmetric matrix $M \in \mathbb{R}^{n \times n}$ such that for all $j \in \{ 1,2, \ldots, n \}$ it holds that $$ v_{j}^{T}Mv_{j} = \delta_{ij},$$ where $\delta_{ij}$ is the Kronecker delta.

Intuitively, I would assume such a Matrix $M$ exists, because we can arbitrarily choose all diagonal elements of $M$ and all the entries above the main diagonal. Thus we have $\frac{n(n+1)}{2}$ degrees of freedom and $n$ only equations that must hold. I tried to diagonalize $M$ to prove that the system has a solution. However, I have not been able to do so. Are there any solutions to the system or do I need more additional conditions on $M$?

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    $\begingroup$ Does $M=I$ not work? The basis being orthonormal and all... maybe I'm very confused. $\endgroup$
    – Tom
    Nov 28, 2023 at 23:42
  • $\begingroup$ @Tom: $i$ is fixed. $\endgroup$ Nov 28, 2023 at 23:47
  • $\begingroup$ Of course, sorry. Then I agree with @MartinArgerami $\endgroup$
    – Tom
    Nov 29, 2023 at 0:19

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Consider first the canonical basis $\{e_1,\ldots,e_n\}$. Then you can consider the matrix $M=e_ie_i^T$, that is the matrix with $1$ in the $i,i$ entry and zeroes everywhere else. Then $e_j^TMe_j=\delta_{ij}$.

Now let $U$ be the linear map induced by $Uv_j=e_j$, $j=1,\ldots,n$. Then $U$ is the unitary change of basis from the canonical basis to your basis. By the first paragraph there exists a matrix a symmetric matrix $N$ with $e_j^TNe_j=\delta_{ij}$. Put $M=U^TNU$. Then $M$ is symmetric and $$ v_j^TMv_j=v_j^TU^TNUv_j=(Uv_j)^TNUv_j=e_j^*Ne_j=\delta_{ij}. $$

Note that the non-diagonal entries play no role in the first paragraph. In other words, one can choose any $M$ with zero diagonal, other than the $1$ in the $i^{\rm th}$ entry, and all the non-diagonal entries are only constrained by the symmetry condition.

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