# Optimal orthonormal basis in $\mathbb{R}^p$

Assume $\{c_i\}_{i=1}^p$ is an orthonormal basis of $\mathbb{R}^p$. $\{\alpha_i\}_{i=1}^p$ and $\{\beta_i\}_{i=1}^p$ are two sets of positive real numbers such that $\alpha_1>\alpha_2>...>\alpha_p$ and $\beta_1>\beta_2>...>\beta_p$.

I want to find the also unit norm, orthonormal vectors $\{v_i\}$ in $\mathbb{R}^p$ that minimize the following expression:

$L = \sum_{i=1}^p \sum_{j=1}^p \frac{\beta_j}{\alpha_i} |v_i^Tc_j|^2$

I believe that the solution is $\{v_i\}_{i=1}^p = \{c_i\}_{i=1}^p$ (in no particular order and perhaps with different signs, e.g. $v_1 = -c_2$, nevermind). I have checked this in the simple case $p=2$, I am looking for the general proof.

EDIT: my proof for the $p=2$ case. Since $\{c_1, c_2\}$ is a basis of $\mathbb{R}^p$, one may write $v_i = k_{i1} \, c_1 + k_{i2} \, c_2$.

When replacing these into the expression to be minimized, it looks like this:

$L = \gamma_{11} \, k_{11}^2 + \gamma_{12} \, k_{12}^2 + \gamma_{21} \, k_{21}^2 + \gamma_{22} \, k_{22}^2$

where $\gamma_{ij} = \frac{\beta_j}{\alpha_i}$ and the orthonormality of $\{c_1, c_2\}$ was used.

Now, the orthonormality of $\{v_1, v_2\}$ implies $k_{11}^2 + k_{12}^2 = 1$ and $k_{21}^2 + k_{22}^2 = 1$ as well as $k_{11} k_{21} + k_{12} k_{22} = 0$. Using these $3$ equations one may write the $k_{ij}$'s coefficients as a function of only one. For example, $k_{11}^2 = k_{22}^2 = 1 - k_{12}^2$ and $k_{21}^2 = k_{12}^2$. Therefore:

$L = (\gamma_{11} + \gamma_{22}) ( 1 - k_{12}^2) + (\gamma_{12} + \gamma_{21})k_{12}^2$ with $0 \leq k_{12}^2 \leq 1$

Now, it is easy to see that $\gamma_{11} + \gamma_{22} < \gamma_{12} + \gamma_{21}$, so the optimal choice is $k_{12}^2 = k_{21}^2 = 0 \implies k_{11}^2 = k_{22}^2 = 1$. In this case $v_1 = \pm c_1$ and $v_2 = \pm c_2$.

I would like to see the proof for arbitrary $p \in \mathbb{N}$.