Given $H_d$ a $d\times d$-Walsh matrix, for matrix $B\in \mathbb{R}^{d\times d}$ satisfying $\mathrm{rank} (B)\leq n<d$, I want to prove
\begin{align*} \inf_B \| B-H_d\|^2_F\geq d(d-n), \end{align*}
My attempt:
\begin{align*} \| B-H_d\|^2_F = \mathrm{tr}((B-H_d)(B-H_d)^T)=\mathrm{tr}(BB^T)-2\mathrm{tr}(BH_d^T)+d^2. \end{align*} By applying SVD, $\mathrm{tr}(BB^T)=\mathrm{tr}(\Sigma\Sigma^T)$, where $\Sigma$ is a non-negative diagonal matrix of at most rank $n$. Although $n$ indeed appears this way, I still can't get the desired term, since $\Sigma$ is quite arbitrary.
We first notice that \begin{align*} \| B-H_d\|^2_F = \mathrm{tr}\bigl((B-H_d)(B-H_d)^T\bigr)=\mathrm{tr}(BB^T)-2\mathrm{tr}(BH_d^T)+d^2 \end{align*} Applying SVD to real matrix $B$, we can decompose it into $B=U\Sigma V^T$, where $U, V$ are orthogonal $d\times d$-matrices and $\Sigma = \mathrm{Diag}_d(\sigma_1,...,\sigma_n,0,...,0)$. Observing that $\bigl(V^TH_d^TU\bigr)\bigl(U^TH_dV\bigr)=d\cdot I_d$, we then denote ${\alpha_i}$ the i-th column of $V^TH_d^TU$ and have \begin{align*} &\mathrm{tr}(BB^T)-2\mathrm{tr}(BH_d^T)=\mathrm{tr}(U\Sigma V^TV\Sigma U^T)-2\mathrm{tr}(U\Sigma V^TH_d^T)\\ =&\mathrm{tr}(\Sigma\Sigma)-2\mathrm{tr}(V^TH_d^TU\Sigma)=\sum_{i=1}^n(\sigma_i^2-2\|{\alpha_i}\|_2\cdot\sigma_i)=\sum_{i=1}^n(\sigma_i^2-2\sqrt{d}\cdot\sigma_i)\\ \geq&-nd \end{align*} Therefore we have $\| B-H_d\|^2_F\geq d(d-n)$.