Show that $\|A(I - vv^T/v^Tv)\|_F^2 = \|A\|_F^2 - \|Av\|^2_2/(v^tv), A \in R^{n \times n}, v \in R^n$
$$ \begin{aligned} \|A(I - vv^T/v^Tv)\|_F^2 &= \sum_{i = 1}^{n}\sum_{j = 1}^{n}(A - Avv^T/c)^2 \\ &= \sum_{i = 1}^{n}\sum_{j = 1}^{n}(A^2 + (Avv^T)^2/c^2 - 2A^2vv^T/c) \\ &= \sum_{i = 1}^{n}\sum_{j = 1}^{n}A^2 - \sum_{i = 1}^{n}\sum_{j = 1}^{n}(vv^TA^TAvv^T/c^2 - 2A^TAvv^T/c)\\ &= \|A\|_F^2 - \sum_{i = 1}^{n}\sum_{j = 1}^{n}(vv^TA^TAvv^T/c^2 - 2A^TAvv^T/c) \end{aligned} $$
you can simplify this by assuming WLOG that $\mathbf v$ has length one.
simple case:
suppose $\mathbf v := \mathbf e_1$, the first standard basis vector. Then by direct computation
$\|A(I - \mathbf{e_1e_1}^T)\|_F^2 =\sum_{j=2}^n\sum_{i=1}^n a_{i,j}^2$ and
$\|A\|_F^2 - \|A\mathbf e_1\|_2^2 =\big(\sum_{j=1}^n\sum_{i=1}^n a_{i,j}^2\big) -\big(\sum_{j=1}^1\sum_{i=1}^n a_{i,j}^2\big)= \sum_{j=2}^n\sum_{i=1}^n a_{i,j}^2$
hence $\|A(I - \mathbf{e_1e_1}^T)\|_F^2=\|A\|_F^2 - \|A\mathbf e_1\|_2^2$
general case:
Real symmetric matrices may be orthogonally diagonalizaed, so $\big(I-\mathbf {vv}^T\big) =Q\big(I -\mathbf e_1\mathbf e_1^T\big)Q^T$, where $Q$ is real orthogonal. Define $B:=Q^TAQ$
$\|A(I - \mathbf{vv}^T)\|_F^2$
$=\|AQ(I - \mathbf e_1\mathbf e_1^T)Q^T\|_F^2$
$=\|Q^TAQ(I - \mathbf e_1\mathbf e_1^T)\|_F^2$
$=\|B(I - \mathbf e_1\mathbf e_1^T)\|_F^2$
$=\|B\|_F^2 - \|B\mathbf e_1\|_2^2$
$=\|Q^TAQ\|_F^2 - \|Q^TAQ\mathbf e_1\|_2^2$
$=\|A\|_F^2 - \|A(Q\mathbf e_1)\|_2^2$
$=\|A\|_F^2 - \|A\mathbf v\|_2^2$