I am finalizing the proof of the spectral theorem and I have to motivate the fact that if $A \in \mathbb R ^{n\times n}$ is symmetric and the columns of $B \in \mathbb R ^{n\times p}$, with $p < n$, make up an orthonormal base of $\mathcal N^\perp (A)$, then the matrix product $$B^T AB x = 0$$ with $x \in \mathbb R ^{p}$, can be satisfied iff $x = 0$.
Since $A = A^T$, $\mathcal N ^{\perp} (A) = \mathcal R (A^T) = \mathcal R (A)$ so the range and the kernel of $A$ are orthogonal and necessarily $\mathcal R (B) = \mathcal R(A)$.
It seems I am stuck in providing a quick explanation for the necessity of $x = 0$.
I guess by $\mathcal{N}(A)$ you mean $\mathrm{ker}A$ and by $\mathcal{N}^{\perp}(A)$ you mean the choice of some subspace of $\mathbb{R}^n$ such that $\mathcal{N}^{\perp}(A)$ and $\mathcal{N}(A)$ are orthogonal and $\mathcal{N}(A) \oplus \mathcal{N}^{\perp}(A) = \mathbb{R}^n$.
In this case, by definition, $AB \colon \mathbb{R}^p \rightarrow \mathbb{R}^n$ is injective and $\mathrm{im}(AB)= \mathrm{im}(A)$. It also follows $\mathrm{rk}(AB)=p$.
Therefore, also $p= \mathrm{rk}((AB)^T)= \mathrm{rk}(B^TA^T)= \mathrm{rk}(B^T A)$. But this means, that $B^T_{|\mathrm{im}(A)} = B^T_{|\mathrm{im}(AB)} $ is an isomorphism which in implies that $B^T A B$ is injective.
Because the proof was deemed too long, here the short version without the unnecessary details:
The map $AB$ is injective and $\mathrm{im}(AB)=\mathrm{im}(A)$. Thus, $p = \mathrm{rk}((AB)^T)= \mathrm{rk}(B^TA) = \mathrm{rk}(B^T A B)$.