How to prove this inverse matrix identity? Let L be an $(n − 1) \times n$ matrix such that LL' $= I_{n−1}$ (where $'$ denotes the transpose) and La = 0 for an $n × 1$ vector a.
Prove that the following result holds for any $n × n$ non-singular symmetric matrix A:
$(LAL')^{-1} = LA^{-1}L' - \frac{LA^{-1}aa'A^{-1}L'}{a'A^{-1}a}$
My attempt: The idea is simple, I just want to show that $LAL'$ times RHS is identity. But I couldn't solve that. How to prove:
$$LAL'\left(LA^{-1}L' - \frac{LA^{-1}aa'A^{-1}L'}{a'A^{-1}a}\right)=I$$
 A: This is definitely false, as $a'A^{-1}a$ can be zero, e.g. when $L=\begin{bmatrix}1&0\end{bmatrix},\,a=\begin{bmatrix}1\\ -1\end{bmatrix}$ and $A=\begin{bmatrix}1\\ &-1\end{bmatrix}$, $LAL'=\begin{bmatrix}1\end{bmatrix}$ is invertible but $a'A^{-1}a=0$.
However, $a'A^{-1}a\ne0$ is sufficient for the equality to hold.

It is clearer to consider a more general setting. Suppose $L\in\mathbb F^{(n-1)\times n},\,A\in\mathbb F^{n\times n},\,R\in\mathbb F^{n\times(n-1)}$ are three matrices and $u,v\in\mathbb F^n$ are two vectors such that $A$ is invertible and
$$
LR=I_n,\quad Lu=0,\quad v'R=0,\quad v'A^{-1}u\ne0.
$$
Since $Lu=0$ but $v'A^{-1}u\ne0$, the row vector $v'A^{-1}$ is not a linear combination of the rows of $L$. Hence it is linearly independent of the rows of $L$. It follows that $\begin{bmatrix}L\\ v'A^{-1}\end{bmatrix}$ is invertible. In fact, we have
$$
\begin{bmatrix}L\\ v'A^{-1}\end{bmatrix}
\begin{bmatrix}\left(I_n-\frac{uv'A^{-1}}{v'A^{-1}u}\right)R& \frac{u}{v'A^{-1}u}\end{bmatrix}
=I_n.\tag{1}
$$
Similarly, $\begin{bmatrix}R&A^{-1}u\end{bmatrix}$ is also invertible:
$$
\begin{bmatrix}L\left(I_n-\frac{A^{-1}uv'}{v'A^{-1}u}\right)\\ \frac{v'}{v'A^{-1}u}\end{bmatrix}
\begin{bmatrix}R&A^{-1}u\end{bmatrix}
=I_n.\tag{2}
$$
Now note that
$$
\begin{bmatrix}L\\ v'A^{-1}\end{bmatrix}
A
\begin{bmatrix}R&A^{-1}u\end{bmatrix}
=\begin{bmatrix}LAR&0\\ 0&v'A^{-1}u\end{bmatrix}.
$$
Since all three matrices on the left are invertible, so is the RHS. In particular, $LAR$ must be invertible. Taking inverses on both sides of the equality above, we obtain
$$
\begin{bmatrix}L\left(I_n-\frac{A^{-1}uv'}{v'A^{-1}u}\right)\\ \frac{v'}{v'A^{-1}u}\end{bmatrix}
A^{-1}
\begin{bmatrix}\left(I_n-\frac{uv'A^{-1}}{v'A^{-1}u}\right)R& \frac{u}{v'A^{-1}u}\end{bmatrix}
=\begin{bmatrix}(LAR)^{-1}&0\\ 0&\frac{1}{v'A^{-1}u}\end{bmatrix}.
$$
Therefore
$$
\begin{aligned}
(LAR)^{-1}
&=
L\left(I_n-\frac{A^{-1}uv'}{v'A^{-1}u}\right)
A^{-1}
\left(I_n-\frac{uv'A^{-1}}{v'A^{-1}u}\right)R\\
&=
L\left(A^{-1}-\frac{A^{-1}uv'A^{-1}}{v'A^{-1}u}\right)R.
\end{aligned}
$$
In your case, simply put $R=L'$ and $u=v=a$. Note that we don't need the assumption that $A$ is symmetric here.

Remark. How can I directly verify that $LAR\cdot
L\left(A^{-1}-\frac{A^{-1}uv'A^{-1}}{v'A^{-1}u}\right)R = I_{n-1}?
$
Let
$$
\Pi_1=I_n-\frac{uv'A^{-1}}{v'A^{-1}u}
\quad\text{and}\quad
\Pi_2=I_n-\frac{A^{-1}uv'}{v'A^{-1}u}.
$$
We want to show that
$$
(LAR)(L\Pi_2A^{-1}R)=I_{n-1}.
$$
The key observation is that
$$
RL\Pi_2=\Pi_2,
$$
because $\Pi_2$ is a projection onto $\operatorname{im}(R)$. It follows that
$$
\begin{align}
(LAR)(L\Pi_2A^{-1}R)
&=LA(RL\Pi_2)A^{-1}R\\
&=LA\Pi_2A^{-1}R\\
&=LA(\Pi_2A^{-1})R\\
&=LA(A^{-1}\Pi_1)R\\
&=(L\Pi_1)R\\
&=LR\\
&=I_{n-1}.
\end{align}
$$
