Determinant of symmetric matrix $A^TA$ Let $A=\begin{pmatrix} I \\ v \end{pmatrix} \in \mathbb R^{n\times n-1}$ with $I\in \mathbb R^{n-1\times n-1}$
identity matrix and $v=(v_1...v_{n-1})$.
Show that $\det(A^TA)=1+v_1^2+...+v_{n-1}^2$ 

I tried multiplying it out
$$A^TA=\begin{pmatrix} I &v^T \\  \end{pmatrix}\begin{pmatrix} I \\ v  \end{pmatrix}=\begin{pmatrix} 1+v_1^2 & v_1v_2 & \dots &v_1v_{n-1} \\ 
  v_1v_2  & 1+v_2^2 & \dots & v_2v_{n-1}\\\
\vdots&\vdots &\ddots &\vdots\\
v_1v_{n-1}& v_2v_{n-1}&\dots &1+v_{n-1}^2 \end{pmatrix}$$ and use induction, but I can't go further.
 A: If $v = 0$, then the answer is immediate from your multiplied-out matrix.
Otherwise, the product may be simplified as $I + v^Tv$. Now, the determinant of a matrix is known to be the product of all eigenvalues (with multiplicity), so we try to see if we can find any:
$$
\det(I+v^Tv - \lambda I) = \det(v^Tv - \eta I)
$$
where $\eta = \lambda - 1$. Thus we've transformed the problem from finding the eigenvalues of $I+v^Tv$ to finding the eigenvalues of $v^Tv$, and then adding $1$ to all of them.
The matrix $v^Tv$ has eigenvalue $0$. The corresponding eigenspace is the space of all vectors orthogonal to $v^T$, which has dimension one less than the space in which $v^T$ lives; this is where I use $v\neq 0$, mostly for simplicity. So the multiplicity of this eigenvalue is $n-2$.
(If you want to nit-pick algebraic vs. geometric multiplicity here, then I say that $v^Tv$ is symmetric, therefore diagonalizable, and the two notions are therefore equal. Alternatively, the algebraic multiplicity is at least as large as the geometric multiplicity. As we yet have one eigenvalue to take care of, the algebraic multiplicity must be the same as the geometric in this case; there is no room for higher multiplicity.)
Also, $v^T$ is an eigenvector of $v^Tv$:
$$
(v^Tv)v^T = v^T(vv^T) = v^T\cdot \|v^T\|^2
$$
and the corresponding eigenvalue, as we can see, is $\|v^T\|^2$.
So that's it. $v^Tv$ has eigenvalues $0$ with multiplicity $n-2$, and $\|v^T\|^2$ with multiplicity $1$. This means that $I+v^Tv$ has eigenvalues $1$ with multiplicity $n-2$ and $\|v^T\|^2 + 1$ with multiplicity $1$. Multiply them all together, and you get your answer.
A: @Arthur
Here is a proof using the matrix-determinant lemma (https://en.wikipedia.org/wiki/Matrix_determinant_lemma):
$$\det(A)=\det(I+vv^T)=(1+v^TI^{-1}v)\det(I)=1+v^Tv.$$
