Let $$ M = zz^{T},$$ be rank $1$ matrix,where $$ z=\begin{bmatrix} \zeta _{1} \\ \zeta _{2} \\ \vdots \\ \zeta _{n} \end{bmatrix} $$ Compute determinant of $M$.
$M$ is obviously symmetric matrix. Is there any "shortcut" to computing determinant of symmetric matrices in general, or is there no other but to solve this using Laplace expansion?
One can see it in a different way :
With your notation for a column vector
$$z=\begin{bmatrix} \zeta_{1} \\ \zeta_{2} \\ \vdots \\ \zeta_{n} \end{bmatrix}$$
Let M be a matrix of the form
$$M=[z|v_2|v_3|\cdots | v_n]$$
such that the columns constitute a basis of $\mathbb{R}^n$ (it is always possible to "complete" the basis of a finite dimensional linear space in this way).
Remark : $M$ is therefore invertible.
Then, we can write $$zz^T=MJM^T \ \ \text{where} \ \ J=diag(1,0,0, \cdots 0)$$
(the "1" coefficient selects $zz^T$ and the 0 zero coefficients eliminate any influence of vectors $v_k$).
We can conclude that $rank(zz^T)=rank(J)=1$ because the rank in preserved by pre- or post- multiplication by invertible matrices.
As a consequence $\det(zz^T)=\underbrace{(1 \times 0 \times 0 \cdots \times 0)}_{\det J}\det(M)^2 = 0. :)$