I'm really stumped on this problem.
Let M be a $n \times n$ matrix such that $M = I_n + uv^T$.
We want to prove that if $v^Tu = -1$ then the matrix is singular (has no inverse).
My attempts:
Observation: $u$ is $n \times 1$ and $v^T$ is $1 \times n$
Matrix M will be singular if $N(M) \neq {0}$.
$$Mx = (I_n + uv^T)x$$ $$Mx = I_nx + uv^Tx$$
Our first term is is just the elements of $x$ along the diagonal while there's zeroes everywhere else. The second term is complicated but I think the important thing is that along the diagonals we have $u_i v_i x_i$ where the subscripts represent the ith element in the vector.
I'm not really sure where to go from here with this approach.
I then tried to gain some insight with a quick example. Let M be a $2 \times 2$ matrix and $$u = \begin{bmatrix}3\\-4 \\ \end{bmatrix} \text{ and } v = \begin{bmatrix}5\\4 \\ \end{bmatrix}$$
We have that $uv^T = \begin{bmatrix}15& 12\\-20 &-16 \end{bmatrix}$. And so $I_n + uv^T = \begin{bmatrix}16& 12\\-20 &-15 \end{bmatrix}$ which does in fact have $\text{determinant} = 0$. The only insight that I gained from this is that ... it worked(?) (doesn't help much I know) and that the only elements that change in $uv^T$ are those in the diagonal by $+1$.
If your condition is true your matrix has eigenvalue $0$ whence it is singular.