Proving the relation $\det(I + xy^T ) = 1 + x^Ty$

Let $x$ and $y$ denote two length-$n$ column vectors. Prove that $$\det(I + xy^T ) = 1 + x^Ty$$

Is Sylvester's determinant theorem an extension of the problem? Is the approach the same?

Solution 1:

Hint: Decomposing $$ \begin{pmatrix} 1 & -y^T\\ x & I \end{pmatrix} $$ as lower $\cdot$ upper and upper $\cdot$ lower gives $$ \begin{pmatrix} 1 & 0\\ x & I\end{pmatrix} \cdot \begin{pmatrix} 1 & -y^T\\ 0 & I + xy^T \end{pmatrix} = \begin{pmatrix} 1 + x^Ty & -y^T\\ 0 & I\end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ x & I \end{pmatrix}. $$

Solution 2:

You can also apply the property

$$\det\begin{pmatrix} A & B\\ C & D \end{pmatrix}=\det(A) \det(D-C A^{-1}B)=\det(D) \det(A-B D^{-1}C)$$

(valid when the inverses exist) to the matrix $$\begin{pmatrix} I & -y\\ x^T & 1 \end{pmatrix}$$

(BTW: here I'm assuming column vectors; the question speaks of "column vectors", but its notation would correspond to row vectors)