Prove $\mathbf{\det(I+xy^T+uv^T)}=(1+\mathbf{y^Tx})(1+\mathbf{v^Tu)-(x^Tv)(y^Tu)}$

Question

I am asked to prove

$$\mathbf{\det(I+xy^T+uv^T)}=(1+\mathbf{y^Tx})(1+\mathbf{v^Tu)-(x^Tv)(y^Tu)}$$

By first proving that $\mathbf{\det(I+xy^T)}=1+\mathbf{y^Tx}$, where $\mathbf{x}$ and $\mathbf{y}$ are $n$ vector.

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Assuming that $x\neq0$, we can find vectors $w_1, w_2, \cdots, w_{n-1}$ such that the matrix $Q$ defined by $$Q=[x,w_1,w_2,\cdots, w_{n-1}]$$ is nonsingular and $x=Qe_1$, where $e_1=(1,0,0,\cdots,0)^T$. If we define $$y^TQ=(z_1,z_2,\cdots,z_n),$$ then $$z_1=y^TQe_1=y^TQ(Q^{-1}x)=y^Tx,$$ and $$\det(I+xy^T)=\det(Q^{-1}(I+xy^T)Q)=\det(I+e_1y^TQ).$$

After here I am not quite sure how to close the argument and then go on to prove that $\mathbf{\det(I+xy^T+uv^T)}=(1+\mathbf{y^Tx})(1+\mathbf{v^Tu)-(x^Tv)(y^Tu)}$.

Answer 1

Hint: First off, for any (compatible) matrices $A$ and $B$, we have $\det(I+AB) = \det(I+BA)$. In your case, $A$, $B$ are just column vectors.

Answer 2

With the same idea as in the proof of the matrix determinant lemma, one finds after a 'short' computation $$ \pmatrix{I&0&0\\ y^T&1&0\\ v^T&0&1} \pmatrix{I+xy^T+uv^T&x&u\\ 0&1&0\\ 0&0&1} \pmatrix{I&0&0\\ -y^T&1&0\\ -v^T&0&1} =\pmatrix{I&x&u\\ 0&y^Tx+1&y^Tu\\ 0&v^Tx& v^Tu+1}. $$ Note that this proof does not rely on any non-singularity condition.