Let be $u,v\in\mathbb{R}^n$, then $\det(I+uv^\intercal)=1+v^\intercal u $
where $I$ denotes the identity matrix of order $n$. How to prove this?
what I did:
let be $A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u \} $, and suppose $A\neq \varnothing $, then for the well-orderer-principle there exists $n_0\in A$ such $n_0\leq n,\;\forall n\in A$. since $1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A $ so: $$\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u $$
Then, let be $u,v\in\mathbb{R}^{n_0}$, so $$\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})$$ where $a_{ij}=\begin{cases} u_iv_j+1 & i=j \\ u_iv_j & i\neq j \end{cases}$,
and $A_{1j}$ is the submatrix of $I+uv^\intercal$ that results deleting the i-file and j-column. Then I try to open the term for $j=1$ and try to use the relation for the new matrix of $n_0-1$ order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?.
Ps. here, in page 2 I've found something similar but I don't understand what it means
PS. I'm taking a course of numerical analisys
Best Answer
You are looking at a special case of the Matrix determinant Lemma. From the Wikipedia page, the proof for the case $A = I$ follows from the equality $$ \begin{bmatrix} I & 0 \\ v^T & 1 \end{bmatrix} \begin{bmatrix} I + uv^T & u \\ 0 &1 \end{bmatrix} \begin{bmatrix} I & 0 \\ -v^T & 1 \end{bmatrix} = \begin{bmatrix} I & u \\ 0 & 1 + v^Tu \end{bmatrix}$$