Ytukyg

Let $u,v in R^n$, How can I find $det(I+uv^*)$? This problem was given as preparatory for final exam and I dont know how to approach it. I dont see some nice ways to express this determinant. This is from Numerical Linear algebra course. Could you please give me some hints?

Suppose $A$ is an invertible square matrix and $u$, $v$ are column vectors. Then the matrix determinant lemma states that
$$
{displaystyle det left({A} +{uv} ^{textsf {T}}right)
=left(1+{v} ^{textsf {T}}{A} ^{-1}{u} right),det left({A} right),.}
$$

In your case, let $A=I$.

The proof is essentially based on the following observation:
$$
{displaystyle {begin{pmatrix}mathbf {I} &0\mathbf {v} ^{textsf {T}}&1end{pmatrix}}{begin{pmatrix}mathbf {I} +mathbf {uv} ^{textsf {T}}&mathbf {u} \0&1end{pmatrix}}{begin{pmatrix}mathbf {I} &0\-mathbf {v} ^{textsf {T}}&1end{pmatrix}}={begin{pmatrix}mathbf {I} &mathbf {u} \0&1+mathbf {v} ^{textsf {T}}mathbf {u} end{pmatrix}}.}
$$

[Added later:]
Alternatively, in the nontrivial case when both $u$ and $v$ are nonzero vecotrs and $n>1$, note that $B:={uv} ^{textsf {T}}$ is a rank one matrix. Moreover, $lambda = {v} ^{textsf {T}}u$ is an eigenvalue of $B$ since:
$$Bu = (uv^T)u=u(v^Tu)=(v^Tu)u.$$ Since $B$ is of rank one, $0$ must be an eigenvalue as well. It is not difficult to show¹ that $0$ has multiplicity $n-1$. So the matrix $B$ is similar to an upper triangle matrix with the diagonal:
$$
(v^Tu,0,cdots,0).
$$
It follows that the matrix $I+B$ is similar to an upper triangle matrix with the diagonal:
$$
(1+v^Tu,1,cdots,1).
$$
But similar matrices has the same determinant. So you have the same result as in the matrix determinant lemma.

¹Note: see also answers to this question: Eigenvalues of the rank one matrix $uv^T$

You can easily convince yourself that one eigenvector is $u$ with eigenvalue $1+v^* u$ and the remaining eigenvalues are 1 (with the eigenspace given by all the vectors $w$ with $v^*w=0$). Thus the determinant is given by $$1+v^*u,.$$

Assuming $uv^*ne0$, then $u$ is an eigenvector, as $(uv^*)u=(v^*u)u$. Since the matrix has rank $1$, the characteristic polynomial is $det(uv^*-XI)=(v^*u-X)(0-X)^{n-1}$. Evaluating this at $X=-1$ yields
$$
det(uv^*+I)=v^*u+1
$$