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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?

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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: $$ {\displaystyle {\begin{pmatrix}\mathbf {I} &0\\\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} +\mathbf {uv} ^{\textsf {T}}&\mathbf {u} \\0&1\end{pmatrix}}{\begin{pmatrix}\mathbf {I} &0\\-\mathbf {v} ^{\textsf {T}}&1\end{pmatrix}}={\begin{pmatrix}\mathbf {I} &\mathbf {u} \\0&1+\mathbf {v} ^{\textsf {T}}\mathbf {u} \end{pmatrix}}.} $$


[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 $uv^T$

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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 $$ \det(uv^*+I)=v^*u+1 $$

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  • $\begingroup$ Very clever short proof. $\endgroup$ – Jean Marie Dec 9 '18 at 23:27
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    $\begingroup$ @JeanMarie And the general formula for $\det(A+uv^*)$ follows in the same fashion, because $A^{-1}u$ is an eigenvector of $A^{-1}uv^*$ (which has rank at most $1$) relative to $v^*A^{-1}u$. $\endgroup$ – egreg Dec 9 '18 at 23:47
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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 $$1+v^*u\,.$$

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