# Determinant of $(I+uv^*)$

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?

## 3 Answers

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

In your case, let $$A=I$$.

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

[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 $$uv^T$$

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

• Very clever short proof. – Jean Marie Dec 9 '18 at 23:27
• @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 $v^*A^{-1}u$. – egreg Dec 9 '18 at 23:47

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 $$1+v^*u\,.$$