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How to prove that $\det(I+ vv^T) = 1 + ||v||^2$?

Here $v$ is vector-column, $||v||^2 = v_1^2 + \dots + v_n^2$, $I$ is the identity matrix.

This determinant arises when computing a surface integral with the parametrization $\phi(x) = (x, f(x))$ (A surface defined as a graph of a real-valued function $f$).

$\det (D \phi)^T D \phi = \det(I + (\nabla f)^T \nabla f)$

J.G.
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Alexander Linderman
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  • This statement holds by Sylvester's determinant identity. – Ben Grossmann Jan 09 '22 at 20:01
  • That said, please note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could [edit] your question to address the following. Where did you encounter this question (that is, what class are you taking or what book are you reading)? What are your thoughts on the problem? What have you tried so far? – Ben Grossmann Jan 09 '22 at 20:06
  • Do you know that the determinant is the product of the eigenvalues? If yes, see what happens if you apply that matrix i) to $v$, ii) to any vector normal to $v$. – Thomas Jan 09 '22 at 20:13
  • @Thomas, thanks! For some reason, I did not think about it in terms of eigenvalues and eigenvectors. Now everything is clear. – Alexander Linderman Jan 09 '22 at 20:29

2 Answers2

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Note that the eigenvalues of $I + vv^T$ are all ones except for one of them which is $1 + v^T v = 1 + \Vert v \Vert^2$. Combine this with the definition of determinant being the product of all eigenvalues.

Ahmad Bazzi
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First we have : $$\det \left(I + v v^{\mathrm{T}}\right) = \chi_{-v v^{\mathrm{T}}} (1)$$

And because we know that for every $n \times m$ matrix $A$ and every $m \times n$ matrix $B$, we have : $$\forall t \in \mathbb{R}, t^n \chi_{A B} (t) = t^m \chi_{BA} (t)$$ then : $$\chi_{-v v^{\mathrm{T}}} (1) = \chi_{-v^{\mathrm{T}} v} (1) = \chi_{-\|v\|^2} (1) = \det (1 + \|v\|^2) = 1 + \|v\|^2$$ $1 + \|v\|^2$ is a scalar and we know that for any scalar $\lambda$, $\det \lambda = \lambda$

Essaidi
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