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I have to solve the determinant $\det(\mathbb{I}+\vec{b}\vec{b}^T)$. (The result shloud be 1.)

We can use the sum rule for rows to get.

$$\det(\mathbb{I}+\vec{b}\vec{b}^T)= \begin{vmatrix} 1 & 0 & \ldots& 0\\ b_1 b_2 & & & \\ \vdots & & (\mathbb{I}+\vec{b}\vec{b}^T)_{(2\ldots n)\times (2\ldots n)} &\\ b_1 b_n & & & \end{vmatrix} + \begin{vmatrix} b_1^2 & b_2 b_1 & \ldots& b_n b_1\\ b_1 b_2 & & & \\ \vdots & & (\mathbb{I}+\vec{b}\vec{b}^T)_{(2\ldots n)\times (2\ldots n)} &\\ b_1 b_n & & & \end{vmatrix}=:\det(A)+\det(B),$$ where $(\mathbb{I}+\vec{b}\vec{b}^T)_{(2\ldots n)\times (2\ldots n)}$ is the submatrix with the first row and column deleted. Using the Laplace decomposition formula on $\det (A)$, we have the same problem with reduced dimension, so $\det(A)=1$ by induction, if $\det(B)=0.$

So we have to show that $\det(B)=0.$ Using the sum rule again, we have $$\det(B)=\begin{vmatrix} b_1^2 & b_2 b_1 & b_3 b_1& \ldots& b_n b_1\\ b_1 b_2 & 1 &0 & \ldots& 0\\ b_1 b_3&b_2 b_3&&\\ \vdots & \vdots & & (\mathbb{I}+\vec{b}\vec{b}^T)_{(3\ldots n)\times (3\ldots n)} &\\ b_1 b_n & b_2 b_n & & \end{vmatrix}+\begin{vmatrix} b_1^2 & b_2 b_1 & b_3 b_1& \ldots& b_n b_1\\ b_1 b_2 & b_2^2 &0 & \ldots& 0\\ b_1 b_3&b_2 b_3&&\\ \vdots & \vdots & & (\mathbb{I}+\vec{b}\vec{b}^T)_{(3\ldots n)\times (3\ldots n)} &\\ b_1 b_n & b_2 b_n & & \end{vmatrix}$$ It is clear that the second term is zero, as the first two columns are linearly dependent (multiples of each other). But I do not see why the first term should vanish.

Did I make a mistake or is the first term in the second equation vanishing?

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I will assume that $b \in \mathbb{R}^n$.

The result is not necessarily $1$. For example, take $b = e_1$. Then $I + e_1e_1^T = \text{diag}(2, 1, 1, \dots, 1)$ so $\det(I + e_1e_1^T) = 2$.

I'll give a hint: To find $\det(I + bb^T)$, it is convenient to diagonalize $I + bb^T$ first. $I + bb^T$ is easily diagonalized: Pick a basis $\{v_2, v_3, \dots, v_n\}$ of $\text{span}(b)^\perp$. Now consider the basis $\{b, v_2, \dots, v_n\}$ of $\mathbb{R}^n$. What is the matrix representation of $I + bb^T$ with respect to this basis?

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  • $\begingroup$ $bb^t$ is not necessarily diagonalizable if $b\in\mathbb C^n$, e.g. $b = (1, i)$ (or $(1,1)\in\mathbb F_2^2$ if finite fields are allowed). But the the last eigenvalue can always be computed using the trace. $\endgroup$ Commented Oct 17, 2021 at 20:20
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    $\begingroup$ @Justauser Yes I am assuming $b \in \mathbb{R}^n$. I will add that to the post. $\endgroup$
    – Mason
    Commented Oct 17, 2021 at 20:26
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    $\begingroup$ This is in fact OK for me, I am working with a matrix in O(n). The result seems to be $1+|b|^2$, as this is the eigenvalue when applied to $b$, the remaining entries are $1$. $\endgroup$ Commented Oct 17, 2021 at 20:33

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