# Determinant $\mathbb{I}+\vec{b}\vec{b}^T$.

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?

• Have you tried a basis in which $b_i=|b|\delta_{i1}$?
– J.G.
Commented Oct 17, 2021 at 8:56
• Commented Oct 17, 2021 at 9:22
• by matrix determinant lemma, $\det(I+bb^T) = 1 + b^Tb$. Commented Oct 17, 2021 at 20:35

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?
• $bb^t$ 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. Commented Oct 17, 2021 at 20:20
• @Justauser Yes I am assuming $b \in \mathbb{R}^n$. I will add that to the post. Commented Oct 17, 2021 at 20:26
• 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 $1$. Commented Oct 17, 2021 at 20:33