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I am trying to show that

$$\det \left(I_{n} + \begin{bmatrix} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{bmatrix} \begin{bmatrix}b_{1}&b_{2}&\cdots&b_n \end{bmatrix} \right) = 1 + \sum_{m=1}^n a_{m}b_{m}$$

I know I have to use the fact that $$\ M=QDQ^{T}$$

I can see that with a diagonal matrix its determinant its equal to its trace, but I am having a hard time figuring out what to do with that identity and that 1?

Melvin Gonzalez
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  • Where I_{n} is the unit identity matrix of dimesion nxn – Melvin Gonzalez Sep 14 '19 at 18:03
  • Apparently, either $\ \begin{bmatrix}a_1\a_2\\vdots\a_n\end{bmatrix}\ $ should be $\ \begin{bmatrix}b_1\b_2\\vdots\b_n\end{bmatrix}\ $, or $\ \begin{bmatrix}a_1&a_2&\dots&a_n\end{bmatrix}\ $ should be $\ \begin{bmatrix}b_1&b_2&\dots&b_n\end{bmatrix}\ $. Which is it? While the answer is the same in either case, its derivation will be slightly different. – lonza leggiera Sep 15 '19 at 03:54
  • I have fixed the second vector, it should be the "b" vector – Melvin Gonzalez Sep 15 '19 at 19:10

1 Answers1

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Hint: Assuming the problem is to find the determinant of the matrix $\ I_n + ab^\top\ $, the determinant can be expressed as a product of that matrix's eigenvalues. One of its eigenvectors is $\ a\ $. What is the corresponding eigenvalue? There are $\ n-1\ $ eigenvectors that are perpendicular to $\ b\ $. What are the corresponding eigenvalues?

lonza leggiera
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  • https://math.stackexchange.com/questions/420198/proving-the-relation-deti-xyt-1-xty – Melvin Gonzalez Sep 15 '19 at 19:12
  • I am not sure how to approach this solving for the eigenvalues. Just trying for a as a 2x1 and b as a 1x2, and adding the identity matrix, it becomes messy, not sure I can zero out the off bottom diagonal. The link above helped me a bit, specially Marc's and leonbloy's suggestions, i just cant put the pieces together. – Melvin Gonzalez Sep 15 '19 at 19:15
  • $$\left(I_n + ab^\top\right)a = \left(1+ b^\top a\right)a\ ,$$ so the corresponding eigenvalue is $\ 1+ b^\top a\ $.

    If $\ v\perp b\ $, then $$\left(I_n + ab^\top\right)v = v\ ,$$ so the corresponding eigenvalue is $1$. The product of the eigenvalues (i.e. the determinant) is therefore $\ 1^{n-1}\left(1+ b^\top a\right)=1+\sum_\limits{m=1}^n a_m b_m\ $.

     – lonza leggiera Sep 16 '19 at 00:53