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I have the following task: Let $K$ be a field, $n \in \mathbb{N}$ and $a,b \in K^n$. Evaluate the determinant of the following matrix:

$$\begin{pmatrix} a_1+b_1 & b_2 & b_3 & \dots & b_n& \\ b_1 & a_2 + b_2 & b_3 & \dots & b_n \\ b_1 & b_2 & a_3 + b_3 & \dots & b_n \\ \vdots & \vdots & \vdots& & \vdots \\ b_1 & b_2 & b_3 &\dots & a_n + b_n \end{pmatrix}$$

What I did was expanding it as follows using the Laplace expansion:

$$\det A =(a_1 + b_1) \det\begin{pmatrix} a_2 + b_2 &\dots& b_n \\ b_2 &\dots& b_n \\ \vdots & & \vdots \\ b_2 & \dots & a_n + b_n \end{pmatrix} - b_2 \det\begin{pmatrix} b_1 & b_3 &\dots& b_n \\ b_1 & a_3 + b_3 &\dots& b_n \\ \vdots & \vdots & & \vdots \\ b_1 & b_3 & \dots & a_n + b_n \end{pmatrix} + \ b_3 \det\begin{pmatrix} \dots \end{pmatrix} - \dots (-1)^{n+1}b_n \det\begin{pmatrix} b_1 &\dots& b_{n-1} \\ b_1 &\dots& b_{n-1} \\ \vdots & & \vdots \\ b_1 & \dots & b_{n-1} \end{pmatrix}$$

And before I expand the rest of those determinants and fill 20 papers with a's and b's I'd like to ask for advice. Is this the right way? And when I think about it I don't really see any simplification that is possible when I have finally expanded everything to a point where I could use Cramers rule. It just came to my mind that I could also expand using the Lapace rule by iterating through the rows instead of the columns. By doing that I'd be able to factor out all of those $b_1$...

NOTE: I am not allowed to use the Sylverster Determinant Theorem

Thank you very much for your help.

FunkyPeanut

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3 Answers 3

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Note that your matrix can be written as $$\text{diag}(a_1,a_2,\ldots,a_n) + \begin{bmatrix} 1\\1\\1\\ \vdots\\1\end{bmatrix} \begin{bmatrix} b_1 & b_2 & \cdots & b_n\end{bmatrix}$$ This is a rank $1$ update to a diagonal matrix, whose determinant can be computed using the Sylvester determinant theorem: $$\det(I+UV^T) = \det(I+V^TU)$$ I will leave the rest to you since the question is tagged as a homework problem.

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    $\begingroup$ Thanks for that. I should note that I am not allowed to use that theorem, since the purpose of that expansion is to proof the theorem in a next step.... $\endgroup$
    – FunkyPeanut
    Apr 13, 2014 at 13:19
  • $\begingroup$ @FunkyPeanut Your question never says that and this will not be a proof of the theorem since the theorem is more general and is applicable any rank $k$ update to the identity matrix. $\endgroup$
    – user141421
    Apr 13, 2014 at 13:21
  • $\begingroup$ @FunkyPeanut That is a reduced rank $1$ version of the Sylvester theorem. $\endgroup$
    – user141421
    Apr 13, 2014 at 13:30
  • $\begingroup$ Sorry I've messed a few things up in the comment system... So if that is some type of the sylverser theorem which I have to prove, how am I supposed to use it for showing something that I want to use for the proof? I've also checked our script which never mentions that theorem unfortunately. I really think that I am not allowed to use it ... Sorry $\endgroup$
    – FunkyPeanut
    Apr 13, 2014 at 13:39
  • $\begingroup$ Sorry I don't get it.. Could you please provide another tip? I don't see how that could help me any further... $\endgroup$
    – FunkyPeanut
    Apr 13, 2014 at 14:52
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$$\begin{array}{ll} D_n&=\begin{vmatrix} a_n+b_n & b_{n-1} & b_{n-2} & \dots & b_1& \\ b_n & a_{n-1} + b_{n-1} & b_{n-2} & \dots & b_1 \\ b_n & b_{n-1} & a_{n-2} + b_{n-2} & \dots & b_1 \\ \vdots & \vdots & \vdots& & \vdots \\ b_n & b_{n-1} & b_{n-2} &\dots & a_1 + b_1 \end{vmatrix}\\ &=\begin{vmatrix} a_n & b_{n-1} & b_{n-2} & \dots & b_1& \\ 0 & a_{n-1} + b_{n-1} & b_{n-2} & \dots & b_1 \\ 0 & b_{n-1} & a_{n-2} + b_{n-2} & \dots & b_1 \\ \vdots & \vdots & \vdots& & \vdots \\ 0 & b_{n-1} & b_{n-2} &\dots & a_1 + b_1 \end{vmatrix}\\ &+\begin{vmatrix} b_n & b_{n-1} & b_{n-2} & \dots & b_1& \\ b_n & a_{n-1} + b_{n-1} & b_{n-2} & \dots & b_1 \\ b_n & b_{n-1} & a_{n-2} + b_{n-2} & \dots & b_1 \\ \vdots & \vdots & \vdots& & \vdots \\ b_n & b_{n-1} & b_{n-2} &\dots & a_1 + b_1 \end{vmatrix}\\ &=a_nD_{n-1}+b_n\begin{vmatrix} 1 & b_{n-1} & b_{n-2} & \dots & b_1& \\ 1 & a_{n-1} + b_{n-1} & b_{n-2} & \dots & b_1 \\ 1 & b_{n-1} & a_{n-2} + b_{n-2} & \dots & b_1 \\ \vdots & \vdots & \vdots& & \vdots \\ 1 & b_{n-1} & b_{n-2} &\dots & a_1 + b_1 \end{vmatrix}\\ &=a_nD_{n-1}+b_n\begin{vmatrix} 1 & 0 & 0 & \dots & 0 & \\ 1 & a_{n-1} & 0 & \dots & 0 \\ 1 & 0 & a_{n-2} & \dots & 0 \\ \vdots & \vdots & \vdots& & \vdots \\ 1 & 0 & 0 &\dots & a_1 \end{vmatrix}\\ &= a_n D_{n-1}+b_n\prod_{k=1}^{n-1}a_k \end{array}$$

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In the case $\Bbb K = \Bbb R$, you can use$$M_n(t)=\begin{vmatrix} a_1+tb_1 & tb_2 & tb_3 & \dots & tb_n& \\ tb_1 & a_2 + tb_2 & tb_3 & \dots & tb_n \\ tb_1 & tb_2 & a_3 + tb_3 & \dots & tb_n \\ \vdots & \vdots & \vdots& & \vdots \\ tb_1 & tb_2 & tb_3 &\dots & a_n + tb_n \end{vmatrix}$$

$$\begin{array}{ll}M_n'(t)&=\begin{vmatrix} b_1 & tb_2 & tb_3 & \dots & tb_n& \\ b_1 & a_2 + tb_2 & tb_3 & \dots & tb_n \\ b_1 & tb_2 & a_3 + tb_3 & \dots & tb_n \\ \vdots & \vdots & \vdots& & \vdots \\ b_1 & tb_2 & tb_3 &\dots & a_n + tb_n \end{vmatrix}\\&+\begin{vmatrix} a_1+tb_1 & b_2 & tb_3 & \dots & tb_n& \\ tb_1 & b_2 & tb_3 & \dots & tb_n \\ tb_1 & b_2 & a_3 + tb_3 & \dots & tb_n \\ \vdots & \vdots & \vdots& & \vdots \\ tb_1 & b_2 & tb_3 &\dots & a_n + tb_n \end{vmatrix}\\ &+\dots\\ &+\begin{vmatrix} a_1+tb_1 & tb_2 & tb_3 & \dots & b_n& \\ tb_1 & a_2 + tb_2 & tb_3 & \dots & b_n \\ tb_1 & tb_2 & a_3 + tb_3 & \dots & b_n \\ \vdots & \vdots & \vdots& & \vdots \\ tb_1 & tb_2 & tb_3 &\dots & b_n \end{vmatrix}\end{array}$$

$$M_n(t)=M_n(0)+\int_0^tM_n'(u)\, du$$

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  • $\begingroup$ Wow - thanks for that. I think I can't assume that $K = \mathbb{R}$... $\endgroup$
    – FunkyPeanut
    Apr 13, 2014 at 14:13

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