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I'm trying to solve a problem from a textbook. This is a listed problem under the first chapter of the textbook (the book starts with introducing determinant without matrix theory) so the only things that are allowed to use are properties of determinant and calculus.

Problem: show that $$D=\begin{vmatrix}a_1&\lambda&\lambda&...&\lambda\\\lambda&a_2&\lambda&...&\lambda\\\lambda&\lambda&a_3&...&\lambda\\...&...&...&...&...\\\lambda&\lambda&\lambda&...&a_n\end{vmatrix}=\varphi(\lambda)-\lambda\frac{d\varphi}{d\lambda}$$ where $\varphi(\lambda)=(a_1-\lambda)(a_2-\lambda)\ ...\ (a_n-\lambda)$

P'bD_KU7B2
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  • Take $A=\operatorname{diag}(a_1-\lambda,\dots,a_n-\lambda)$ and $v=w=(1,\dots,1)^T$ in this question. – Hans Lundmark May 23 '21 at 11:16
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    Using the formula searching tool https://approach0.xyz/ , I found different places where this question was treated, in particular, I found an straightforward derivation (in the question itself, in blue) in https://math.stackexchange.com/q/3476433 It uses only determinantal properties. – Jean Marie May 23 '21 at 12:30
  • No comment on my comment ? Besides, a little remark: the formula you obtain has the form of a Wronskian $W(\lambda,\varphi(\lambda))$. I have no idea if this form is useful... maybe in connection with a treatment by a differential equation... – Jean Marie May 23 '21 at 22:59
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    @JeanMarie, I think you are right, the book which this problem comes from treats differential equations as a prerequisites and devote couple sections on system of linear differential equation – P'bD_KU7B2 May 24 '21 at 11:24

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Edit: after checking through my steps several times, I got the answer:

[Step 1]: Let $a_1=\lambda(1+\sigma_1),\ a_2=\lambda(1+\sigma_2),\ ...,\ a_n=\lambda(1+\sigma_n)$ $$D=\begin{vmatrix}\lambda(1+\sigma_1)&\lambda&\lambda&...&\lambda\\\lambda&\lambda(1+\sigma_2)&\lambda&...&\lambda\\\lambda&\lambda&\lambda(1+\sigma_3)&...&\lambda\\...&...&...&...&...\\\lambda&\lambda&\lambda&...&\lambda(1+\sigma_n)\end{vmatrix}\\=\lambda^n\begin{vmatrix}(1+\sigma_1)&1&1&...&1\\1&(1+\sigma_2)&1&...&1\\1&1&(1+\sigma_3)&...&1\\...&...&...&...&...\\1&1&1&...&(1+\sigma_n)\end{vmatrix}\\=\lambda^n\begin{vmatrix}(1+\sigma_1)&1&1&...&1\\-\sigma_1&\sigma_2&0&...&0\\-\sigma_1&0&\sigma_3&...&0\\...&...&...&...&...\\-\sigma_1&0&0&...&\sigma_n\end{vmatrix}\\=-\sigma_1\lambda^n\begin{vmatrix}\frac{(1+\sigma_1)}{-\sigma_1}&1&1&...&1\\1&\sigma_2&0&...&0\\1&0&\sigma_3&...&0\\...&...&...&...&...\\1&0&0&...&\sigma_n\end{vmatrix}$$ [Step 2]: apply Laplace expansion for the 1st column $$D=-\sigma_1\lambda^n[(-1)^{1+1}(\sigma_2\sigma_3...\sigma_n)\frac{(1+\sigma_1)}{-\sigma_1}+(-1)^{1+2}(-1)^{1+1}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_2}+(-1)^{1+3}(-1)^{1+2}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_3}+(-1)^{1+4}(-1)^{1+3}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_4}+...+(-1)^{1+n}(-1)^{n}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_n}]\\=\lambda^n(1+\frac{1}{\sigma_1}+\frac{1}{\sigma_2}+\frac{1}{\sigma_3}+...+\frac{1}{\sigma_n})(\sigma_1\sigma_2\sigma_3...\sigma_n)$$

[Step 3]: Because $a_1=\lambda(1+\sigma_1),\ a_2=\lambda(1+\sigma_2),\ ...,\ a_n=\lambda(1+\sigma_n)$ $$\varphi(\lambda)=\lambda^n(\sigma_1\sigma_2...\sigma_n)\ \ \ \textbf{[1]}$$ $$\lambda\frac{d\varphi(\lambda)}{d\lambda}=\lambda\frac{d}{d\lambda}e^{ln(\varphi(\lambda))}=\lambda(\frac{1}{\lambda-a_1}+\frac{1}{\lambda-a_1}+...+\frac{1}{\lambda-a_n})\varphi(\lambda)=(-\frac{1}{\sigma_1}-\frac{1}{\sigma_2}-...-\frac{1}{\sigma_n})\varphi(\lambda)\ \ \ \textbf{[2]}$$ $$\varphi(\lambda)-\lambda\frac{d\varphi}{d\lambda}=\lambda^n(1+\frac{1}{\sigma_1}+\frac{1}{\sigma_2}+...+\frac{1}{\sigma_n})(\sigma_1\sigma_2...\sigma_n)\ \ \ \textbf{[3]}$$

P'bD_KU7B2
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