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 Post subject: Determinant of a $n\times{n}$ matrixPosted: Thu Jun 09, 2016 6:38 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 442
Location: Ioannina, Greece
Calculate the determinant of the $n\times{n}$ real matrix $A=\left({\begin{array}{cccccc} x & a_1 & a_2 & \cdots & a_{n-2} & a_{n-1}\\ a_1 & x & a_2 & \cdots & a_{n-2} & a_{n-1}\\ a_1 & a_2 & x & \cdots & a_{n-2} & a_{n-1}\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a_1 & a_2 & a_3 & \cdots & x & a_{n-1}\\ a_1 & a_2 & a_3 & \cdots & a_{n-1} & x \end{array}}\right)\,.$

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Grigorios Kostakos

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 Post subject: Re: Determinant of a $n\times{n}$ matrixPosted: Thu Jun 09, 2016 6:39 am

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
We observe that if $x = a_1$ then the first row of $A$ is identical to the second row and so in this case the determinant is 0. This implies that $x-a_1$ is a factor of the determinant. Similarly, $x-a_2$ is also a factor since in the case that $x=a_2$ the second row of $A$ is identical to the third row and so on.

We deduce that $\det(A) = C(x-a_1) \cdots (x-a_{n-1})(x-k)$ for some constants $C,k$. Clearly, the coefficient of $x^n$ is equal to 1, and so $C=1$. Furthermore, the coefficient of $x^{n-1}$ must be equal to $0$. To see this just observe that we cannot choose $n$ entries of $A$ such that we have exactly one entry from each row, exactly one from each column and with exactly $n-1$ of them to be equal to the $x$'s. This means that $k = -(a_1+\cdots + a_{n-1})$ and so $\det(A) = (x-a_1) \cdots (x-a_{n-1})(x+a_1 + \cdots + a_{n-1}).$ Strictly speaking the above proof only works in the cases that the $a_i$'s are distinct as otherwise we need to be more careful with the multiplicities of the roots. However, given a matrix $A$ in which some of the $a_i$'s are equal we can find a sequence $A_m$ of matrices of the above form having distinct $a_i$'s and such as the entries of $A_m$ tend to the entries of $A$ as $m \to \infty$. Since the result is true for the $A_m$'s and since the determinant is computed using just the operations of addition and multiplication of its elements we deduce that the result must be true for $A$ as well.

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 Post subject: Re: Determinant of a $n\times{n}$ matrixPosted: Thu Jun 09, 2016 6:42 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 442
Location: Ioannina, Greece
Thank you Demetres.
A second solution is \begin{align*}
|{A}|&=\left|{\begin{array}{cccccc}
x & a_1 & a_2 & \cdots & a_{n-2} & a_{n-1}\\
a_1 & x & a_2 & \cdots & a_{n-2} & a_{n-1}\\
a_1 & a_2 & x & \cdots & a_{n-2} & a_{n-1}\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
a_1 & a_2 & a_3 & \cdots & x & a_{n-1}\\
a_1 & a_2 & a_3 & \cdots & a_{n-1} & x
\end{array}}\right|{\begin{array}{l}
{{\rm{r}}_1\rightarrow{\rm{r}}_1-{\rm{r}}_{2}}\\
{{\rm{r}}_2\rightarrow{\rm{r}}_2-{\rm{r}}_{3}}\\
{{\rm{r}}_3\rightarrow{\rm{r}}_3-{\rm{r}}_{4}}\\
{\;\vdots\hspace{1.2cm}\vdots}\\
{{\rm{r}}_{n-1}\rightarrow{\rm{r}}_{n-1}-{\rm{r}}_{n}}\\
{}\end{array}}\\
& =\left|{\begin{array}{cccccc}
x-a_1 & a_1-x & 0 & \cdots & 0 & 0\\
0 & x-a_2 & a_2-x & \cdots & 0 & 0\\
0 & 0 & x-a_3 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & x-a_{n-1} & a_{n-1}-x\\
a_1 & a_2 & a_3 & \cdots & a_{n-1} & x
\end{array}}\right|\\
& =(x-a_1)\,(x-a_2)(x-a_3)\cdots(x-a_{n-1})\left|{\begin{array}{rrrccr}
1 & -1 & 0 & \cdots & 0 & 0\\
0 & 1 & -1 & \cdots & 0 & 0\\
0 & 0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 1 & -1\\
a_1 & a_2 & a_3 & \cdots & a_{n-1} & x
\end{array}}\right|\\
& =\displaystyle\mathop{\prod}\limits_{i=1}^{n-1}(x-a_i)\,(-1)^{n+1}\,a_1\left|{\begin{array}{rrrcrr}
-1 & 0 & 0 & \cdots & 0 & 0\\
1 & -1 & 0 &\cdots & 0 & 0\\
0 & 1 & -1 &\cdots & 0 & 0\\
\vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\cdots & -1 & 0\\
0 & 0 & 0 &\cdots & 1 & -1
\end{array}}\right|\,+\\
1 & 0 & 0 & \cdots & 0 & 0\\
0 & -1 & 0 &\cdots & 0 & 0\\
0 & 1 & -1 &\cdots & 0 & 0\\
\vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\cdots & -1 & 0\\
0 & 0 & 0 &\cdots & 1 & -1
\end{array}}\right|+\ldots\,+\\
1 & -1 & 0 & \cdots & 0 & 0\\
0 & 1 & -1 &\cdots & 0 & 0\\
0 & 0 & 1 &\cdots & 0 & 0\\
\vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\cdots & 1 & 0\\
0 & 0 & 0 &\cdots & 0 & -1
\end{array}}\right|\,+\\
1 & -1 & 0 & \cdots & 0 & 0\\
0 & 1 & -1 &\cdots & 0 & 0\\
0 & 0 & 1 &\cdots & 0 & 0\\
\vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 &\cdots & 1 & -1\\
0 & 0 & 0 &\cdots & 0 & 1
\end{array}}\right|\\
&=\mathop{\prod}\limits_{i=1}^{n-1}(x-a_i)\,(-1)^{n+1}\,a_1\,(-1)^{n-1}+\mathop{\prod}\limits_{i=1}^{n-1}(x-a_i)\,(-1)^{n+2}\,a_2\,(-1)^{n-2}+\ldots\,+\\
&=\mathop{\prod}\limits_{i=1}^{n-1}(x-a_i)\,\bigl({a_1\,(-1)^{2n}+(-1)^{2n}\,a_2+\ldots+(-1)^{2n}\,a_{n-1}+(-1)^{2n}\,x}\bigr)\\
&=\mathop{\prod}\limits_{i=1}^{n-1}(x-a_i)\,\bigl({a_1+a_2+\ldots+a_{n-1}+x}\bigr)\\
\end{align*}

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Grigorios Kostakos

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