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$A \in \mathbb R^{n,n}$

$A_n=\begin{bmatrix} 1&2&3&\dots&n\\x&1&2&\dots&n-1\\x&x&1&\dots&n-2\\\vdots&\vdots&\vdots&\ddots&\vdots\\x&x&x&\dots&1\end{bmatrix}$

How to calculate $\det A_n$ for $x\in\mathbb R$?

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Can you see a pattern? Focus on how you would manually compute the determinant of a $3 X 3$ matrix. Then extend forwards. – Torsten Hĕrculĕ Cärlemän Dec 19 '13 at 19:28
For $3x3$ it is $3x^2-3x+1$, for $4x4 -4x^3+6x^2-4x+1$. – Josh Dec 19 '13 at 19:40
You can generalize: $A_n=(1-x)^n-(-x)^n$. But I am still looking for the proof :-) One thing is obvious though: $A_n$ is a polynomial of degree $(n-1)$ in $x$, with leading coefficient $n$. – Jean-Claude Arbaut Dec 19 '13 at 19:42
@Josh: seeing that, investigate what successive powers of $(x-1)$ look like. – John Dec 19 '13 at 19:43
Have you looked at determinants of minors? – user116017 Dec 19 '13 at 20:02
up vote 6 down vote accepted

I denote $A_n$ for the determinant of your matrix of size $n\times n$, rather than the matrix itself:

$$A_n=\begin{vmatrix} 1&2&3&\dots&n\\x&1&2&\dots&n-1\\x&x&1&\dots&n-2\\\vdots&\vdots&\vdots&\ddots&\vdots\\x&x&x&\dots&1\end{vmatrix}$$

And we will prove that


First use row operations to write

$$A_n=\begin{vmatrix} 1-x&1&1&\dots&1\\ 0&1-x&1&\dots&1\\ 0&0&1-x&\dots&1\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ x&x&x&\dots&1\end{vmatrix}$$

To achieve this, you subtract the second row from the first, then the third from the second, etc. Only the last line remains untouched.

From this new determinant, you get immediately by developping along the first column:


Where $B_n$ is this determinant (of size $n \times n$):

$$B_n=\begin{vmatrix} 1&1&1&\dots&1&1\\ 1-x&1&1&\dots&1&1\\ 0&1-x&1&\dots&1&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&1-x&1\end{vmatrix}$$

Now, use column operations to rewrite your determinant$B_n$ as

$$B_n=\begin{vmatrix} 0&0&0&\dots&0&1\\ -x&0&0&\dots&0&1\\ 0&-x&0&\dots&0&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&-x&1\end{vmatrix}$$

To achieve this, subtract the last column from the butlast, and continue likewise to the left.

This is almost the determinant of a companion matrix, you can rewrite (factoring out $(-x)$ from the first $n-1$ columns):

$$B_n=(-x)^{n-1}\begin{vmatrix} 0&0&0&\dots&0&1\\ 1&0&0&\dots&0&1\\ 0&1&0&\dots&0&1\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\dots&1&1\end{vmatrix}$$

The characteristic polynomial of this companion matrix is $(-1)^n(-1-t-\dots+t^n)$, and it's determinant is $(-1)^{n-1}$. You have thus $B_n=(-1)^{n-1}(-x)^{n-1}=x^{n-1}$.



Now, we can prove by induction our claim, that is


If it's true for $n-1$, then


Then, using our previous relation,

$$A_n=(1-x)\left((1-x)^{n-1}-(-x)^{n-1}\right) + (-x)^{n-1}$$ $$=(1-x)^n-(1-x)(-x)^{n-1}+(-x)^{n-1}$$ $$=(1-x)^n-(-x)^n$$

Then the property is true for $n$. Since it's trivially true for $n=1$ (then $A_1=1$), it's true for all $n$.

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induction rules! – Torsten Hĕrculĕ Cärlemän Dec 20 '13 at 2:50

The derivation is a bit easier if you know the formula for the determinant of a circulant matrix. First, by applying appropriate row operations, we get $$ \det(A_n)=\begin{vmatrix} 1-x&1&1&\dots&1&1\\ 0&1-x&1&\dots&1&1\\ 0&0&1-x&\dots&1&1\\ \vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\ 0&0&\cdots&0&1-x&1\\ x&x&x&\dots&x&1\end{vmatrix}. $$ Then, by applying some column operations, we get $$ \det(A_n)=\begin{vmatrix} 1-x&x\\ &1-x&x\\ &&\ddots&\ddots\\ &&&1-x&x\\ x&&&&1-x\end{vmatrix}. $$ This is the determinant of a circulant matrix. If we put $\omega=e^{2\pi i/n}$, we have $$ \det(A_n) =\prod_{j=0}^{n-1}(1-x+x\omega^j) =(-x)^n\prod_{j=0}^{n-1}\left(\frac{1-x}{-x}-\omega^j\right) =(-x)^n\left[\left(\frac{1-x}{-x}\right)^n-1\right], $$ i.e. $\det(A_n)=(1-x)^n-(-x)^n$.

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