Calculating a determinant.

Question

$D_n$=\begin{vmatrix} a & 0 & 0 & \cdots &0&0& n-1 \\ 0 & a & 0 & \cdots &0&0& n-2\\ 0 & 0 & a & \ddots &0&0& n-3 \\ \vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\ 0 & \cdots & \cdots & \cdots &0&a&1 \\ n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\ \end{vmatrix}

I tried getting the eigenvalues for A =

\begin{vmatrix} 0 & 0 & 0 & \cdots &0&0& n-1 \\ 0 & 0 & 0 & \cdots &0&0& n-2\\ 0 & 0 & 0 & \ddots &0&0& n-3 \\ \vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\ \vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\ 0 & \cdots & \cdots & \cdots &0&0&1 \\ n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\ \end{vmatrix}

For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $

$m(0)>=n-2$

However, I was not able to determine the other eigenvalues.

Testing for different values of n :

for $n=2$ :

$D_2 = a^2-1$

for $n=3$ :

$D_3 = a^3 -5a$

$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .

However I'm aware that testing for different values of $n$ is not enough to generalize the formula.

Thanks in advance.

Answer 1

Expand with respect to the first line: the term obtained with $a$ is $aD_{n-1}$. For the second one, we get $(-1)^{n+1}$ times a determinant that can be expanded with respect to the first column. This lead to the recurrence relation $$ D_n=aD_{n-1}-\left(n-1\right)^2a^{n-2}. $$ Letting $b_n:=a^{-n}D_n$ for $a\neq 0$ allows to derive and easier recurrence relation whose resolution shows that the formula mentioned in the opening post, namely, $$ D_n=a^n - a^{n-2}\sum_{i=1}^{n-1}i^2, $$ is correct.

Answer 2

Develop with respect to the first column. Then $$ \begin{aligned} D_n &= aD_{n-1} -(n-1)\cdot (n-1)\cdot a^{n-2} \\ &=aD_{n-1}-(n-1)^2a^{n-2}\ . \end{aligned} $$ This recursion, together with $D_1=a$ gives for $n\ge 2$ the solution $$ D_n= (a^2-(1^2+2^2+\dots+(n-1)^2)a^{n-2}\ . $$ (The sum in the first factor has a closed formula.)

Answer 3

To find other values of $x$ besides $0$ for which $$ \det \begin{bmatrix} x & 0 & 0 & \cdots & n-1 \\ 0 & x & 0 & \cdots & n-2 \\ 0 & 0 & x & \cdots & n-3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ n-1 & n-2 & n-3 & \cdots & x \end{bmatrix} = 0 $$ consider taking $\frac{n-1}{x}$ times the first row, plus $\frac{n-2}{x}$ times the second row, plus $\frac{n-3}{x}$ times the third, and so on, on the grounds that you will get a row vector which is equal to the last row in its first $n-1$ entries.

Its last entry is going to be $\frac{(n-1)^2 + (n-2)^2 + \dots + 2^2 + 1^2}{x} = \frac{n(n-1)(2n-1)}{6x}$. So if this happens to be equal to $x$, then the last row is a linear combination of the other rows, which means the determinant is $0$.

There are two values of $x$ for which this works: the two square roots of $\frac{n(n-1)(2n-1)}{6}$. This gives us the last two eigenvalues: the whole list is $$ a, a, \dots, a, a - \sqrt{\frac{n(n-1)(2n-1)}{6}}, a + \sqrt{\frac{n(n-1)(2n-1)}{6}} $$ and their product is $$ a^{n-2}\left(a^2 - \frac{n(n-1)(2n-1)}{6}\right) $$ as you conjectured.

Answer 4

You can just go for calculating the characteristic polynomial $\chi_{-A}$ of minus the matrix $A$ (the one at $a=0$), then your determinant will be $\chi_{-A}[a]$. As you already found that the rank of $A$ is$~2$ (if $n\geq2$; otherwise it is $0$) the coefficients of $\chi_{-A}$ in all degrees less than $n-2$ are zero (as its coefficient of degree $n-r$ is the sum of all principal $r$-minors of$~A$. Since $A$ has zero trace, one has $$\chi_{-A}=X^n+0x^{n-1}+c_nX^{n-1}$$ where $c_n$ is the sum of all principal $2$-minors of$~A$, which is easily seen to be $$ c_n=-\sum_{k=0}^{n-1}k^2=-\frac{2n^3-3n^2+n}6.$$ Therefore $\det(D_n)=a^n+c_na^{n-2}=a^{n-2}(a^2+c_n)$, as you guessed (with $c_n\leq0$, so the roots of $\chi_{-A}$ are real: $\pm\sqrt{-c_n}$ and $0$ with multiplicity $n-2$).

Answer 5

It is maybe a bit easier to use the matrix determinant formula. If you rewrite your matrix as

$$ M = aI +UV^T $$ where $UV^T$ is $$ \begin{bmatrix} (n-1) &0\\ (n-2) &0\\ \vdots&0\\ 1 &0\\ 0&1 \end{bmatrix} \begin{bmatrix} 0&0&\cdots&1\\ (n-1)&(n-2)&\cdots&0 \end{bmatrix} $$

Then, from $$ \operatorname{det}({\mathbf{A}}+{\mathbf {UV}}^{{\mathrm {T}}})=\operatorname {det}({\mathbf{I}}+{\mathbf {V}}^{{\mathrm {T}}}{\mathbf {A}}^{{-1}}{\mathbf {U}})\operatorname {det}({\mathbf {A}}). $$ and $\mathbf{A}^{-1}=\frac1aI$, we have only a $2\times 2$ determinant to evaluate: $$ \det(M) = \left|\begin{matrix}1&\frac{1}{a}\\\displaystyle\frac{1}{a}\sum_{i=1}^{n-1}i^2&1\end{matrix}\right|a^n = (1-\frac{1}{a^2}\sum_{i=1}^{n-1}i^2)a^n $$