2

I have the following matrix with dimensions $n \times n$ ($z\in\mathbb R$ and $n\in\mathbb N$):

$$\begin{bmatrix} z & 1 & 1 & \cdots & 1 \\ 1 & z & 1 & \cdots & 1 \\ 1 & 1 & z & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & z \\ \end{bmatrix}$$

I need to find the determinant of it. I came up with the term $(z-1)^{n-1} (z+(n-1))$ after finding out the determinants for $n=1, n=2, n=3, n=4$, etc. Now my question is, how can I proof it that this

Thanks in advance.

Lorenzo B.
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jamyolive
  • 57

5 Answers5

4

There is several ways of doing it:

  • Using an induction.

  • Using a trick, let: $$J_n=\begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 1 \\ \end{bmatrix}$$ then $J_n$ is of rank $1$. So there exists $\lambda$ such that: $$J_n \sim \begin{bmatrix} \lambda & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \\ \end{bmatrix}$$ More over $\operatorname{tr}(J_n)=n$, so: $$J_n \sim \begin{bmatrix} n & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \\ \end{bmatrix}$$ And: $$J_n+(z-1)I_n \sim \begin{bmatrix} n+(z-1) & 0 & 0 & \cdots & 0 \\ 0 & (z-1) & 0 & \cdots & 0 \\ 0 & 0 & (z-1) & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & (z-1) \\ \end{bmatrix}$$ and in particular: $$\det(J_n+(z-1)I_n))=(z-1)^{n-1} (z+(n-1))$$

Delta-u
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1

Your matrix is of the form $(z-1) I_n +\boldsymbol{1}^{\rm T}_n \boldsymbol{1}_n$ (a diagonal matrix plus a rank-one matrix.) There is information available for the eigenvalues.

Rócherz
  • 3,976
1

Subtract the first from the rest

$$\begin{bmatrix} z & 1-z & 1-z & \cdots & 1-z \\ 1 & z-1 & 0 & \cdots & 0 \\ 1 & 0 & z-1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots & z-1 \\ \end{bmatrix}$$ Then expand along the top row, and then go expand the submatrix along the left side.

1

Let's give an alternative solution for fun.

It's a circulant matrix, so its determinant is

\begin{align} \begin{vmatrix} z & 1 & 1 & \cdots & 1 \\ 1 & z & 1 & \cdots & 1 \\ 1 & 1 & z & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & z \\ \end{vmatrix} &= \prod_{k=0}^{n-1}(z+\zeta^k+\zeta^{2k}+\cdots+\zeta^{k(n-1)}) \\ &= \prod_{k=0}^{n-1}((z-1) + 1+\zeta^k+\zeta^{2k}+\cdots+\zeta^{k(n-1)}) \\ &= ((z-1)+n) (z-1)^{n-1} \end{align}

where $\zeta$ is the $n$-th root of unity.

GNUSupporter 8964民主女神 地下教會
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1

Here is another trick:

  • Add every column to the 1st column, so you get: $$\begin{bmatrix} z + n-1 & 1 & 1 & \cdots & 1 \\ z + n-1 & z & 1 & \cdots & 1 \\ z + n-1 & 1 & z & \cdots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ z + n-1 & 1 & 1 & \cdots & z \\ \end{bmatrix}$$
  • Take the first line away from all the lines: $$\begin{bmatrix} z + n-1 & 1 & 1 & \cdots & 1 \\ 0 & z-1 & 0 & \cdots & 0 \\ 0 & 0 & z-1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & z-1 \\ \end{bmatrix}$$

Now the resultant matrix is upper triangular, so the determinant is simply the product of the elements on the diagonal. Since all we did was add lines and columns, it is the determinant we're looking for.

Quantaliinuxite
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