Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For $n \geq 2$, find the determinant of $A_{n}=\begin{bmatrix} n & 1 & 1 &\ldots &1 \\ 1 & n & 1 &\ldots &1 \\ 1 & 1 & n &\ldots &1 \\ \vdots & \vdots &\vdots & \ddots & \vdots\\ 1 & 1 & 1 &\ldots & n \end{bmatrix}_{n \times n} $

One can deduce the determinant of $A_{n}=(n-1)^{n-1}(2n-1)$ by taking $n=2,3,4...$. I solve it as follows but I found it is not elegant. Any better approach?

Here is my attempt:

Let's find the eigenvalues of $A_{n}$ by solving $det(\lambda I-A_{n})=0$ because $det(A_{n})=\lambda_{1}\lambda_{2}...\lambda_{n}$ where $\lambda_{i}$ are eigenvalues of $A_{n}$. Express $A_{n}=(n-1)I+J$ where $I$ is the n-by-n identity matrix while $J$ is an n-by-n matrix with all entries equal to 1. Denote $B=(n-1)I$ so that $$\lambda I-A_{n}=\lambda I-(B+J)=(\lambda I - B)(I-(\lambda I - B)^{-1}J)$$ Since $det(XY)=det(X)det(Y)$ for square matrices $X$ and $Y$ $$det(\lambda I-A_{n})=det(\lambda I - B)det(I-(\lambda I - B)^{-1}J)$$ Since $det(kX)=k^{n}det(X)$ for a scalar k and $det(I)=1$ $$det(\lambda I-B)=\det(\lambda I - (n-1)I)=(\lambda-(n-1))^{n}det(I)=(\lambda-(n-1))^{n}$$ If $u$ be a column vector of one's in $\mathbb{R}^{n}$, then $uu^{T}=J$ so that: $$det(I-(\lambda I - B)^{-1}J)=det(I-(\lambda I - B)^{-1}uu^{T}) $$ By Sylvester's Determinant Theorem: \begin{equation*} \begin{split} det(I-(\lambda I - B)^{-1}J)&=&det(I-(\lambda I - B)^{-1}uu^{T})=det(I-u^{T}(\lambda I - B)^{-1}u)\\ &=&1-u^{T}(\lambda I - B)^{-1}u=1-u^{T}(\lambda-(n-1))^{-1}Iu\\ &=&1-(\lambda-(n-1))^{-1}u^{T}Iu\\ &=&1-\frac{u^{T}u}{\lambda-(n-1)}=1-\frac{n}{\lambda-(n-1)}\\ &=&\frac{\lambda-(2n-1)}{\lambda-(n-1)} \end{split} \end{equation*} This yields $$det(\lambda I-A_{n})=(\lambda-(n-1))^{n}\frac{\lambda-(2n-1)}{\lambda-(n-1)}=(\lambda-(n-1))^{n-1}(\lambda -(2n-1))=0$$ The eigenvalues of $A_{n}$ are $n-1$ (with algebraic multiplicity of $n-1$) and $2n-1$ (with algebraic multiplicity of 1). Thus, $det(A_{n}=(n-1)^{n-1}(2n-1)$

I wonder if it is possible to obtain the determinant without solving for the eigenvalues. Or at least without using the Sylvester's determinant theorem....

share|improve this question
2  
This question is a specialisation of math.stackexchange.com/questions/86644/… –  Colin McQuillan Jul 25 '12 at 21:08

5 Answers 5

up vote 4 down vote accepted

Assuming we are working over a field of characteristic 0:

Add all other rows to the bottom row, so that the bottom row has $2n-1$ in each spot. Multiply the bottom row by $\frac{1}{2n-1}$ so it is all $1$s, then subtract it from each other row. This gives us the following matrix:

$$\begin{pmatrix} n-1 & 0 & \cdots & 0 \\ 0 & n-1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 2n-1 & 2n-1 & \cdots & 2n-1 \end{pmatrix}$$

This is lower triangular, so it has determinant equal to the product of the diagonal entries, or $(n-1)^{n-1}(2n-1)$.

share|improve this answer

One can find the eigenvalues much more easily than this:

Note that $A = B + (n-1)I$, where $B$ is the matrix all of whose entries are $1$, and $I$ is the identity.

The rank of $B$ is $1$ so the nullspace is of dimension $n-1$, and since $B^2=nB$, then $B$ satisfies the polynomial $x(x-n)=0$.

Therefore, the characteristic polynomial must be $x^{n-1}(x-n)$; if the characteristic of the ground field divides $n$, then this is $x^n$.

If the characteristic of the ground field does not divide $n$, then eigenvalues of $B$ are $0$, with multiplicity $n-1$, and $n$, with multiplicity $1$. Hence, the eigenvalues of $A$ are $0+(n-1) = n-1$, with multiplicity $(n-1)$, and $n+(n-1) = 2n-1$, with multiplicity $1$. So the determinant is $(n-1)^{n-1}(2n-1)$, as you obtained.

If we are working over a field whose characteristic does divide $n$, then $A = B-I$. Thus, the minimal polynomial of $A$ is $(x-1)^2$, so the characteristic polynomial is $(x-1)^n$. Again we obtain that the determinant is just $(-1)^n$, same as we would get above.

share|improve this answer
    
Thanks for the answer.What is "characteristic of the ground field"? (Please forgive my math deficiency). –  Yongkai Jul 25 '12 at 21:43
    
The characteristic of a field is the smallest positive integer $n$, if one exists, such that $$\underbrace{1+1+\cdots+1}_{n\text{ summands}} = 0.$$If you are working over $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$, you don't need to worry about characteristic; it's $0$ and no problems occur. But when one works over more general fields (e..g., if you work modulo $2$, or modulo $7$), then certain pathologies can occur. If you don't know what "characteristic of the ground field" means, chances are it doesn't matter to you what it means. –  Arturo Magidin Jul 26 '12 at 0:18

Let $B$ be the matrix with all $-1's$. Since the $rank(B)=1$, it follows that $\lambda=0$ is an eigenvalue of $B$ of multiplicity $n-1$.

Also, it is trivial to see that $\lambda=-n$ is an eigenvalue of $B$, since in this case the rows add to $0$. This proved that the characteristic polynomial of $B$ is

$$\det (xI-B)= x^{n-1}(x+n) \,.$$

Set $x=n-1$ and you are done....

Sigh, Arturo beat me to the answer :)

share|improve this answer

Write $A = (n-1)I+uu'$ instead, where $u'=[1 1 \dots1].$ Using the same Sylvester's Determinant Theorem you can write $\det(A)=\det((n-1)I)(1+u'u/(n-1) = (n-1)^n(1-\frac{n}{n-1})=(n-1)^{n-1}(2n-1)$

share|improve this answer
    
well, this uses Sylvester's theorem... –  karakfa Jul 25 '12 at 21:31

For this matrix you can quite easily see what all the eigenvalues and eigenvectors are. If $e_i$ is the vector with entry $1$ in position $i$ and $0$ elsewhere then $e_1 + \ldots + e_n$ is an eigenvector with eigenvalue $2n-1$ and the vectors of the form $e_i-e_j$ span an $n-1$ dimensional eigenspace with eigenvector $n-1$.

share|improve this answer
    
You have "eigenvalue" and "eigenvector" mixed up; $e_1+\cdots+e_n$ is a vector, hence cannot be an eigen value. And $2n-1$ is a scalar, so it is the eigenvalue, not the eigenvector. –  Arturo Magidin Jul 25 '12 at 21:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.