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.

How do I calculate the determinant of the following $n\times n$ matrices

$ \left[ \begin {matrix} 0 & 1 & \ldots & 1 \\ 1 & 0 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & ... & 0 \end {matrix} \right] $

and the same matrix but one of columns replaced only with 1s.

In the above matrix all off-diagonal elements are 1 and diagonal elements are 0.

share|improve this question
1  
i.e., all the off-diagonal elements are $1$? –  J. M. Nov 21 '11 at 12:12
    
yeah, all off-diagonal elements are 1 and diagonal elements are 0. –  Mohan Nov 21 '11 at 12:14
    
Did you try the first few cases, $n=1,2,3,4$? Here's $n=4$: wolframalpha.com/input/…. –  lhf Nov 21 '11 at 12:15
    
An approach similar to that taken here might work. –  J. M. Nov 21 '11 at 12:18
3  
If we call your first matrix as $\mathbf M$, then your second matrix can be written as $\mathbf M+\mathbf e_k\mathbf e_k^\top$, where $\mathbf e_k$ is the $k$-th column of the identity. You can now use this. –  J. M. Nov 21 '11 at 12:38
show 2 more comments

7 Answers

up vote 6 down vote accepted

$D_n(a,b)= \begin{vmatrix} a & b & b & b \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$ ($n\times n$-matrix).

$D_n(a,b)= \begin{vmatrix} a & b & b & b \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$

$=[a+(n-1)b] \begin{vmatrix} 1 & 1 & 1 & 1 \\ b & a & b & b \\ b & b & a & b \\ b & b & b & a \end{vmatrix}$ $=[a+(n-1)b] \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & a-b & 0 & 0 \\ 0 & 0 & a-b & 0 \\ 0 & 0 & 0 & a-b \end{vmatrix}$ $=[a+(n-1)b](a-b)^{n-1} $

(In the first step we added the remaining rows to the first row and then "pulled out" constant out of the determinant. Then we subtracted $b$-multiple of the first row from each of the remaining rows.)

You're asking about $D_n(0,1)=(-1)^{n-1}(n-1)$.


If you replace one column by 1's, you can use this result to get the following. (I've computed it for $n=4$, but I guess you can generalize this for arbitrary $n$.)

$ \begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} = $ $ \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} + \begin{vmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix}= $ $ \begin{vmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ \end{vmatrix} + \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} $

Note that both these determinants are of the type you already handled in the first part.

share|improve this answer
    
Shouldn't it be $D_n(0,1)=(-1)^{n-1}(n-1)$? –  Mohan Nov 22 '11 at 5:08
    
@user774025 Thanks for noticing, I've corrected the mistake. –  Martin Sleziak Nov 22 '11 at 7:55
add comment

Let us denote $A_n$ to be that matrix of order $ n \times n $.

Subtract each of the columns multiplied by $1/(n-2)$ from the first column. You determinant then becomes:

$ |A_n|= \left| \begin {matrix} \frac {n-1}{n-2} & 1 & \ldots & 1 \\ 0 & 0 & \ldots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & ... & 0 \end {matrix} \right| = \frac{n-1}{n-2}|A_{n-1}| $

Inductively you get that $|A_n| = \frac{n-1}{n-2} \frac{n-2}{n-3}...\frac{2}{1}|A_2|=(n-1)|A_2| $.

Yet $|A_2|=-1 $ so you get that $|A_n|=1-n$.

share|improve this answer
1  
$|A_3| = 2$. I think the answer is $(-1)^n(1-n)$. –  lhf Nov 21 '11 at 12:41
    
True... thanks for your adjustment –  Shai Deshe Nov 21 '11 at 12:43
    
@Shai Note lhf's comment. I think you should have $\left(-\frac{n-1}{n-2}\right)$ instead of $\frac{n-1}{n-2}$. –  Martin Sleziak Nov 21 '11 at 12:43
add comment

Replace the first 0 with $x$ and the other zeros with $y$.

Now, your determinant is a polynomial in $x$ and $y$ with dominant term $xy^{n-1}$.

If $y=1$, then $n-1$ rows are equal which immediately gives $n-2$ independent kernel vectors, so $(y-1)^{n-2}$ divides the determinant. If $x=(n-1)/(y+n-2)$, then the sum of the $n-1$ bottom lines is a multiple of the first line, so $(x(y+n-2)-(n-1)$ divides the determinant.

Therefore, the determinant is $(x(y+n-2)-(n-1))(y-1)^{n-2}$.

Let $x=y=0$ to get $(-1)^{n-1}(n-1)$.

Let $x=1$ and $y=0$ to get $(-1)^{n-1}$.

share|improve this answer
add comment

Let $E$ be the $(n\times n)$-matrix with all ones and put $f_1:=(1,\ldots,1)$. Then $Ex=(f_1\cdot x)f_1$ for all $x\in{\mathbb R}^n$. Since $A=E-I$ we therefore have

$$Ax\ =\ (f_1\cdot x)f_1 -x\qquad(x\in{\mathbb R}^n)\ .$$

Let $(f_2,\ldots, f_n)$ be a basis of the orthogonal complement of $\langle f_1\rangle$. One has

$$Af_1=(f_1\cdot f_1)f_1 - f_1=(n-1)f_1$$

and

$$Af_i=(f_1\cdot f_i)f_1- f_i=-f_i\qquad(2\leq i\leq n)\ .$$

Therefore the matrix of $A$ with respect to the basis $(f_1,f_2,\ldots, f_n)$ is the diagonal matrix ${\rm diag}(n-1,-1,-1,\ldots,-1)\phantom{\Bigl|}$ and has determinant $(-1)^{n-1}(n-1)$.

share|improve this answer
add comment

I will compute the determinant of the matrix $$ A = \left( \begin {matrix} b & a & \ldots & a \\ a & b & \ddots & \vdots \\ \vdots & \ddots & \ddots & a \\ a & \cdots & a & b \end {matrix} \right), $$ where $a, b \in \mathbb{K}$. To obtain your case, put $a=1$ and $b=1$.

First proof. This works if $\mathrm{char}(\mathbb{K}) = 0$ or $n$ is prime to $\mathrm{char}(\mathbb{K}) > 0$. If $a =0$, then $\det A = b^n$. Suppose $a \neq 0$ and consider the vector $v = (1, \dots, 1) \in \mathbb{K}^n$; it is clear that $v$ is an eigenvector of $A$ with eigenvalue $\alpha = (n-1) a + b$. Now consider $\beta = b-a$. $\beta$ is an eigenvalue of $A$ because $$ B = A - \beta I_n = \left( \begin {matrix} a & a & \ldots & a \\ a & a & \ldots & a \\ \vdots & \vdots & \ddots & \vdots \\ a & a & ... & a \end {matrix} \right) $$ has rank $1$.

Let $E_\alpha, E_\beta \subseteq \mathbb{K}^n$ the eigenspaces of $A$ of the eigenvalues $\alpha, \beta$. We have $\alpha \neq \beta$, $E_\alpha \cap E_\beta = \{ 0 \}$ and $\dim E_\beta = n-1$, thus $\mathbb{K}^n = E_\alpha \oplus E_\beta$ and $E_\alpha = \langle v \rangle$. This proves that $A$ is similar to the matrix $\mathrm{diag}(\alpha, \beta, \dots, \beta)$, therefore $\det A = \alpha \beta^{n-1} = [(n-1)a +b] (b-a)^{n-1}$. (Notice that this formula holds also when $a=0$.)

Second proof. The characteristic polynomial of the matrix $B = A - (b-a) I$ is $$ \chi_B(t) = (-t)^n + c_{n-1} (-t)^{n-1} + \cdots + c_1(-t) + c_0, $$ where $c_i$ is the sum of the principal $(n-i)$-minors of $B$. It is clear that all principal minors of $B$ are zero, except on $1$-minors. Thus $$ \chi_B(t) = (-t)^n + na (-t)^{n-1}. $$ From $A = B + (b-a) I$, we have $\chi_A(t) = \chi_B(t-b+a)$. Thus $\det A = \chi_A(0) = \chi_B(a-b) = (b-a)^n + na (b-a)^{n-1}$.

Now consider the matrix $$ C = \left( \begin {matrix} a & a & \ldots & a \\ a & b & \ddots & \vdots \\ \vdots & \ddots & \ddots & a \\ a & \cdots & a & b \end {matrix} \right). $$ For every $i=2,\dots,n$, replace the $i$th row $C_i$ of $C$ with $C_i - C_1$, where $C_1$ is the first row of $C$. Obtain $$ \det C = \det \left( \begin{matrix} a & a & a & \cdots & a \\ 0 & b-a & 0 & \cdots & 0 \\ 0 & 0 & b-a & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & 0 \\ 0 & 0 & \cdots & 0 & b-a \end{matrix} \right) = a (b-a)^{n-1}. $$

share|improve this answer
add comment

Here's an approach using Sylvester's determinant theorem, which says that for any rectangular matrices of mutually transposed shapes $A\in\mathrm M_{n,m}(K)$ and $B\in \mathrm M_{m,n}(K)$ one has $$\det(I_n+AB)=\det(I_m+BA).$$

If $N$ is your matrix then $-N=I_n-AB$ where $A\in\mathrm M_{n,1}(K)$ is a one column all-one matrix and $B$ is its transpose. Then $$ \det(N)=(-1)^n\det(-N)=(-1)^n\det(I_1-BA)=(-1)^n(1-n). $$

share|improve this answer
add comment

This is very close to Christian Blatter's solution:

Let $E$ be the $n$ by $n$ matrix with all coefficients equal to $1$.

Let $H\subset K^n$ be the hyperplane formed by the vectors whose coordinates add up to $0$, and set $v:=(1,\dots,1)$.

Then $H=\ker E$ and $Ev=nv$. This implies
$$ \det(E-X)=(-1)^n\ X^{n-1}\ (X-n), $$ and thus $$ \det(E-1)=(-1)^n\ (1-n). $$

share|improve this answer
add comment

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.