2
$\begingroup$

I want to find the rank of this $n\times n$ matrix \begin{pmatrix} 1 & a & a & \cdots & \cdots & a \\ a & 1 & a & \cdots & \cdots & a \\ a & a & 1 & a & \cdots & a \\ \vdots & \vdots & a& \ddots & & \vdots\\ \vdots & \vdots & \vdots & & \ddots & \vdots \\ a & a & a & \cdots &\cdots & 1 \end{pmatrix}

that is, the matrix whose diagonals are $1's$ and $a$ otherwise, where $a$ is any real number.

My first observation is when $a=0$ the rank is $n$ and when $a=1$ the rank is $1.$ Then I can assume $a\neq 0, 1$ and proceed row reduction to find its pivot rows. I obtain

\begin{pmatrix} 1 & a & a & \cdots & a \\ 0 & 1+a & a & \cdots & a \\ 0 & a & 1+a & \cdots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & a & a & \cdots & 1+a \end{pmatrix} by subtracting the first row multiplied $a$ for each row below the first, and then divides the factor $(1-a)$, and stuck there. Any hints/helps?

$\endgroup$
  • $\begingroup$ Hint: in the rest of the rows subtract the top from it and that will give you the identity with all save for the first row $\endgroup$ – Sentinel135 Mar 11 '17 at 4:02
  • 1
    $\begingroup$ Without doing anything you can at least say that, if $A$ is your given matrix, then $\det A=(1+(n-1)a)(1-a)^{(n-1)}$ as is shown here. Then $\det A=0\Rightarrow \mathrm{rank}(A)<n$, and $\quad\quad\det A\ne0\Rightarrow \mathrm{rank}(A)=n$ $\endgroup$ – StubbornAtom Mar 11 '17 at 5:38
  • $\begingroup$ Thanks all. And the determinant approach is amazing! $\endgroup$ – Karma Mar 11 '17 at 6:07
1
$\begingroup$

If you haven't figured it out yet here's the solution: if $a=1$ $\mathrm{Rank}(A)=1$ otherwise $\mathrm{Rank}(A)=n$ where $$A:= \begin{pmatrix} 1&a&a&\cdots&a\\ a&1&a&\cdots&a\\ a&a&1&\cdots&a\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ a&a&1&\cdots&1 \end{pmatrix}$$

You've already shown that you know if $a\in\{0,1\}$. So as for the rest, your close to a solution. The next set of Row equations are as follows $R_i-R_1 \to R_i$ such that $1<i\leq n$. This gives the matrix,

$$A_2= \begin{pmatrix} 1&a&a&\cdots&a\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

The next Row equation is $\frac{R_1-\sum_{i=2}^n aR_i}{1+a(n-1)}\to R_1$. This gives the matrix

$$A_3= \begin{pmatrix} 1&0&0&\cdots&0\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

The next Row equations are $R_i+R_1\to R_i$ where $1<i\leq n$. This gives $$A_4= \begin{pmatrix} 1&0&0&\cdots&0\\ 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1 \end{pmatrix}=I$$

Note that $\mathrm{Rank}(A)=\mathrm{Rank}(I)=n$ as desired.

Karma made note that if $a=\frac{1}{1-n}$ then $$A_3= \begin{pmatrix} 0&0&0&\cdots&0\\ -1&1&0&\cdots&0\\ -1&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ -1&0&0&\cdots&1 \end{pmatrix}$$

This implies that for $a=\frac{1}{1-n}$ the $\mathrm{Rank}(A)=n-1$.

$\endgroup$
  • $\begingroup$ Thank you! Now I'm wondering the step $\frac{R_1-\sum_{i=2}^n aR_i}{1+a^n}\to R_1$, can you explain a bit since I obtain $(1+(n-1)a$ in the $(1,1)$ entry, if I calculate it right $\endgroup$ – Karma Mar 11 '17 at 5:28
  • $\begingroup$ @karma yeah your right I need to fix that sorry $\endgroup$ – Sentinel135 Mar 11 '17 at 5:43
  • $\begingroup$ So we need to consider the case when $a=\frac{1}{1-n}$ too? If $a=\frac{1}{1-n}$ then the first row is a linear combination of all the other rows, with coefficient $=\frac{1}{1-n}$ for each row. Then the rank of $A$ should be $n-1$ in this case. $\endgroup$ – Karma Mar 11 '17 at 5:57
  • $\begingroup$ If $a=\frac{1}{1-n}$ then the denominator is zero, and with this value $\det(A)=0$, so rank$A\neq n$. $\endgroup$ – Karma Mar 11 '17 at 17:08
  • $\begingroup$ ok yes i I was misreading your $a$ I'm going to delete my comment and edit the answer given. Thanks $\endgroup$ – Sentinel135 Mar 11 '17 at 17:15
2
$\begingroup$

Let

$$\mathrm M_n (a) := \begin{bmatrix} 1 & a & a & \dots & a & a\\ a & 1 & a & \dots & a & a\\ a & a & 1 & \dots & a & a\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a & a & a & \dots & 1 & a\\ a & a & a & \dots & a & 1\end{bmatrix} = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$$

The eigenvalues of rank-$1$ matrix $a 1_n 1_n^{\top}$ are

  • $\color{blue}{0}$ with multiplicity $n-1$.
  • $\color{blue}{n a}$ with multiplicity $1$.

Thus, the eigenvalues of $\mathrm M_n (a) = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$ are

  • $\color{blue}{1-a}$ with multiplicity $n-1$.
  • $(1-a) + na = \color{blue}{(n-1) \, a + 1}$ with multiplicity $1$.

We could also have arrived at this conclusion computing the characteristic polynomial of $\mathrm M_n (a)$

$$\begin{array}{rl} \det ( s \mathrm I_n - \mathrm M_n (a) ) &= \det \left( (s-(1-a)) \mathrm I_n - a 1_n 1_n^{\top} \right)\\ &= \det \left( (s-(1-a)) \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right) \right)\\ &= (s-(1-a))^n \cdot \det \left( \mathrm I_n - \frac{a}{s-(1-a)} 1_n 1_n^{\top} \right)\\ &= (s-(1-a))^n \cdot \left(1 - \frac{n a}{s-(1-a)}\right)\\ &= (s-(1-a))^{n-1} \cdot \left(s-(1-a) - n a\right)\\ &= (s-(1-a))^{n-1} \cdot \left( s - ((n-1) \, a + 1) \right)\end{array}$$

where the matrix determinant lemma was used. If

$$a \in \left\{ -\frac{1}{n-1}, 1 \right\}$$

then $\mathrm M_n (a)$ is singular. Thus, using the rank-nullity theorem, we conclude that

$$\boxed{\mbox{rank} (\mathrm M_n (a)) = \begin{cases} 1 & \text{if } a = 1\\ n-1 & \text{if } a = -\frac{1}{n-1}\\ n & \text{otherwise}\end{cases}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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