What are the eigenvalues of the given matrix $M$












2












$begingroup$


I am learning linear algebra and now I'm in eigenvalues and eigenvectors part of it. there is a question that I can't solve it or any idea that I have is hard and nasty. I think this question must have a trick that I am not familiar with it because I'm new to eigenvalues and eigenvectors.



the question is this:




Let $M in Bbb R^{ntimes n}$ and real numbers $a_1$ to $a_n$
and every $m_{ij} = frac{a_i}{a_j}$, so:
$$ M = begin{pmatrix}1&cdots&frac{a_1}{a_n}\vdots&ddots&vdots\frac{a_n}{a_1}&cdots&1end{pmatrix} $$
find all eigenvalues.




any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
    $endgroup$
    – José Carlos Santos
    Dec 20 '18 at 19:51










  • $begingroup$
    A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
    $endgroup$
    – Mark
    Dec 20 '18 at 19:52












  • $begingroup$
    sorryyyy! I'm going to edit it
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 19:53






  • 9




    $begingroup$
    Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
    $endgroup$
    – Robert Israel
    Dec 20 '18 at 19:57






  • 1




    $begingroup$
    @Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
    $endgroup$
    – Bungo
    Dec 20 '18 at 20:18
















2












$begingroup$


I am learning linear algebra and now I'm in eigenvalues and eigenvectors part of it. there is a question that I can't solve it or any idea that I have is hard and nasty. I think this question must have a trick that I am not familiar with it because I'm new to eigenvalues and eigenvectors.



the question is this:




Let $M in Bbb R^{ntimes n}$ and real numbers $a_1$ to $a_n$
and every $m_{ij} = frac{a_i}{a_j}$, so:
$$ M = begin{pmatrix}1&cdots&frac{a_1}{a_n}\vdots&ddots&vdots\frac{a_n}{a_1}&cdots&1end{pmatrix} $$
find all eigenvalues.




any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
    $endgroup$
    – José Carlos Santos
    Dec 20 '18 at 19:51










  • $begingroup$
    A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
    $endgroup$
    – Mark
    Dec 20 '18 at 19:52












  • $begingroup$
    sorryyyy! I'm going to edit it
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 19:53






  • 9




    $begingroup$
    Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
    $endgroup$
    – Robert Israel
    Dec 20 '18 at 19:57






  • 1




    $begingroup$
    @Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
    $endgroup$
    – Bungo
    Dec 20 '18 at 20:18














2












2








2


0



$begingroup$


I am learning linear algebra and now I'm in eigenvalues and eigenvectors part of it. there is a question that I can't solve it or any idea that I have is hard and nasty. I think this question must have a trick that I am not familiar with it because I'm new to eigenvalues and eigenvectors.



the question is this:




Let $M in Bbb R^{ntimes n}$ and real numbers $a_1$ to $a_n$
and every $m_{ij} = frac{a_i}{a_j}$, so:
$$ M = begin{pmatrix}1&cdots&frac{a_1}{a_n}\vdots&ddots&vdots\frac{a_n}{a_1}&cdots&1end{pmatrix} $$
find all eigenvalues.




any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question











$endgroup$




I am learning linear algebra and now I'm in eigenvalues and eigenvectors part of it. there is a question that I can't solve it or any idea that I have is hard and nasty. I think this question must have a trick that I am not familiar with it because I'm new to eigenvalues and eigenvectors.



the question is this:




Let $M in Bbb R^{ntimes n}$ and real numbers $a_1$ to $a_n$
and every $m_{ij} = frac{a_i}{a_j}$, so:
$$ M = begin{pmatrix}1&cdots&frac{a_1}{a_n}\vdots&ddots&vdots\frac{a_n}{a_1}&cdots&1end{pmatrix} $$
find all eigenvalues.




any help would be appreciated.






linear-algebra matrices eigenvalues-eigenvectors




linear-algebra matrices eigenvalues-eigenvectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 20:17









mechanodroid

28.9k62648




28.9k62648










asked Dec 20 '18 at 19:49









Peyman mohseni kiasariPeyman mohseni kiasari

13711




13711












  • $begingroup$
    But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
    $endgroup$
    – José Carlos Santos
    Dec 20 '18 at 19:51










  • $begingroup$
    A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
    $endgroup$
    – Mark
    Dec 20 '18 at 19:52












  • $begingroup$
    sorryyyy! I'm going to edit it
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 19:53






  • 9




    $begingroup$
    Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
    $endgroup$
    – Robert Israel
    Dec 20 '18 at 19:57






  • 1




    $begingroup$
    @Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
    $endgroup$
    – Bungo
    Dec 20 '18 at 20:18


















  • $begingroup$
    But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
    $endgroup$
    – José Carlos Santos
    Dec 20 '18 at 19:51










  • $begingroup$
    A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
    $endgroup$
    – Mark
    Dec 20 '18 at 19:52












  • $begingroup$
    sorryyyy! I'm going to edit it
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 19:53






  • 9




    $begingroup$
    Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
    $endgroup$
    – Robert Israel
    Dec 20 '18 at 19:57






  • 1




    $begingroup$
    @Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
    $endgroup$
    – Bungo
    Dec 20 '18 at 20:18
















$begingroup$
But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 19:51




$begingroup$
But you are saying from the start that the eigenvalues are the $lambda_i$'s. Are you sure about that?
$endgroup$
– José Carlos Santos
Dec 20 '18 at 19:51












$begingroup$
A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
$endgroup$
– Mark
Dec 20 '18 at 19:52






$begingroup$
A matrix of size $ntimes n$ can't have more than $n$ distinct eigenvalues. So your question makes no sense as it is written right now.
$endgroup$
– Mark
Dec 20 '18 at 19:52














$begingroup$
sorryyyy! I'm going to edit it
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 19:53




$begingroup$
sorryyyy! I'm going to edit it
$endgroup$
– Peyman mohseni kiasari
Dec 20 '18 at 19:53




9




9




$begingroup$
Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
$endgroup$
– Robert Israel
Dec 20 '18 at 19:57




$begingroup$
Note that all columns are scalar multiples of the first column. Thus this matrix has rank $1$, and there is only one nonzero eigenvalue.
$endgroup$
– Robert Israel
Dec 20 '18 at 19:57




1




1




$begingroup$
@Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
$endgroup$
– Bungo
Dec 20 '18 at 20:18




$begingroup$
@Damien Seems more fruitful to look at the vector $(a_1, a_2, ldots, a_n)$.
$endgroup$
– Bungo
Dec 20 '18 at 20:18










3 Answers
3






active

oldest

votes


















3












$begingroup$

Another way to do this is by noting that if $mathbf{a} = (a_1, dots, a_n)^top$ and $mathbf{b} = (1/a_1, dots, 1/a_n)^top$, then
$$
M = mathbf{a} mathbf{b}^top,
$$
where $mathbf{b}^top$ denotes the transpose of $mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering
$$
M mathbf{a} = (mathbf{a} mathbf{b}^top) mathbf{a} = mathbf{a} (mathbf{b}^top mathbf{a}) = n mathbf{a},
$$
i.e. the final eigenvalue is $n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:28










  • $begingroup$
    @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
    $endgroup$
    – Decaf-Math
    Dec 20 '18 at 20:30












  • $begingroup$
    You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
    $endgroup$
    – ekkilop
    Dec 20 '18 at 20:32





















2












$begingroup$

To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.



$$0= det(M - lambda I) = begin{vmatrix} 1-lambda &frac{a_1}{a_2} & cdots & frac{a_1}{a_n} \
frac{a_2}{a_1} & 1-lambda & cdots & frac{a_2}{a_n} \
vdots & vdots & ddots & vdots \
frac{a_n}{a_1} & frac{a_n}{a_2} & cdots & 1-lambda
end{vmatrix}$$



Since $a_1, ldots, a_n ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, ldots, n$ to obtain:



$$0 = begin{vmatrix} a_1(1-lambda) & a_1 & cdots & a_1 \
a_2 & a_2(1-lambda) & cdots & a_2 \
vdots & vdots & ddots & vdots \
a_n & a_n & cdots & a_n(1-lambda)
end{vmatrix}$$



Now divide $i$-th row by $a_i$ for $i =1, ldots, n$ to obtain



begin{align}
0 &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
1 & 1-lambda & cdots & 1 \
vdots & vdots & ddots & vdots \
1 & 1 & cdots & 1-lambda
end{vmatrix} \
&= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
lambda & -lambda & cdots & 0 \
vdots & vdots & ddots & vdots \
lambda & 0 & cdots & -lambda
end{vmatrix} \
&= begin{vmatrix} n-lambda & 1 & cdots & 1 \
0 & -lambda & cdots & 0 \
vdots & vdots & ddots & vdots \
0& 0 & cdots & -lambda
end{vmatrix} \
&= (n-lambda)(-lambda)^{n-1}
end{align}



so the eigenvalues are $0$ and $n$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:10










  • $begingroup$
    @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:15



















1












$begingroup$

We can write your matrix as $M = DJD^{-1}$, where
$$
D = pmatrix{a_1\ & ddots \ && a_n}, quad
J = pmatrix{1 & cdots & 1\ vdots & ddots & vdots \ 1 & cdots & 1}
$$
So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $operatorname{tr}(J) = n$.



We could also recognize that $M$ has rank $1$ as Robert did in his comment.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Another way to do this is by noting that if $mathbf{a} = (a_1, dots, a_n)^top$ and $mathbf{b} = (1/a_1, dots, 1/a_n)^top$, then
    $$
    M = mathbf{a} mathbf{b}^top,
    $$
    where $mathbf{b}^top$ denotes the transpose of $mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering
    $$
    M mathbf{a} = (mathbf{a} mathbf{b}^top) mathbf{a} = mathbf{a} (mathbf{b}^top mathbf{a}) = n mathbf{a},
    $$
    i.e. the final eigenvalue is $n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:28










    • $begingroup$
      @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
      $endgroup$
      – Decaf-Math
      Dec 20 '18 at 20:30












    • $begingroup$
      You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
      $endgroup$
      – ekkilop
      Dec 20 '18 at 20:32


















    3












    $begingroup$

    Another way to do this is by noting that if $mathbf{a} = (a_1, dots, a_n)^top$ and $mathbf{b} = (1/a_1, dots, 1/a_n)^top$, then
    $$
    M = mathbf{a} mathbf{b}^top,
    $$
    where $mathbf{b}^top$ denotes the transpose of $mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering
    $$
    M mathbf{a} = (mathbf{a} mathbf{b}^top) mathbf{a} = mathbf{a} (mathbf{b}^top mathbf{a}) = n mathbf{a},
    $$
    i.e. the final eigenvalue is $n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:28










    • $begingroup$
      @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
      $endgroup$
      – Decaf-Math
      Dec 20 '18 at 20:30












    • $begingroup$
      You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
      $endgroup$
      – ekkilop
      Dec 20 '18 at 20:32
















    3












    3








    3





    $begingroup$

    Another way to do this is by noting that if $mathbf{a} = (a_1, dots, a_n)^top$ and $mathbf{b} = (1/a_1, dots, 1/a_n)^top$, then
    $$
    M = mathbf{a} mathbf{b}^top,
    $$
    where $mathbf{b}^top$ denotes the transpose of $mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering
    $$
    M mathbf{a} = (mathbf{a} mathbf{b}^top) mathbf{a} = mathbf{a} (mathbf{b}^top mathbf{a}) = n mathbf{a},
    $$
    i.e. the final eigenvalue is $n$.






    share|cite|improve this answer











    $endgroup$



    Another way to do this is by noting that if $mathbf{a} = (a_1, dots, a_n)^top$ and $mathbf{b} = (1/a_1, dots, 1/a_n)^top$, then
    $$
    M = mathbf{a} mathbf{b}^top,
    $$
    where $mathbf{b}^top$ denotes the transpose of $mathbf{b}$. The rank of $M$ is therefore 1 (can you see why?), meaning that only one eigenvalue is non-zero. This eigenvalue is found by considering
    $$
    M mathbf{a} = (mathbf{a} mathbf{b}^top) mathbf{a} = mathbf{a} (mathbf{b}^top mathbf{a}) = n mathbf{a},
    $$
    i.e. the final eigenvalue is $n$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 20 '18 at 20:35

























    answered Dec 20 '18 at 20:23









    ekkilopekkilop

    1,736519




    1,736519












    • $begingroup$
      short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:28










    • $begingroup$
      @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
      $endgroup$
      – Decaf-Math
      Dec 20 '18 at 20:30












    • $begingroup$
      You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
      $endgroup$
      – ekkilop
      Dec 20 '18 at 20:32




















    • $begingroup$
      short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:28










    • $begingroup$
      @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
      $endgroup$
      – Decaf-Math
      Dec 20 '18 at 20:30












    • $begingroup$
      You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
      $endgroup$
      – ekkilop
      Dec 20 '18 at 20:32


















    $begingroup$
    short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:28




    $begingroup$
    short and nice. sorry if my question is silly but how do you mult two vectors in this way? I just know that mult of two vectors is a number. this way of mult is not in the book yet.
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:28












    $begingroup$
    @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
    $endgroup$
    – Decaf-Math
    Dec 20 '18 at 20:30






    $begingroup$
    @Peyman, you may interpret a vector as an $ntimes 1$ matrix.
    $endgroup$
    – Decaf-Math
    Dec 20 '18 at 20:30














    $begingroup$
    You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
    $endgroup$
    – ekkilop
    Dec 20 '18 at 20:32






    $begingroup$
    You multiply them by normal matrix multiplication. Note that the number of columns of $mathbf{a}$ (i.e. one) is equal to the number of rows of $mathbf{b}^top$ (also one).
    $endgroup$
    – ekkilop
    Dec 20 '18 at 20:32













    2












    $begingroup$

    To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.



    $$0= det(M - lambda I) = begin{vmatrix} 1-lambda &frac{a_1}{a_2} & cdots & frac{a_1}{a_n} \
    frac{a_2}{a_1} & 1-lambda & cdots & frac{a_2}{a_n} \
    vdots & vdots & ddots & vdots \
    frac{a_n}{a_1} & frac{a_n}{a_2} & cdots & 1-lambda
    end{vmatrix}$$



    Since $a_1, ldots, a_n ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, ldots, n$ to obtain:



    $$0 = begin{vmatrix} a_1(1-lambda) & a_1 & cdots & a_1 \
    a_2 & a_2(1-lambda) & cdots & a_2 \
    vdots & vdots & ddots & vdots \
    a_n & a_n & cdots & a_n(1-lambda)
    end{vmatrix}$$



    Now divide $i$-th row by $a_i$ for $i =1, ldots, n$ to obtain



    begin{align}
    0 &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    1 & 1-lambda & cdots & 1 \
    vdots & vdots & ddots & vdots \
    1 & 1 & cdots & 1-lambda
    end{vmatrix} \
    &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    lambda & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    lambda & 0 & cdots & -lambda
    end{vmatrix} \
    &= begin{vmatrix} n-lambda & 1 & cdots & 1 \
    0 & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    0& 0 & cdots & -lambda
    end{vmatrix} \
    &= (n-lambda)(-lambda)^{n-1}
    end{align}



    so the eigenvalues are $0$ and $n$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:10










    • $begingroup$
      @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
      $endgroup$
      – mechanodroid
      Dec 20 '18 at 20:15
















    2












    $begingroup$

    To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.



    $$0= det(M - lambda I) = begin{vmatrix} 1-lambda &frac{a_1}{a_2} & cdots & frac{a_1}{a_n} \
    frac{a_2}{a_1} & 1-lambda & cdots & frac{a_2}{a_n} \
    vdots & vdots & ddots & vdots \
    frac{a_n}{a_1} & frac{a_n}{a_2} & cdots & 1-lambda
    end{vmatrix}$$



    Since $a_1, ldots, a_n ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, ldots, n$ to obtain:



    $$0 = begin{vmatrix} a_1(1-lambda) & a_1 & cdots & a_1 \
    a_2 & a_2(1-lambda) & cdots & a_2 \
    vdots & vdots & ddots & vdots \
    a_n & a_n & cdots & a_n(1-lambda)
    end{vmatrix}$$



    Now divide $i$-th row by $a_i$ for $i =1, ldots, n$ to obtain



    begin{align}
    0 &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    1 & 1-lambda & cdots & 1 \
    vdots & vdots & ddots & vdots \
    1 & 1 & cdots & 1-lambda
    end{vmatrix} \
    &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    lambda & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    lambda & 0 & cdots & -lambda
    end{vmatrix} \
    &= begin{vmatrix} n-lambda & 1 & cdots & 1 \
    0 & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    0& 0 & cdots & -lambda
    end{vmatrix} \
    &= (n-lambda)(-lambda)^{n-1}
    end{align}



    so the eigenvalues are $0$ and $n$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:10










    • $begingroup$
      @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
      $endgroup$
      – mechanodroid
      Dec 20 '18 at 20:15














    2












    2








    2





    $begingroup$

    To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.



    $$0= det(M - lambda I) = begin{vmatrix} 1-lambda &frac{a_1}{a_2} & cdots & frac{a_1}{a_n} \
    frac{a_2}{a_1} & 1-lambda & cdots & frac{a_2}{a_n} \
    vdots & vdots & ddots & vdots \
    frac{a_n}{a_1} & frac{a_n}{a_2} & cdots & 1-lambda
    end{vmatrix}$$



    Since $a_1, ldots, a_n ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, ldots, n$ to obtain:



    $$0 = begin{vmatrix} a_1(1-lambda) & a_1 & cdots & a_1 \
    a_2 & a_2(1-lambda) & cdots & a_2 \
    vdots & vdots & ddots & vdots \
    a_n & a_n & cdots & a_n(1-lambda)
    end{vmatrix}$$



    Now divide $i$-th row by $a_i$ for $i =1, ldots, n$ to obtain



    begin{align}
    0 &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    1 & 1-lambda & cdots & 1 \
    vdots & vdots & ddots & vdots \
    1 & 1 & cdots & 1-lambda
    end{vmatrix} \
    &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    lambda & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    lambda & 0 & cdots & -lambda
    end{vmatrix} \
    &= begin{vmatrix} n-lambda & 1 & cdots & 1 \
    0 & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    0& 0 & cdots & -lambda
    end{vmatrix} \
    &= (n-lambda)(-lambda)^{n-1}
    end{align}



    so the eigenvalues are $0$ and $n$.






    share|cite|improve this answer











    $endgroup$



    To find the eigenvalues, we are calculating the zeroes of the characteristic polynomial of $M$.



    $$0= det(M - lambda I) = begin{vmatrix} 1-lambda &frac{a_1}{a_2} & cdots & frac{a_1}{a_n} \
    frac{a_2}{a_1} & 1-lambda & cdots & frac{a_2}{a_n} \
    vdots & vdots & ddots & vdots \
    frac{a_n}{a_1} & frac{a_n}{a_2} & cdots & 1-lambda
    end{vmatrix}$$



    Since $a_1, ldots, a_n ne 0$, we can multiply $j$-th column by $a_j$ for $j = 1, ldots, n$ to obtain:



    $$0 = begin{vmatrix} a_1(1-lambda) & a_1 & cdots & a_1 \
    a_2 & a_2(1-lambda) & cdots & a_2 \
    vdots & vdots & ddots & vdots \
    a_n & a_n & cdots & a_n(1-lambda)
    end{vmatrix}$$



    Now divide $i$-th row by $a_i$ for $i =1, ldots, n$ to obtain



    begin{align}
    0 &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    1 & 1-lambda & cdots & 1 \
    vdots & vdots & ddots & vdots \
    1 & 1 & cdots & 1-lambda
    end{vmatrix} \
    &= begin{vmatrix} 1-lambda & 1 & cdots & 1 \
    lambda & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    lambda & 0 & cdots & -lambda
    end{vmatrix} \
    &= begin{vmatrix} n-lambda & 1 & cdots & 1 \
    0 & -lambda & cdots & 0 \
    vdots & vdots & ddots & vdots \
    0& 0 & cdots & -lambda
    end{vmatrix} \
    &= (n-lambda)(-lambda)^{n-1}
    end{align}



    so the eigenvalues are $0$ and $n$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 20 '18 at 20:13

























    answered Dec 20 '18 at 20:07









    mechanodroidmechanodroid

    28.9k62648




    28.9k62648








    • 1




      $begingroup$
      I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:10










    • $begingroup$
      @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
      $endgroup$
      – mechanodroid
      Dec 20 '18 at 20:15














    • 1




      $begingroup$
      I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
      $endgroup$
      – Peyman mohseni kiasari
      Dec 20 '18 at 20:10










    • $begingroup$
      @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
      $endgroup$
      – mechanodroid
      Dec 20 '18 at 20:15








    1




    1




    $begingroup$
    I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:10




    $begingroup$
    I think it is wrong. let a1=1 and a2 =1 then M is [[1,1],[1,1]]. but the eigenvalues of M are 0 and 2
    $endgroup$
    – Peyman mohseni kiasari
    Dec 20 '18 at 20:10












    $begingroup$
    @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:15




    $begingroup$
    @Peyman Whoops, you are right. The eigenvalues are $0$ and $n$.
    $endgroup$
    – mechanodroid
    Dec 20 '18 at 20:15











    1












    $begingroup$

    We can write your matrix as $M = DJD^{-1}$, where
    $$
    D = pmatrix{a_1\ & ddots \ && a_n}, quad
    J = pmatrix{1 & cdots & 1\ vdots & ddots & vdots \ 1 & cdots & 1}
    $$
    So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $operatorname{tr}(J) = n$.



    We could also recognize that $M$ has rank $1$ as Robert did in his comment.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      We can write your matrix as $M = DJD^{-1}$, where
      $$
      D = pmatrix{a_1\ & ddots \ && a_n}, quad
      J = pmatrix{1 & cdots & 1\ vdots & ddots & vdots \ 1 & cdots & 1}
      $$
      So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $operatorname{tr}(J) = n$.



      We could also recognize that $M$ has rank $1$ as Robert did in his comment.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        We can write your matrix as $M = DJD^{-1}$, where
        $$
        D = pmatrix{a_1\ & ddots \ && a_n}, quad
        J = pmatrix{1 & cdots & 1\ vdots & ddots & vdots \ 1 & cdots & 1}
        $$
        So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $operatorname{tr}(J) = n$.



        We could also recognize that $M$ has rank $1$ as Robert did in his comment.






        share|cite|improve this answer









        $endgroup$



        We can write your matrix as $M = DJD^{-1}$, where
        $$
        D = pmatrix{a_1\ & ddots \ && a_n}, quad
        J = pmatrix{1 & cdots & 1\ vdots & ddots & vdots \ 1 & cdots & 1}
        $$
        So, $M$ is similar to $J$. $J$ is a rank $1$ symmetric matrix, so it's only non-zero eigenvalue will be $operatorname{tr}(J) = n$.



        We could also recognize that $M$ has rank $1$ as Robert did in his comment.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 20 '18 at 20:26









        OmnomnomnomOmnomnomnom

        129k794188




        129k794188






























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