Finding Eigenvalues of matrix with sum of all rows being equal











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$begin{bmatrix}
101 & 2 &3 &4 &5 \
1 & 102 &3 &4 &5 \
1 & 2 & 103 &4 &5 \
1& 2 &3 &104 &5 \
1 & 2 &3 &4 & 105
end{bmatrix}$



Sum of all rows of this matrix is 115 hence $exists lambda = 115$



Now is there any property that can be used to find out remaining eigenvalues of this matrix?



remaining 4 eigenvalues are 100.






linear-algebra matrices eigenvalues-eigenvectors







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  • another property is that the trace is the sum of all eigenvalues.
    – pointguard0
    12 hours ago










  • that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
    – Mk Utkarsh
    12 hours ago















up vote
1
down vote

favorite












$begin{bmatrix}
101 & 2 &3 &4 &5 \
1 & 102 &3 &4 &5 \
1 & 2 & 103 &4 &5 \
1& 2 &3 &104 &5 \
1 & 2 &3 &4 & 105
end{bmatrix}$



Sum of all rows of this matrix is 115 hence $exists lambda = 115$



Now is there any property that can be used to find out remaining eigenvalues of this matrix?



remaining 4 eigenvalues are 100.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question
























  • another property is that the trace is the sum of all eigenvalues.
    – pointguard0
    12 hours ago










  • that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
    – Mk Utkarsh
    12 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











$begin{bmatrix}
101 & 2 &3 &4 &5 \
1 & 102 &3 &4 &5 \
1 & 2 & 103 &4 &5 \
1& 2 &3 &104 &5 \
1 & 2 &3 &4 & 105
end{bmatrix}$



Sum of all rows of this matrix is 115 hence $exists lambda = 115$



Now is there any property that can be used to find out remaining eigenvalues of this matrix?



remaining 4 eigenvalues are 100.






linear-algebra matrices eigenvalues-eigenvectors







share|cite|improve this question















$begin{bmatrix}
101 & 2 &3 &4 &5 \
1 & 102 &3 &4 &5 \
1 & 2 & 103 &4 &5 \
1& 2 &3 &104 &5 \
1 & 2 &3 &4 & 105
end{bmatrix}$



Sum of all rows of this matrix is 115 hence $exists lambda = 115$



Now is there any property that can be used to find out remaining eigenvalues of this matrix?



remaining 4 eigenvalues are 100.






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 12 hours ago









José Carlos Santos

138k18110200




138k18110200










asked 12 hours ago









Mk Utkarsh

7719




7719












  • another property is that the trace is the sum of all eigenvalues.
    – pointguard0
    12 hours ago










  • that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
    – Mk Utkarsh
    12 hours ago


















  • another property is that the trace is the sum of all eigenvalues.
    – pointguard0
    12 hours ago










  • that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
    – Mk Utkarsh
    12 hours ago
















another property is that the trace is the sum of all eigenvalues.
– pointguard0
12 hours ago




another property is that the trace is the sum of all eigenvalues.
– pointguard0
12 hours ago












that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
12 hours ago




that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
12 hours ago










1 Answer
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Let $A$ be your matrix and let$$B=begin{bmatrix}1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5end{bmatrix};$$then $A=B+100operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $ainmathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.






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    Let $A$ be your matrix and let$$B=begin{bmatrix}1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5end{bmatrix};$$then $A=B+100operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $ainmathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.






    share|cite|improve this answer

























      up vote
      3
      down vote













      Let $A$ be your matrix and let$$B=begin{bmatrix}1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5end{bmatrix};$$then $A=B+100operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $ainmathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        Let $A$ be your matrix and let$$B=begin{bmatrix}1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5end{bmatrix};$$then $A=B+100operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $ainmathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.






        share|cite|improve this answer












        Let $A$ be your matrix and let$$B=begin{bmatrix}1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5\1&2&3&4&5end{bmatrix};$$then $A=B+100operatorname{Id}$. Since the dimension of the space spanned by the columns of $B$ is $1$ and since $B$ is a $5times5$ matrix, the characteristic polynomial of $B$ is of the type $x^4(a-x)=ax^4-x^5$, for some $ainmathbb R$. On the other hand, by the same argument that you used, $15$ is an eigenvalue of $B$. Therefore, the eigenvalues of $B$ are $0$ ($4$ times) and $15$. So, the eigenvalues of $A$ are $100$ ($4$ times) and $115$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        José Carlos Santos

        138k18110200




        138k18110200






























             

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