Finding Eigenvalues of matrix with sum of all rows being equal
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1
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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
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
add a comment |
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
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06
add a comment |
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
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
$begin{bmatrix}
101 & 2 &3 &4 &5 \
1 & 102 &3 &4 &5 \
1 & 2 & 103 &4 &5 \
1& 2 &3 &104 &5 \
1 & 2 &3 &4 & 105
end{bmatrix}$
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
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
edited Nov 15 at 9:06
José Carlos Santos
141k19111207
141k19111207
asked Nov 15 at 8:50
Mk Utkarsh
89110
asked Nov 15 at 8:50
Mk Utkarsh
89110
asked Nov 15 at 8:50
Mk Utkarsh
89110
89110
another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06
add a comment |
another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06
another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06
add a comment |
1 Answer
1
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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 $115$.
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $115$.
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
add a comment |
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 $115$.
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
add a comment |
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 $115$.
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
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 $115$.
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
answered Nov 15 at 9:05
José Carlos Santos
141k19111207
141k19111207
add a comment |
add a comment |
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another property is that the trace is the sum of all eigenvalues.
– pointguard0
Nov 15 at 9:03
that tells us sum of remaining eigenvalues is 400 but i can't guess from that the remaining all are 100
– Mk Utkarsh
Nov 15 at 9:06