A left stochastic matrix implies the solution to AX=X is non-trivial.
Theorem:
Let $A in mathbb{R}^{n times n}$ whose (each) column sums to one.
Then, $A vec{X} = vec{X}$ has non-trivial solution.
Proof:
By definition, A is a left stochastic matrix.
Observe:
$Avec{X}=vec{X}$ is a transformation under an identify map.
$Rightarrow vec{X} = A^{-1}vec{X}$
But recall that for any matrix $A in mathbb{n times n}$: $A$ is invertible so the inverse of $A, A^{-1}$ exists.
By the uniqueness theorem, any solution $vec{X}$ to $Avec{X}=vec{b}$ is unique.
Clearly, for $vec{b} = vec{0}$, $Avec{X}=vec{0}$ can be satisfied by $vec{X} = vec{0}$ and only $vec{X} = vec{0}$.
Since, $A$ is not the zero matrix and $vec{X} neq vec{0}$, $vec{X} = A^{-1}vec{X}$ is not the zero vector.
Indeed, no trivial solution exists.
Am I going around in circles in my above attempt?
Any help is greatly appreciated.
linear-algebra proof-verification
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
add a comment |
Theorem:
Let $A in mathbb{R}^{n times n}$ whose (each) column sums to one.
Then, $A vec{X} = vec{X}$ has non-trivial solution.
Proof:
By definition, A is a left stochastic matrix.
Observe:
$Avec{X}=vec{X}$ is a transformation under an identify map.
$Rightarrow vec{X} = A^{-1}vec{X}$
But recall that for any matrix $A in mathbb{n times n}$: $A$ is invertible so the inverse of $A, A^{-1}$ exists.
By the uniqueness theorem, any solution $vec{X}$ to $Avec{X}=vec{b}$ is unique.
Clearly, for $vec{b} = vec{0}$, $Avec{X}=vec{0}$ can be satisfied by $vec{X} = vec{0}$ and only $vec{X} = vec{0}$.
Since, $A$ is not the zero matrix and $vec{X} neq vec{0}$, $vec{X} = A^{-1}vec{X}$ is not the zero vector.
Indeed, no trivial solution exists.
Am I going around in circles in my above attempt?
Any help is greatly appreciated.
linear-algebra proof-verification
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33
add a comment |
Theorem:
Let $A in mathbb{R}^{n times n}$ whose (each) column sums to one.
Then, $A vec{X} = vec{X}$ has non-trivial solution.
Proof:
By definition, A is a left stochastic matrix.
Observe:
$Avec{X}=vec{X}$ is a transformation under an identify map.
$Rightarrow vec{X} = A^{-1}vec{X}$
But recall that for any matrix $A in mathbb{n times n}$: $A$ is invertible so the inverse of $A, A^{-1}$ exists.
By the uniqueness theorem, any solution $vec{X}$ to $Avec{X}=vec{b}$ is unique.
Clearly, for $vec{b} = vec{0}$, $Avec{X}=vec{0}$ can be satisfied by $vec{X} = vec{0}$ and only $vec{X} = vec{0}$.
Since, $A$ is not the zero matrix and $vec{X} neq vec{0}$, $vec{X} = A^{-1}vec{X}$ is not the zero vector.
Indeed, no trivial solution exists.
Am I going around in circles in my above attempt?
Any help is greatly appreciated.
linear-algebra proof-verification
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
Theorem:
Let $A in mathbb{R}^{n times n}$ whose (each) column sums to one.
Then, $A vec{X} = vec{X}$ has non-trivial solution.
Proof:
By definition, A is a left stochastic matrix.
Observe:
$Avec{X}=vec{X}$ is a transformation under an identify map.
$Rightarrow vec{X} = A^{-1}vec{X}$
But recall that for any matrix $A in mathbb{n times n}$: $A$ is invertible so the inverse of $A, A^{-1}$ exists.
By the uniqueness theorem, any solution $vec{X}$ to $Avec{X}=vec{b}$ is unique.
Clearly, for $vec{b} = vec{0}$, $Avec{X}=vec{0}$ can be satisfied by $vec{X} = vec{0}$ and only $vec{X} = vec{0}$.
Since, $A$ is not the zero matrix and $vec{X} neq vec{0}$, $vec{X} = A^{-1}vec{X}$ is not the zero vector.
Indeed, no trivial solution exists.
Am I going around in circles in my above attempt?
Any help is greatly appreciated.
linear-algebra proof-verification
linear-algebra proof-verification
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
asked Nov 28 '18 at 5:32
MathematicingMathematicing
2,44321854
2,44321854
Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33
add a comment |
Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33
Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33
Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33
add a comment |
1 Answer
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Be careful: the inverse of $A$ need not exist. For instance,
$$left(begin{matrix}0.4 & 0.4\0.6&0.6\ end{matrix}right)$$
is a stochastic matrix but it is non-invertible.
What you can see is that since for any stochastic matrix $A$ the columns add up to $1$, then the columns of $A-I=A-1I$ add up to $0$, that is the sum of the rows gives the null vector. This means $A-1I$ is non invertible, so $1$ is an eigenvalue of $A$. By definition there exists a vector $xneq 0$ such that $Ax=1x$, that is, a non-trivial solution to $Ax=x$.
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
add a comment |
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1 Answer
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Be careful: the inverse of $A$ need not exist. For instance,
$$left(begin{matrix}0.4 & 0.4\0.6&0.6\ end{matrix}right)$$
is a stochastic matrix but it is non-invertible.
What you can see is that since for any stochastic matrix $A$ the columns add up to $1$, then the columns of $A-I=A-1I$ add up to $0$, that is the sum of the rows gives the null vector. This means $A-1I$ is non invertible, so $1$ is an eigenvalue of $A$. By definition there exists a vector $xneq 0$ such that $Ax=1x$, that is, a non-trivial solution to $Ax=x$.
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
add a comment |
Be careful: the inverse of $A$ need not exist. For instance,
$$left(begin{matrix}0.4 & 0.4\0.6&0.6\ end{matrix}right)$$
is a stochastic matrix but it is non-invertible.
What you can see is that since for any stochastic matrix $A$ the columns add up to $1$, then the columns of $A-I=A-1I$ add up to $0$, that is the sum of the rows gives the null vector. This means $A-1I$ is non invertible, so $1$ is an eigenvalue of $A$. By definition there exists a vector $xneq 0$ such that $Ax=1x$, that is, a non-trivial solution to $Ax=x$.
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
add a comment |
Be careful: the inverse of $A$ need not exist. For instance,
$$left(begin{matrix}0.4 & 0.4\0.6&0.6\ end{matrix}right)$$
is a stochastic matrix but it is non-invertible.
What you can see is that since for any stochastic matrix $A$ the columns add up to $1$, then the columns of $A-I=A-1I$ add up to $0$, that is the sum of the rows gives the null vector. This means $A-1I$ is non invertible, so $1$ is an eigenvalue of $A$. By definition there exists a vector $xneq 0$ such that $Ax=1x$, that is, a non-trivial solution to $Ax=x$.
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
Be careful: the inverse of $A$ need not exist. For instance,
$$left(begin{matrix}0.4 & 0.4\0.6&0.6\ end{matrix}right)$$
is a stochastic matrix but it is non-invertible.
What you can see is that since for any stochastic matrix $A$ the columns add up to $1$, then the columns of $A-I=A-1I$ add up to $0$, that is the sum of the rows gives the null vector. This means $A-1I$ is non invertible, so $1$ is an eigenvalue of $A$. By definition there exists a vector $xneq 0$ such that $Ax=1x$, that is, a non-trivial solution to $Ax=x$.
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
edited Nov 28 '18 at 5:51
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
answered Nov 28 '18 at 5:44
Alejandro Nasif SalumAlejandro Nasif Salum
4,409118
4,409118
add a comment |
add a comment |
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Take $X = [1,1,1,...,1]$, to see that $A^TX = X$ has a solution. Now, what is the relationship between the eigenvalues of $A$ and $A^T$? Also, why should $A^{-1}$ exist? The determinant has to be non-zero, but this is not clear from the condition given.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 5:33