Does $exists$ a matrix $A$ so that ${A^p:pgeq 1}$ spans $M_n(mathbb R)?$
$begingroup$
The main question is :
(TRUE/FALSE)
For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$
I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.
Any help is appreciated. Thank you
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
The main question is :
(TRUE/FALSE)
For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$
I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.
Any help is appreciated. Thank you
linear-algebra matrices
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4
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
2
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38
add a comment |
$begingroup$
The main question is :
(TRUE/FALSE)
For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$
I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.
Any help is appreciated. Thank you
linear-algebra matrices
$endgroup$
The main question is :
(TRUE/FALSE)
For any $ngeq 2,$ there exists an $ntimes n$ real matrix $A$ such thatthe set ${A^p|pgeq 1}$ spans the $mathbb R$ vector space $M_n(mathbb R).$
I tried whether we can find such $A$ that the above set spans $left(begin{array}{cc} 1 & 0\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 1\ 0 & 0 end{array}right)$, $left(begin{array}{cc} 0 & 0\ 1 & 0 end{array}right)$ and $left(begin{array}{cc} 0 & 0\ 0 & 1 end{array}right)$. But I could not go further.
Any help is appreciated. Thank you
linear-algebra matrices
linear-algebra matrices
asked Jan 6 at 13:28
nurun neshanurun nesha
9952623
9952623
4
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
2
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38
add a comment |
4
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
2
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38
4
4
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
2
2
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.
$endgroup$
add a comment |
$begingroup$
In fact it is false for all $nge2$ (an even stronger statement).
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.
So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.
This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.
So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.
But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.
$endgroup$
add a comment |
$begingroup$
Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.
$endgroup$
add a comment |
$begingroup$
It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.
$endgroup$
add a comment |
$begingroup$
It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.
$endgroup$
It is already false for $n=2$. By Cayley-Hamilton, $A^2-tr(A)A+det(A)I=0$ for all $Ain M_2(K)$, so that $I,A,A^2,ldots,$ do not span the $4$-dimensional vector space $M_2(K)$ for any $A$.
answered Jan 6 at 13:44
Dietrich BurdeDietrich Burde
78.3k64386
78.3k64386
add a comment |
add a comment |
$begingroup$
In fact it is false for all $nge2$ (an even stronger statement).
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.
So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.
This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.
So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.
But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.
$endgroup$
add a comment |
$begingroup$
In fact it is false for all $nge2$ (an even stronger statement).
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.
So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.
This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.
So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.
But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.
$endgroup$
add a comment |
$begingroup$
In fact it is false for all $nge2$ (an even stronger statement).
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.
So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.
This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.
So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.
But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.
$endgroup$
In fact it is false for all $nge2$ (an even stronger statement).
By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial. This is an $n$th degree polynomial, $C_A(x)=operatorname{det}(xI-A)=x^n+a_{n-1}A^{n-1}+dots +a_0$.
So $C_A(A)=A^n+a_{n-1}A^{n-1}dots+a_0=0$.
This shows that $A^ninoperatorname{span}{A,dots,A^{n-1}}$. Thus, $operatorname{span}{A^pmid pge1}=operatorname{span}{A,dots,A^{n-1}}$.
So, $operatorname{dim}(operatorname{span}{A^pmid pge1})lt n$.
But the dimension of $M_n(Bbb R)$ is $n^2$. Therefore the answer is no, since $nge2implies n^2gt n$.
edited Jan 6 at 15:47
answered Jan 6 at 14:56
Chris CusterChris Custer
11.2k3824
11.2k3824
add a comment |
add a comment |
$begingroup$
Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.
$endgroup$
add a comment |
$begingroup$
Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.
$endgroup$
add a comment |
$begingroup$
Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.
$endgroup$
Here is another proof. If ${A^p: pge1}$ spans $M_n(mathbb R)$, then all matrices are polynomials in $A$ and hence they commute with $A$. Now it is a standard exercise to show that if $A$ that commutes with all matrices, $A$ must be a scalar matrix. Yet, clearly, when $nge2$, if $A$ is a scalar matrix, the linear span of ${A^p: pge1}$ is at most the set of all scalar matrices. It cannot possibly be the set of all square matrices.
answered Jan 7 at 16:20
user1551user1551
72.2k566127
72.2k566127
add a comment |
add a comment |
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4
$begingroup$
Hint: use the Cayley-Hamilton theorem.
$endgroup$
– Mindlack
Jan 6 at 13:35
2
$begingroup$
Show that $text{span}{A^k ;|;kge 0}$ is of dimension $le n$.
$endgroup$
– Song
Jan 6 at 13:37
$begingroup$
Yes we can't find $n^2$ linearly independent matrices.
$endgroup$
– nurun nesha
Jan 6 at 13:38