Mixing
If $A$ be a matrix with $n$ distinct eigenvalues then $u, Au , A^2u ,cdots, A^{n-1} u$ will be independent.
$begingroup$
If $A$ be a matrix with $n$ distinct eigenvalues then there exists a vector $u$ such that $u,Au,A^2u,cdots, A^{n-1} u$ will be independent.
Can anyone give me a hint to solve this problem?
I know that if $A$ has $n$ distinct eigen values then it will have $n$ independent eigen vectors..
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Jan 27 at 1:18
cmicmi
1,136312
$endgroup$
add a comment |
$begingroup$
If $A$ be a matrix with $n$ distinct eigenvalues then there exists a vector $u$ such that $u,Au,A^2u,cdots, A^{n-1} u$ will be independent.
Can anyone give me a hint to solve this problem?
I know that if $A$ has $n$ distinct eigen values then it will have $n$ independent eigen vectors..
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Jan 27 at 1:18
cmicmi
1,136312
$endgroup$
$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
2
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21
add a comment |
$begingroup$
If $A$ be a matrix with $n$ distinct eigenvalues then there exists a vector $u$ such that $u,Au,A^2u,cdots, A^{n-1} u$ will be independent.
Can anyone give me a hint to solve this problem?
I know that if $A$ has $n$ distinct eigen values then it will have $n$ independent eigen vectors..
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Jan 27 at 1:18
cmicmi
1,136312
$endgroup$
If $A$ be a matrix with $n$ distinct eigenvalues then there exists a vector $u$ such that $u,Au,A^2u,cdots, A^{n-1} u$ will be independent.
Can anyone give me a hint to solve this problem?
I know that if $A$ has $n$ distinct eigen values then it will have $n$ independent eigen vectors..
linear-algebra vector-spaces eigenvalues-eigenvectors
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Jan 27 at 1:18
cmicmi
1,136312
asked Jan 27 at 1:18
cmicmi
1,136312
edited Jan 27 at 1:57
cmi
asked Jan 27 at 1:18
cmicmi
1,136312
asked Jan 27 at 1:18
cmicmi
1,136312
asked Jan 27 at 1:18
cmicmi
1,136312
1,136312
$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
2
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21
add a comment |
$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
2
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21
$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
2
2
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21
add a comment |
1 Answer
1
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oldest
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$begingroup$
Hint:
You must prove there exists $u$ such that $u,Au,A^2u,cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!
Added: Start with $$a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$ where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$
Edit: Now $$Au=Au_1+Au_2+cdots+Au_n=lambda_1u_1+lambda_2u_2+cdots+lambda_nu_n$$
$$A^2u=lambda_1^2u_1+lambda_2^2u_2+cdots+lambda_n^2u_n$$
$$vdots$$
$$A^{n-1}u=lambda_1^{n-1}u_1+lambda_2^{n-1}u_2+cdots+lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$. That is,
$a_0(u_1+cdots+u_n)+a_1(lambda_1u_1+cdots+lambda_nu_n)+cdots+a_{n-1}(lambda_1^{n-1}u_1+cdots+lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1})u_1$$ $$+(a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-2})u_2$$ $$+cdots+(a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1})u_n=0$$
since ${u_i}$ are linearly independent(How?), $$a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1}=0$$ $$a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-1}=0$$ $$vdots$$ $$a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1}=0$$
The above can be written as $$begin{pmatrix} 1& lambda_1 &lambda_1^2& cdots&lambda_1^{n-1}\ 1& lambda_2 &lambda_2^2& cdots&lambda_2^{n-1} \ vdots\ 1& lambda_n &lambda_n^2& cdots&lambda_n^{n-1} end{pmatrix} begin{pmatrix} a_0\a_1\vdots\a_{n-1}end{pmatrix}=begin{pmatrix} 0\0\vdots\0end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $lambda_i$'s are distinct, so its determinant is non zero, so......?
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint:
You must prove there exists $u$ such that $u,Au,A^2u,cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!
Added: Start with $$a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$ where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$
Edit: Now $$Au=Au_1+Au_2+cdots+Au_n=lambda_1u_1+lambda_2u_2+cdots+lambda_nu_n$$
$$A^2u=lambda_1^2u_1+lambda_2^2u_2+cdots+lambda_n^2u_n$$
$$vdots$$
$$A^{n-1}u=lambda_1^{n-1}u_1+lambda_2^{n-1}u_2+cdots+lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$. That is,
$a_0(u_1+cdots+u_n)+a_1(lambda_1u_1+cdots+lambda_nu_n)+cdots+a_{n-1}(lambda_1^{n-1}u_1+cdots+lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1})u_1$$ $$+(a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-2})u_2$$ $$+cdots+(a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1})u_n=0$$
since ${u_i}$ are linearly independent(How?), $$a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1}=0$$ $$a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-1}=0$$ $$vdots$$ $$a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1}=0$$
The above can be written as $$begin{pmatrix} 1& lambda_1 &lambda_1^2& cdots&lambda_1^{n-1}\ 1& lambda_2 &lambda_2^2& cdots&lambda_2^{n-1} \ vdots\ 1& lambda_n &lambda_n^2& cdots&lambda_n^{n-1} end{pmatrix} begin{pmatrix} a_0\a_1\vdots\a_{n-1}end{pmatrix}=begin{pmatrix} 0\0\vdots\0end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $lambda_i$'s are distinct, so its determinant is non zero, so......?
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
add a comment |
$begingroup$
Hint:
You must prove there exists $u$ such that $u,Au,A^2u,cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!
Added: Start with $$a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$ where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$
Edit: Now $$Au=Au_1+Au_2+cdots+Au_n=lambda_1u_1+lambda_2u_2+cdots+lambda_nu_n$$
$$A^2u=lambda_1^2u_1+lambda_2^2u_2+cdots+lambda_n^2u_n$$
$$vdots$$
$$A^{n-1}u=lambda_1^{n-1}u_1+lambda_2^{n-1}u_2+cdots+lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$. That is,
$a_0(u_1+cdots+u_n)+a_1(lambda_1u_1+cdots+lambda_nu_n)+cdots+a_{n-1}(lambda_1^{n-1}u_1+cdots+lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1})u_1$$ $$+(a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-2})u_2$$ $$+cdots+(a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1})u_n=0$$
since ${u_i}$ are linearly independent(How?), $$a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1}=0$$ $$a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-1}=0$$ $$vdots$$ $$a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1}=0$$
The above can be written as $$begin{pmatrix} 1& lambda_1 &lambda_1^2& cdots&lambda_1^{n-1}\ 1& lambda_2 &lambda_2^2& cdots&lambda_2^{n-1} \ vdots\ 1& lambda_n &lambda_n^2& cdots&lambda_n^{n-1} end{pmatrix} begin{pmatrix} a_0\a_1\vdots\a_{n-1}end{pmatrix}=begin{pmatrix} 0\0\vdots\0end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $lambda_i$'s are distinct, so its determinant is non zero, so......?
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
add a comment |
$begingroup$
Hint:
You must prove there exists $u$ such that $u,Au,A^2u,cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!
Added: Start with $$a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$ where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$
Edit: Now $$Au=Au_1+Au_2+cdots+Au_n=lambda_1u_1+lambda_2u_2+cdots+lambda_nu_n$$
$$A^2u=lambda_1^2u_1+lambda_2^2u_2+cdots+lambda_n^2u_n$$
$$vdots$$
$$A^{n-1}u=lambda_1^{n-1}u_1+lambda_2^{n-1}u_2+cdots+lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$. That is,
$a_0(u_1+cdots+u_n)+a_1(lambda_1u_1+cdots+lambda_nu_n)+cdots+a_{n-1}(lambda_1^{n-1}u_1+cdots+lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1})u_1$$ $$+(a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-2})u_2$$ $$+cdots+(a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1})u_n=0$$
since ${u_i}$ are linearly independent(How?), $$a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1}=0$$ $$a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-1}=0$$ $$vdots$$ $$a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1}=0$$
The above can be written as $$begin{pmatrix} 1& lambda_1 &lambda_1^2& cdots&lambda_1^{n-1}\ 1& lambda_2 &lambda_2^2& cdots&lambda_2^{n-1} \ vdots\ 1& lambda_n &lambda_n^2& cdots&lambda_n^{n-1} end{pmatrix} begin{pmatrix} a_0\a_1\vdots\a_{n-1}end{pmatrix}=begin{pmatrix} 0\0\vdots\0end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $lambda_i$'s are distinct, so its determinant is non zero, so......?
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
$endgroup$
Hint:
You must prove there exists $u$ such that $u,Au,A^2u,cdots,A^{n-1}u$ is linearly independent
Take $u=u_1+u_2+cdots+u_n$ where $u_i$'s are eigenvectors corresponding to the distinct $n$ eigenvalues and then show this $u$ works!
Added: Start with $$a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$$
and substitute $u$ to expand this and do some calculations to end with $Bx=0$ where $B$ is a Vandermonde matrix with distinct entries and $x=(a_0,a_1,...,a_{n-1})^t$ and then conclude $x=B^{-1}0=0$
Edit: Now $$Au=Au_1+Au_2+cdots+Au_n=lambda_1u_1+lambda_2u_2+cdots+lambda_nu_n$$
$$A^2u=lambda_1^2u_1+lambda_2^2u_2+cdots+lambda_n^2u_n$$
$$vdots$$
$$A^{n-1}u=lambda_1^{n-1}u_1+lambda_2^{n-1}u_2+cdots+lambda_n^{n-1}u_n$$
Suppose $a_0u+a_1Au+cdots+a_{n-1}A^{n-1}u=0$. That is,
$a_0(u_1+cdots+u_n)+a_1(lambda_1u_1+cdots+lambda_nu_n)+cdots+a_{n-1}(lambda_1^{n-1}u_1+cdots+lambda_n^{n-1}u_n)=0$
The above can be re written as $$(a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1})u_1$$ $$+(a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-2})u_2$$ $$+cdots+(a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1})u_n=0$$
since ${u_i}$ are linearly independent(How?), $$a_0+a_1lambda_1+cdots+a_{n-1}lambda_1^{n-1}=0$$ $$a_0+a_1lambda_2+cdots+a_{n-1}lambda_2^{n-1}=0$$ $$vdots$$ $$a_0+a_1lambda_n+cdots+a_{n-1}lambda_n^{n-1}=0$$
The above can be written as $$begin{pmatrix} 1& lambda_1 &lambda_1^2& cdots&lambda_1^{n-1}\ 1& lambda_2 &lambda_2^2& cdots&lambda_2^{n-1} \ vdots\ 1& lambda_n &lambda_n^2& cdots&lambda_n^{n-1} end{pmatrix} begin{pmatrix} a_0\a_1\vdots\a_{n-1}end{pmatrix}=begin{pmatrix} 0\0\vdots\0end{pmatrix}$$ Since the coefficient matrix is Vandermonde, and $lambda_i$'s are distinct, so its determinant is non zero, so......?
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
edited Jan 29 at 7:05
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
answered Jan 27 at 1:28
Chinnapparaj RChinnapparaj R
5,8082928
5,8082928
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
add a comment |
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
$begingroup$
I did not get how to get a vandermonde Matrix $B$? Can you show the calculation?@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:31
1
1
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
@cmi: Just Hold on.....I'm typing.....!
$endgroup$
– Chinnapparaj R
Jan 29 at 6:39
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
I got it on my own...You may not type any more...@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 6:49
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
$B$ seems to be the sum of n Vandermonde Matrices..@Chinnapparaj R
$endgroup$
– cmi
Jan 29 at 7:04
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
$begingroup$
@cmi: Anyway, see my edit !
$endgroup$
– Chinnapparaj R
Jan 29 at 7:06
add a comment |
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$begingroup$
See here: math.stackexchange.com/questions/2627894/…
$endgroup$
– Math1000
Jan 27 at 1:20
2
$begingroup$
This is certainly not true for any $u$, just pick an eigenvector...
$endgroup$
– Klaus
Jan 27 at 1:21