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Convergence of unit vector in dir. of $M^k*v$ to the principal eigenvector of M when $kto infty$ and M is symmetric

0

$begingroup$

It is a standard fact that a square matrix $M$ of dimension $n$ has at most $n$ distinct eigenvalues, each of them a real number, and the sum of their multiplicities is exactly $n$. We will denote these eigenvalues by $λ_1(M), λ_2(M),...,λ_n(M)$, indexed in order of decreasing absolute value, and with each eigenvalue listed a number of times equal to its multiplicity. For each distinct eigenvalue, we
choose an orthonormal basis of its eigenspace; considering the vectors in all these bases, we obtain a set of eigenvectors $ω_1(M), ω_2(M),...,ω_n(M)$ that we can index in such a way that $ω_i(M)$ belongs to the eigenspace of $λ_i(M)$.
For the sake of simplicity, we will make the following technical assumption about all the
matrices we deal with:
|$λ_1(M)$| > |$λ_2(M)$|.
When this assumption holds, we refer to $ω_1(M)$ as the principal eigenvector, and all other $ω_i(M)$ as non-principal eigenvectors.

If $M$ is a symmetric $n × n$ matrix, and $v$ is a vector not orthogonal to the principal eigenvector $ω_1(M)$, then the unit vector in the direction of $M^kv$ converges to $ω_1(M)$ as $k$ increases without bound.

I have been trying to prove the above theorem but haven't been able to. Please help.

linear-algebra matrices limits

share|cite|improve this question

edited Mar 20 at 8:54

Jeevesh Juneja

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

$endgroup$

add a comment | 

0

$begingroup$

It is a standard fact that a square matrix $M$ of dimension $n$ has at most $n$ distinct eigenvalues, each of them a real number, and the sum of their multiplicities is exactly $n$. We will denote these eigenvalues by $λ_1(M), λ_2(M),...,λ_n(M)$, indexed in order of decreasing absolute value, and with each eigenvalue listed a number of times equal to its multiplicity. For each distinct eigenvalue, we
choose an orthonormal basis of its eigenspace; considering the vectors in all these bases, we obtain a set of eigenvectors $ω_1(M), ω_2(M),...,ω_n(M)$ that we can index in such a way that $ω_i(M)$ belongs to the eigenspace of $λ_i(M)$.
For the sake of simplicity, we will make the following technical assumption about all the
matrices we deal with:
|$λ_1(M)$| > |$λ_2(M)$|.
When this assumption holds, we refer to $ω_1(M)$ as the principal eigenvector, and all other $ω_i(M)$ as non-principal eigenvectors.

If $M$ is a symmetric $n × n$ matrix, and $v$ is a vector not orthogonal to the principal eigenvector $ω_1(M)$, then the unit vector in the direction of $M^kv$ converges to $ω_1(M)$ as $k$ increases without bound.

I have been trying to prove the above theorem but haven't been able to. Please help.

linear-algebra matrices limits

share|cite|improve this question

edited Mar 20 at 8:54

Jeevesh Juneja

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

$endgroup$

add a comment | 

$begingroup$

It is a standard fact that a square matrix $M$ of dimension $n$ has at most $n$ distinct eigenvalues, each of them a real number, and the sum of their multiplicities is exactly $n$. We will denote these eigenvalues by $λ_1(M), λ_2(M),...,λ_n(M)$, indexed in order of decreasing absolute value, and with each eigenvalue listed a number of times equal to its multiplicity. For each distinct eigenvalue, we
choose an orthonormal basis of its eigenspace; considering the vectors in all these bases, we obtain a set of eigenvectors $ω_1(M), ω_2(M),...,ω_n(M)$ that we can index in such a way that $ω_i(M)$ belongs to the eigenspace of $λ_i(M)$.
For the sake of simplicity, we will make the following technical assumption about all the
matrices we deal with:
|$λ_1(M)$| > |$λ_2(M)$|.
When this assumption holds, we refer to $ω_1(M)$ as the principal eigenvector, and all other $ω_i(M)$ as non-principal eigenvectors.

If $M$ is a symmetric $n × n$ matrix, and $v$ is a vector not orthogonal to the principal eigenvector $ω_1(M)$, then the unit vector in the direction of $M^kv$ converges to $ω_1(M)$ as $k$ increases without bound.

I have been trying to prove the above theorem but haven't been able to. Please help.

linear-algebra matrices limits

share|cite|improve this question

edited Mar 20 at 8:54

Jeevesh Juneja

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

$endgroup$

share|cite|improve this question

edited Mar 20 at 8:54

Jeevesh Juneja

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

share|cite|improve this question

edited Mar 20 at 8:54

Jeevesh Juneja

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34

asked Mar 20 at 8:02

Jeevesh JunejaJeevesh Juneja

34