Mixing
Calculating the null space of a $n^textth$ power of a $2times 2$ matrix with an unkown variable.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$$
left(beginmatrix
0 & 1 \
x & 1 \
endmatrixright)
$$
Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
matrices
edited Aug 6 at 10:56
Davide Morgante
1,938220
asked Aug 6 at 6:36
king jigg
112
add a comment |Â
up vote
2
down vote
favorite
$$
left(beginmatrix
0 & 1 \
x & 1 \
endmatrixright)
$$
Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
matrices
edited Aug 6 at 10:56
Davide Morgante
1,938220
asked Aug 6 at 6:36
king jigg
112
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title does not reflect the question.
– lhf
Aug 6 at 11:02
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$
left(beginmatrix
0 & 1 \
x & 1 \
endmatrixright)
$$
Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
matrices
edited Aug 6 at 10:56
Davide Morgante
1,938220
asked Aug 6 at 6:36
king jigg
112
$$
left(beginmatrix
0 & 1 \
x & 1 \
endmatrixright)
$$
Suppose I have a matrix above to the power of n, is it possible to use eigendecomposition in order to find all the eigenvectors in terms of x?
matrices
edited Aug 6 at 10:56
Davide Morgante
1,938220
asked Aug 6 at 6:36
king jigg
112
edited Aug 6 at 10:56
Davide Morgante
1,938220
edited Aug 6 at 10:56
Davide Morgante
1,938220
edited Aug 6 at 10:56
Davide Morgante
1,938220
1,938220
asked Aug 6 at 6:36
king jigg
112
asked Aug 6 at 6:36
king jigg
112
asked Aug 6 at 6:36
king jigg
112
112
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title does not reflect the question.
– lhf
Aug 6 at 11:02
add a comment |Â
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title does not reflect the question.
– lhf
Aug 6 at 11:02
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title does not reflect the question.
– lhf
Aug 6 at 11:02
The title does not reflect the question.
– lhf
Aug 6 at 11:02
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Hint: An eigenvector of $A$ for eigenvalue $lambda$ is an eigenvector of $A^n$ for eigenvalue $lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
answered Aug 6 at 7:36
Robert Israel
304k22201443
add a comment |Â
up vote
1
down vote
A more general result is true. It is a well known result that if $A in M_n(mathbbC)$, there exists $P in GL_n(mathbbC)$ such that $PAP^-1$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B in M_n(mathbbC)$ are upper triangular and $i in [n]$,
$$
(lambda A)_ii = lambda A_ii
$$
and
$$
(AB)_ii = sum_m= 1^ma_imb_mi = a_iib_ii
$$
so if $q in mathbbC[X]$, then $q(A)_ii = q(A_ii)$ for all $i$ (again, induction should do the trick). Moreover, if $q = sum_j = 1^dc_jX^j$, then
$$
q(PAP^-1) = sum_j = 1^dc_j(PAP^-1)^j = sum_j = 1^dc_jPA^jP^-1 = Pleft(sum_j = 1^dc_jA^jright)P^-1 = Pq(A)P^-1
$$
To sum up, since
$$
q(A) = q(P^-1PAP^-1P) = P^-1q(PAP^-1)P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^-1)$. Since the eigenavalues of $q(PAP^-1)$ are the eigenvalues of $PAP^-1$ via $q$ and $PAP^-1$ has the same eigenvalues than $A$,
$$
operatornameSpec(q(A)) = q(operatornameSpec(A)).
$$
In your particular case, $q = X^n$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
answered Aug 6 at 7:41
Guido A.
3,851624
add a comment |Â
up vote
1
down vote
Hint
$$A^n = P D^n P^-1$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
answered Aug 6 at 7:36
Davide Morgante
1,938220
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: An eigenvector of $A$ for eigenvalue $lambda$ is an eigenvector of $A^n$ for eigenvalue $lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
answered Aug 6 at 7:36
Robert Israel
304k22201443
add a comment |Â
up vote
1
down vote
Hint: An eigenvector of $A$ for eigenvalue $lambda$ is an eigenvector of $A^n$ for eigenvalue $lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
answered Aug 6 at 7:36
Robert Israel
304k22201443
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: An eigenvector of $A$ for eigenvalue $lambda$ is an eigenvector of $A^n$ for eigenvalue $lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
answered Aug 6 at 7:36
Robert Israel
304k22201443
Hint: An eigenvector of $A$ for eigenvalue $lambda$ is an eigenvector of $A^n$ for eigenvalue $lambda^n$.
The process of finding eigenvalues and eigenvectors is pretty much the same with a symbolic parameter $x$.
The only tricky part is if $x = -1/4$.
answered Aug 6 at 7:36
Robert Israel
304k22201443
answered Aug 6 at 7:36
Robert Israel
304k22201443
answered Aug 6 at 7:36
Robert Israel
304k22201443
answered Aug 6 at 7:36
Robert Israel
304k22201443
304k22201443
add a comment |Â
add a comment |Â
up vote
1
down vote
A more general result is true. It is a well known result that if $A in M_n(mathbbC)$, there exists $P in GL_n(mathbbC)$ such that $PAP^-1$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B in M_n(mathbbC)$ are upper triangular and $i in [n]$,
$$
(lambda A)_ii = lambda A_ii
$$
and
$$
(AB)_ii = sum_m= 1^ma_imb_mi = a_iib_ii
$$
so if $q in mathbbC[X]$, then $q(A)_ii = q(A_ii)$ for all $i$ (again, induction should do the trick). Moreover, if $q = sum_j = 1^dc_jX^j$, then
$$
q(PAP^-1) = sum_j = 1^dc_j(PAP^-1)^j = sum_j = 1^dc_jPA^jP^-1 = Pleft(sum_j = 1^dc_jA^jright)P^-1 = Pq(A)P^-1
$$
To sum up, since
$$
q(A) = q(P^-1PAP^-1P) = P^-1q(PAP^-1)P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^-1)$. Since the eigenavalues of $q(PAP^-1)$ are the eigenvalues of $PAP^-1$ via $q$ and $PAP^-1$ has the same eigenvalues than $A$,
$$
operatornameSpec(q(A)) = q(operatornameSpec(A)).
$$
In your particular case, $q = X^n$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
answered Aug 6 at 7:41
Guido A.
3,851624
add a comment |Â
up vote
1
down vote
A more general result is true. It is a well known result that if $A in M_n(mathbbC)$, there exists $P in GL_n(mathbbC)$ such that $PAP^-1$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B in M_n(mathbbC)$ are upper triangular and $i in [n]$,
$$
(lambda A)_ii = lambda A_ii
$$
and
$$
(AB)_ii = sum_m= 1^ma_imb_mi = a_iib_ii
$$
so if $q in mathbbC[X]$, then $q(A)_ii = q(A_ii)$ for all $i$ (again, induction should do the trick). Moreover, if $q = sum_j = 1^dc_jX^j$, then
$$
q(PAP^-1) = sum_j = 1^dc_j(PAP^-1)^j = sum_j = 1^dc_jPA^jP^-1 = Pleft(sum_j = 1^dc_jA^jright)P^-1 = Pq(A)P^-1
$$
To sum up, since
$$
q(A) = q(P^-1PAP^-1P) = P^-1q(PAP^-1)P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^-1)$. Since the eigenavalues of $q(PAP^-1)$ are the eigenvalues of $PAP^-1$ via $q$ and $PAP^-1$ has the same eigenvalues than $A$,
$$
operatornameSpec(q(A)) = q(operatornameSpec(A)).
$$
In your particular case, $q = X^n$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
answered Aug 6 at 7:41
Guido A.
3,851624
add a comment |Â
up vote
1
down vote
up vote
1
down vote
A more general result is true. It is a well known result that if $A in M_n(mathbbC)$, there exists $P in GL_n(mathbbC)$ such that $PAP^-1$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B in M_n(mathbbC)$ are upper triangular and $i in [n]$,
$$
(lambda A)_ii = lambda A_ii
$$
and
$$
(AB)_ii = sum_m= 1^ma_imb_mi = a_iib_ii
$$
so if $q in mathbbC[X]$, then $q(A)_ii = q(A_ii)$ for all $i$ (again, induction should do the trick). Moreover, if $q = sum_j = 1^dc_jX^j$, then
$$
q(PAP^-1) = sum_j = 1^dc_j(PAP^-1)^j = sum_j = 1^dc_jPA^jP^-1 = Pleft(sum_j = 1^dc_jA^jright)P^-1 = Pq(A)P^-1
$$
To sum up, since
$$
q(A) = q(P^-1PAP^-1P) = P^-1q(PAP^-1)P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^-1)$. Since the eigenavalues of $q(PAP^-1)$ are the eigenvalues of $PAP^-1$ via $q$ and $PAP^-1$ has the same eigenvalues than $A$,
$$
operatornameSpec(q(A)) = q(operatornameSpec(A)).
$$
In your particular case, $q = X^n$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
answered Aug 6 at 7:41
Guido A.
3,851624
A more general result is true. It is a well known result that if $A in M_n(mathbbC)$, there exists $P in GL_n(mathbbC)$ such that $PAP^-1$ is upper triangular. In particular, all eigenvalues of $A$ are in this matrices' diagonal, counted with multiplicity (if not inmediately clear, recall that eigenvalues are preserved by conjugation, and write the expression of a characteristic polynomial for a general upper triangular matrix).
Now, if $A, B in M_n(mathbbC)$ are upper triangular and $i in [n]$,
$$
(lambda A)_ii = lambda A_ii
$$
and
$$
(AB)_ii = sum_m= 1^ma_imb_mi = a_iib_ii
$$
so if $q in mathbbC[X]$, then $q(A)_ii = q(A_ii)$ for all $i$ (again, induction should do the trick). Moreover, if $q = sum_j = 1^dc_jX^j$, then
$$
q(PAP^-1) = sum_j = 1^dc_j(PAP^-1)^j = sum_j = 1^dc_jPA^jP^-1 = Pleft(sum_j = 1^dc_jA^jright)P^-1 = Pq(A)P^-1
$$
To sum up, since
$$
q(A) = q(P^-1PAP^-1P) = P^-1q(PAP^-1)P
$$
the eigenvalues of $q(A)$ are the same of those of $q(PAP^-1)$. Since the eigenavalues of $q(PAP^-1)$ are the eigenvalues of $PAP^-1$ via $q$ and $PAP^-1$ has the same eigenvalues than $A$,
$$
operatornameSpec(q(A)) = q(operatornameSpec(A)).
$$
In your particular case, $q = X^n$. Thus, it will suffice to find the eigenvalues of $A$ and take their $n$-th power.
answered Aug 6 at 7:41
Guido A.
3,851624
answered Aug 6 at 7:41
Guido A.
3,851624
answered Aug 6 at 7:41
Guido A.
3,851624
answered Aug 6 at 7:41
Guido A.
3,851624
3,851624
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint
$$A^n = P D^n P^-1$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
answered Aug 6 at 7:36
Davide Morgante
1,938220
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
add a comment |Â
up vote
1
down vote
Hint
$$A^n = P D^n P^-1$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
answered Aug 6 at 7:36
Davide Morgante
1,938220
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint
$$A^n = P D^n P^-1$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
answered Aug 6 at 7:36
Davide Morgante
1,938220
Hint
$$A^n = P D^n P^-1$$ where $P$ is the matrix of the eigenvectors and $D$ the matrix of the eigenvalues
answered Aug 6 at 7:36
Davide Morgante
1,938220
edited Aug 6 at 7:42
answered Aug 6 at 7:36
Davide Morgante
1,938220
answered Aug 6 at 7:36
Davide Morgante
1,938220
answered Aug 6 at 7:36
Davide Morgante
1,938220
1,938220
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
add a comment |Â
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
Hey, thanks for the help but I have one more inquiry about solving for the eigenvectors. I calculated the eigenvalues to be (x+(1+4x)^.5)/2 and (x-(1+4x)^.5)/2. I substituted that back into the (A-(lambda)(I)) matrix for lambda but could not solve the corresponding eigenvectors. Is it possible to solve it and if so, how?
– king jigg
Aug 6 at 13:24
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
To find eigenvectors you either solve the equation $$Amathbfv=lambdamathbfv$$ for a generic $mathbfv$ or $$ker(A-lambda mathcalI)=0$$
– Davide Morgante
Aug 6 at 14:40
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873634%2fcalculating-the-null-space-of-a-n-textth-power-of-a-2-times-2-matrix-wi%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
The title says $n times n$ matrix, but you're asking about a $2 times 2$ matrix to the power $n$, which is still a $2 times 2$ matrix.
– Robert Israel
Aug 6 at 7:34
The title does not reflect the question.
– lhf
Aug 6 at 11:02