Mixing
Understanding repeated Eigen values
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I am trying to understand the method of finding eigen vectors in the case of repeated eigen values. My question is based on section 3.5.2 of this link.
In the first example (Example 3.5.4),
$A = beginbmatrix2&0\0&2endbmatrix$
Here, the $lambda = 2$ (repeated eigen value). If $(A - lambda I)$ is calculated, it is zero matrix of dimension 2. So, the geometric multiplicity is 2, which means there must be two linearly independent eigen vectors.
1) Am I correct in understanding that these can be any two linearly independent vectors since $(A - lambda I)$ is a zero matrix? Or is there a reason for picking $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as shown in the link?
Now, consider the second example (Example 3.5.5):
$A = beginbmatrix5&1\-4&1endbmatrix$
In this case, $lambda = 3$ (repeated eigen value) and
$A - lambda I = beginbmatrix2&1\-4&-2endbmatrix$
Here geometric multiplicity of $lambda$ is 1. So, there is only one linearly independent eigen vector.
2) What is the idea behind using $(A - lambda I)v_2 = v_1$ to find second eigen vector?
3) Is this technique used only when geometric multiplicity is less than algebraic multiplicity? Otherwise, do we just use logic to find all independent eigen vectors as we did in the first example?
4) Since the matrix is not diagonalizable, is the idea to minimize the error, rather than solve the system of equations? I am trying to understand the need to find a second eigen vector in practical situations.
eigenvalues-eigenvectors diagonalization
asked Jul 30 at 19:55
skr_robo
1013
add a comment |Â
up vote
0
down vote
favorite
I am trying to understand the method of finding eigen vectors in the case of repeated eigen values. My question is based on section 3.5.2 of this link.
In the first example (Example 3.5.4),
$A = beginbmatrix2&0\0&2endbmatrix$
Here, the $lambda = 2$ (repeated eigen value). If $(A - lambda I)$ is calculated, it is zero matrix of dimension 2. So, the geometric multiplicity is 2, which means there must be two linearly independent eigen vectors.
1) Am I correct in understanding that these can be any two linearly independent vectors since $(A - lambda I)$ is a zero matrix? Or is there a reason for picking $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as shown in the link?
Now, consider the second example (Example 3.5.5):
$A = beginbmatrix5&1\-4&1endbmatrix$
In this case, $lambda = 3$ (repeated eigen value) and
$A - lambda I = beginbmatrix2&1\-4&-2endbmatrix$
Here geometric multiplicity of $lambda$ is 1. So, there is only one linearly independent eigen vector.
2) What is the idea behind using $(A - lambda I)v_2 = v_1$ to find second eigen vector?
3) Is this technique used only when geometric multiplicity is less than algebraic multiplicity? Otherwise, do we just use logic to find all independent eigen vectors as we did in the first example?
4) Since the matrix is not diagonalizable, is the idea to minimize the error, rather than solve the system of equations? I am trying to understand the need to find a second eigen vector in practical situations.
eigenvalues-eigenvectors diagonalization
asked Jul 30 at 19:55
skr_robo
1013
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
1
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
1
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to understand the method of finding eigen vectors in the case of repeated eigen values. My question is based on section 3.5.2 of this link.
In the first example (Example 3.5.4),
$A = beginbmatrix2&0\0&2endbmatrix$
Here, the $lambda = 2$ (repeated eigen value). If $(A - lambda I)$ is calculated, it is zero matrix of dimension 2. So, the geometric multiplicity is 2, which means there must be two linearly independent eigen vectors.
1) Am I correct in understanding that these can be any two linearly independent vectors since $(A - lambda I)$ is a zero matrix? Or is there a reason for picking $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as shown in the link?
Now, consider the second example (Example 3.5.5):
$A = beginbmatrix5&1\-4&1endbmatrix$
In this case, $lambda = 3$ (repeated eigen value) and
$A - lambda I = beginbmatrix2&1\-4&-2endbmatrix$
Here geometric multiplicity of $lambda$ is 1. So, there is only one linearly independent eigen vector.
2) What is the idea behind using $(A - lambda I)v_2 = v_1$ to find second eigen vector?
3) Is this technique used only when geometric multiplicity is less than algebraic multiplicity? Otherwise, do we just use logic to find all independent eigen vectors as we did in the first example?
4) Since the matrix is not diagonalizable, is the idea to minimize the error, rather than solve the system of equations? I am trying to understand the need to find a second eigen vector in practical situations.
eigenvalues-eigenvectors diagonalization
asked Jul 30 at 19:55
skr_robo
1013
I am trying to understand the method of finding eigen vectors in the case of repeated eigen values. My question is based on section 3.5.2 of this link.
In the first example (Example 3.5.4),
$A = beginbmatrix2&0\0&2endbmatrix$
Here, the $lambda = 2$ (repeated eigen value). If $(A - lambda I)$ is calculated, it is zero matrix of dimension 2. So, the geometric multiplicity is 2, which means there must be two linearly independent eigen vectors.
1) Am I correct in understanding that these can be any two linearly independent vectors since $(A - lambda I)$ is a zero matrix? Or is there a reason for picking $v_1 = (1, 0)$ and $v_2 = (0, 1)$ as shown in the link?
Now, consider the second example (Example 3.5.5):
$A = beginbmatrix5&1\-4&1endbmatrix$
In this case, $lambda = 3$ (repeated eigen value) and
$A - lambda I = beginbmatrix2&1\-4&-2endbmatrix$
Here geometric multiplicity of $lambda$ is 1. So, there is only one linearly independent eigen vector.
2) What is the idea behind using $(A - lambda I)v_2 = v_1$ to find second eigen vector?
3) Is this technique used only when geometric multiplicity is less than algebraic multiplicity? Otherwise, do we just use logic to find all independent eigen vectors as we did in the first example?
4) Since the matrix is not diagonalizable, is the idea to minimize the error, rather than solve the system of equations? I am trying to understand the need to find a second eigen vector in practical situations.
eigenvalues-eigenvectors diagonalization
asked Jul 30 at 19:55
skr_robo
1013
asked Jul 30 at 19:55
skr_robo
1013
asked Jul 30 at 19:55
skr_robo
1013
asked Jul 30 at 19:55
skr_robo
1013
1013
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
1
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
1
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18
add a comment |Â
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
1
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
1
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
1
1
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
1
1
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f2867360%2funderstanding-repeated-eigen-values%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
What they are doing is finding the Jordan Normal Form. The point of that is writing some matrix $J = D + N,$ in this case $D$ would be $3I$ and $N$ would satisfy $N^2 = 0,$ crucial that $DN=ND.$ So $e^J$ and $e^Jt$ are fairl easy and concrete
– Will Jagy
Jul 30 at 20:01
1
anyway, as the minimal polynomial is the same as the characteristic, you take any vector such that $(A-lambda I)^2 w = 0$ BUT $(A-lambda I) w neq 0.$ Then take $v = (A-lambda I) w$ satisfies $(A-lambda I) v = 0$ so is a genuine eigenvector
– Will Jagy
Jul 30 at 20:04
en.wikipedia.org/wiki/Jordan_normal_form
– Will Jagy
Jul 30 at 20:06
I believe you are addressing question 2 here. I haven't quite understood the explanation. But I will get back after some more reading.
– skr_robo
Jul 30 at 20:11
1
alright. Given your interest, I recommend getting a fairly applied linear algebra book, one that emphasizes the real numbers and complexes. I have answered a dozen Jordan form questions on this site, when I get back from grocery shopping I will figure out some that you can read profitably. You can also search for questions on Jordan form yourself.
– Will Jagy
Jul 30 at 20:18