Constructing matrices that are diagonalizable or notDiagonalizable matrices with complex values are dense in set of $ntimes n$ complex matrices.Conditions for a Matrix to be DiagonalizableDiagonalizable vs full rank vs nonsingular (square matrix)Any Nilpotent Matrix is not DiagonalizableWhich matrices are permutation diagonalizable?I need diagonalizable and non-diagonalizable matrices which are easy to calculateNon-Negative Diagonalizable MatricesShow that this matrix is not diagonalizableDiagonalizable matricies and eigenvaluesFind the eigenvalue of the non-diagonalizable matrix
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Constructing matrices that are diagonalizable or not
Diagonalizable matrices with complex values are dense in set of $ntimes n$ complex matrices.Conditions for a Matrix to be DiagonalizableDiagonalizable vs full rank vs nonsingular (square matrix)Any Nilpotent Matrix is not DiagonalizableWhich matrices are permutation diagonalizable?I need diagonalizable and non-diagonalizable matrices which are easy to calculateNon-Negative Diagonalizable MatricesShow that this matrix is not diagonalizableDiagonalizable matricies and eigenvaluesFind the eigenvalue of the non-diagonalizable matrix
$begingroup$
What are some methods where I can construct a matrix that is diagonalizable or not quickly for some constraints such as it being singular or not or it having complex eigenvalues for 2x2 and 3x3 matrix. So far what i have done to check is to find the eigenvalues and use that to find the number of eigenvectors to see if they equal n columns of the matrix, but this is time consuming.
linear-algebra matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
$endgroup$
|
show 1 more comment
$begingroup$
What are some methods where I can construct a matrix that is diagonalizable or not quickly for some constraints such as it being singular or not or it having complex eigenvalues for 2x2 and 3x3 matrix. So far what i have done to check is to find the eigenvalues and use that to find the number of eigenvectors to see if they equal n columns of the matrix, but this is time consuming.
linear-algebra matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
$endgroup$
$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
1
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago
|
show 1 more comment
$begingroup$
What are some methods where I can construct a matrix that is diagonalizable or not quickly for some constraints such as it being singular or not or it having complex eigenvalues for 2x2 and 3x3 matrix. So far what i have done to check is to find the eigenvalues and use that to find the number of eigenvectors to see if they equal n columns of the matrix, but this is time consuming.
linear-algebra matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
$endgroup$
What are some methods where I can construct a matrix that is diagonalizable or not quickly for some constraints such as it being singular or not or it having complex eigenvalues for 2x2 and 3x3 matrix. So far what i have done to check is to find the eigenvalues and use that to find the number of eigenvectors to see if they equal n columns of the matrix, but this is time consuming.
linear-algebra matrices
linear-algebra matrices
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
13k41960
asked 2 days ago
Samurai BaleSamurai Bale
494
asked 2 days ago
Samurai BaleSamurai Bale
494
asked 2 days ago
Samurai BaleSamurai Bale
494
494
$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
1
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago
|
show 1 more comment
$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
1
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
1
1
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If you want to construct an $ntimes n$ matrix $A_lambda$ with a single eigenvalue $lambdainmathbbC$ which is not diagonalizable then use a Jordan block:
$$A = beginpmatrixlambda & 1 & & \&lambda&1&\&&lambda&1\&&&ddotsendpmatrix.$$
The Jordan block is not diagonalizable since if it was, $A - lambda I$ would be too. But $A - lambda I$ is an upper triangular matrix with zeros on the diagonal, which implies it has one eigenvalue: zero. But if $A - lambda I = P D P^-1$ where $D$ is the zero matrix, this implies $A = lambda I$, a contradiction. Thus we can conclude that $A$ is not diagonalizable.
If you'd like $A$ to be diagonalizable, just use a diagonal matrix:
$$A = beginpmatrixlambda&&&\&lambda&&\&&lambda&\&&&ddotsendpmatrix.$$
answered 2 days ago
cdipaolocdipaolo
650313
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
If you want to construct an $ntimes n$ matrix $A_lambda$ with a single eigenvalue $lambdainmathbbC$ which is not diagonalizable then use a Jordan block:
$$A = beginpmatrixlambda & 1 & & \&lambda&1&\&&lambda&1\&&&ddotsendpmatrix.$$
The Jordan block is not diagonalizable since if it was, $A - lambda I$ would be too. But $A - lambda I$ is an upper triangular matrix with zeros on the diagonal, which implies it has one eigenvalue: zero. But if $A - lambda I = P D P^-1$ where $D$ is the zero matrix, this implies $A = lambda I$, a contradiction. Thus we can conclude that $A$ is not diagonalizable.
If you'd like $A$ to be diagonalizable, just use a diagonal matrix:
$$A = beginpmatrixlambda&&&\&lambda&&\&&lambda&\&&&ddotsendpmatrix.$$
answered 2 days ago
cdipaolocdipaolo
650313
$endgroup$
add a comment |
$begingroup$
If you want to construct an $ntimes n$ matrix $A_lambda$ with a single eigenvalue $lambdainmathbbC$ which is not diagonalizable then use a Jordan block:
$$A = beginpmatrixlambda & 1 & & \&lambda&1&\&&lambda&1\&&&ddotsendpmatrix.$$
The Jordan block is not diagonalizable since if it was, $A - lambda I$ would be too. But $A - lambda I$ is an upper triangular matrix with zeros on the diagonal, which implies it has one eigenvalue: zero. But if $A - lambda I = P D P^-1$ where $D$ is the zero matrix, this implies $A = lambda I$, a contradiction. Thus we can conclude that $A$ is not diagonalizable.
If you'd like $A$ to be diagonalizable, just use a diagonal matrix:
$$A = beginpmatrixlambda&&&\&lambda&&\&&lambda&\&&&ddotsendpmatrix.$$
answered 2 days ago
cdipaolocdipaolo
650313
$endgroup$
add a comment |
$begingroup$
If you want to construct an $ntimes n$ matrix $A_lambda$ with a single eigenvalue $lambdainmathbbC$ which is not diagonalizable then use a Jordan block:
$$A = beginpmatrixlambda & 1 & & \&lambda&1&\&&lambda&1\&&&ddotsendpmatrix.$$
The Jordan block is not diagonalizable since if it was, $A - lambda I$ would be too. But $A - lambda I$ is an upper triangular matrix with zeros on the diagonal, which implies it has one eigenvalue: zero. But if $A - lambda I = P D P^-1$ where $D$ is the zero matrix, this implies $A = lambda I$, a contradiction. Thus we can conclude that $A$ is not diagonalizable.
If you'd like $A$ to be diagonalizable, just use a diagonal matrix:
$$A = beginpmatrixlambda&&&\&lambda&&\&&lambda&\&&&ddotsendpmatrix.$$
answered 2 days ago
cdipaolocdipaolo
650313
$endgroup$
If you want to construct an $ntimes n$ matrix $A_lambda$ with a single eigenvalue $lambdainmathbbC$ which is not diagonalizable then use a Jordan block:
$$A = beginpmatrixlambda & 1 & & \&lambda&1&\&&lambda&1\&&&ddotsendpmatrix.$$
The Jordan block is not diagonalizable since if it was, $A - lambda I$ would be too. But $A - lambda I$ is an upper triangular matrix with zeros on the diagonal, which implies it has one eigenvalue: zero. But if $A - lambda I = P D P^-1$ where $D$ is the zero matrix, this implies $A = lambda I$, a contradiction. Thus we can conclude that $A$ is not diagonalizable.
If you'd like $A$ to be diagonalizable, just use a diagonal matrix:
$$A = beginpmatrixlambda&&&\&lambda&&\&&lambda&\&&&ddotsendpmatrix.$$
answered 2 days ago
cdipaolocdipaolo
650313
edited 2 days ago
answered 2 days ago
cdipaolocdipaolo
650313
answered 2 days ago
cdipaolocdipaolo
650313
answered 2 days ago
cdipaolocdipaolo
650313
650313
add a comment |
add a comment |
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$begingroup$
You can construct a diagonizable matrix by using a diagonal matrix and multiplying it with by rotation matrices on both sides.
$endgroup$
– lightxbulb
2 days ago
1
$begingroup$
Have you learnt about Jordan forms? To get a non-diagonal matrix, you could just take a non-diagonal Jordan block or Jordan matrix, for example. And you can easily select the eigenvalues of such a matrix (they are just the diagonal elements of it), so you can easily make it singular (by giving it a $0$ on the diagonal) or have a complex eigenvalue (if you're happy to have complex entries).
$endgroup$
– Minus One-Twelfth
2 days ago
$begingroup$
no its not taught in this class, but i did hear a little about them , but i don't full understand it
$endgroup$
– Samurai Bale
2 days ago
$begingroup$
Then how did you learn about diagonalization?
$endgroup$
– enedil
2 days ago
$begingroup$
Well for $2times 2$ case, taking a matrix of the form $beginbmatrixlambda & 1 \ 0 & lambdaendbmatrix$ (for some scalar $lambda$) will give you a non-diagonalisable matrix eigenvalue $lambda$.
$endgroup$
– Minus One-Twelfth
2 days ago