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Sufficient condition for a matrix to be diagonalizable and similar matrices
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my question is about diagonalizable matrices and similar matrices.
I have a trouble proving a matrix is diagonalizable.
I know some options to do that:
Matrix $A$ $(n times n)$, is diagonalizable if:
- Number of eigenvectors equals to number of eigenvalues.
- There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB$.
But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB?$
I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?
Thanks for help!!
linear-algebra matrices diagonalization
asked Jul 26 at 8:46
D.Rotnemer
11216
add a comment |Â
up vote
5
down vote
favorite
my question is about diagonalizable matrices and similar matrices.
I have a trouble proving a matrix is diagonalizable.
I know some options to do that:
Matrix $A$ $(n times n)$, is diagonalizable if:
- Number of eigenvectors equals to number of eigenvalues.
- There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB$.
But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB?$
I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?
Thanks for help!!
linear-algebra matrices diagonalization
asked Jul 26 at 8:46
D.Rotnemer
11216
3
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
my question is about diagonalizable matrices and similar matrices.
I have a trouble proving a matrix is diagonalizable.
I know some options to do that:
Matrix $A$ $(n times n)$, is diagonalizable if:
- Number of eigenvectors equals to number of eigenvalues.
- There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB$.
But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB?$
I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?
Thanks for help!!
linear-algebra matrices diagonalization
asked Jul 26 at 8:46
D.Rotnemer
11216
my question is about diagonalizable matrices and similar matrices.
I have a trouble proving a matrix is diagonalizable.
I know some options to do that:
Matrix $A$ $(n times n)$, is diagonalizable if:
- Number of eigenvectors equals to number of eigenvalues.
- There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB$.
But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^-1AB?$
I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?
Thanks for help!!
linear-algebra matrices diagonalization
asked Jul 26 at 8:46
D.Rotnemer
11216
edited Jul 26 at 11:36
asked Jul 26 at 8:46
D.Rotnemer
11216
asked Jul 26 at 8:46
D.Rotnemer
11216
asked Jul 26 at 8:46
D.Rotnemer
11216
11216
3
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52
add a comment |Â
3
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52
3
3
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $mathbb R$ or $mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=beginpmatrix
0 & 0\
1 & 0
endpmatrix$$ is an example. The only eigenspace is $mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable
- The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
- $A$ is similar to a diagonal matrix.
- Its minimal polynomial is a product of distinct linear factors.
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
add a comment |Â
up vote
5
down vote
If a $ntimes n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^−1AB$â€, well… That's basically what being a diagonalizable matrix means.
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
add a comment |Â
up vote
3
down vote
The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
answered Jul 26 at 9:02
gimusi
65k73583
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $mathbb R$ or $mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=beginpmatrix
0 & 0\
1 & 0
endpmatrix$$ is an example. The only eigenspace is $mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable
- The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
- $A$ is similar to a diagonal matrix.
- Its minimal polynomial is a product of distinct linear factors.
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
add a comment |Â
up vote
7
down vote
accepted
First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $mathbb R$ or $mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=beginpmatrix
0 & 0\
1 & 0
endpmatrix$$ is an example. The only eigenspace is $mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable
- The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
- $A$ is similar to a diagonal matrix.
- Its minimal polynomial is a product of distinct linear factors.
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $mathbb R$ or $mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=beginpmatrix
0 & 0\
1 & 0
endpmatrix$$ is an example. The only eigenspace is $mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable
- The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
- $A$ is similar to a diagonal matrix.
- Its minimal polynomial is a product of distinct linear factors.
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
First a comment
The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $mathbb R$ or $mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.
If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?
The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true.
The matrix
$$A=beginpmatrix
0 & 0\
1 & 0
endpmatrix$$ is an example. The only eigenspace is $mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $mathbb F$ the field of the vector space).
Some equivalent conditions for a matrix $A$ to be diagonalizable
- The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
- $A$ is similar to a diagonal matrix.
- Its minimal polynomial is a product of distinct linear factors.
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
edited Jul 26 at 12:12
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
answered Jul 26 at 8:57
mathcounterexamples.net
23.5k21652
23.5k21652
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
add a comment |Â
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
2
2
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=beginpmatrix0&0\0&1endpmatrix$.
– Toffomat
Jul 26 at 11:42
1
1
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
@Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer.
– mathcounterexamples.net
Jul 26 at 12:09
add a comment |Â
up vote
5
down vote
If a $ntimes n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^−1AB$â€, well… That's basically what being a diagonalizable matrix means.
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
add a comment |Â
up vote
5
down vote
If a $ntimes n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^−1AB$â€, well… That's basically what being a diagonalizable matrix means.
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
add a comment |Â
up vote
5
down vote
up vote
5
down vote
If a $ntimes n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^−1AB$â€, well… That's basically what being a diagonalizable matrix means.
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
If a $ntimes n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizabel if and only if the sum of these dimensions is $n$.
Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^−1AB$â€, well… That's basically what being a diagonalizable matrix means.
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
edited Jul 27 at 6:03
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
answered Jul 26 at 8:53
José Carlos Santos
113k1696174
113k1696174
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
add a comment |Â
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
"If a matrix has a row of zeros then, yes, it can be diagonalizable." But it seems the OP is asking whether having a row of zeroes implies being diagonalisable.
– Vim
Jul 27 at 3:23
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
You are right. I've edited my answer.
– José Carlos Santos
Jul 27 at 6:04
add a comment |Â
up vote
3
down vote
The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
answered Jul 26 at 9:02
gimusi
65k73583
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
 |Â
show 1 more comment
up vote
3
down vote
The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
answered Jul 26 at 9:02
gimusi
65k73583
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
 |Â
show 1 more comment
up vote
3
down vote
up vote
3
down vote
The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
answered Jul 26 at 9:02
gimusi
65k73583
The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.
We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.
When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.
answered Jul 26 at 9:02
gimusi
65k73583
edited Jul 26 at 9:31
answered Jul 26 at 9:02
gimusi
65k73583
answered Jul 26 at 9:02
gimusi
65k73583
answered Jul 26 at 9:02
gimusi
65k73583
65k73583
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
 |Â
show 1 more comment
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
1
1
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $mathbb F$ over which the vector space is defined.
– mathcounterexamples.net
Jul 26 at 9:25
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
@mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye!
– gimusi
Jul 26 at 9:27
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist.
– mathcounterexamples.net
Jul 26 at 9:29
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
@mathcounterexamples.net Ah ok, thanks again!
– gimusi
Jul 26 at 9:30
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $mathbbF$) is that they have the same Rational Canonical Form.
– ancientmathematician
Jul 26 at 9:34
 |Â
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3
The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable.
– Arthur
Jul 26 at 8:53
@Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable.
– Meni Rosenfeld
Jul 26 at 9:57
@MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is.
– Arthur
Jul 26 at 10:52