Diagonalizability of $A in M_{n}(Bbb{R})$ (Upper traingular matrix) with all diagonal entries 1 and $A neq...
$begingroup$
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
$endgroup$
add a comment |
$begingroup$
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
$endgroup$
$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
$begingroup$
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
$endgroup$
If $A in M_{n}(Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that $A neq I$, then what can we say about the diagonalizability of $A$ ?
I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.
And that the eigenvalues of the triangular matrices are given by the diagonal elements like here and here, but they work nocely if we had distinct elements on the main diagonal.
But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
linear-algebra matrices diagonalization
linear-algebra matrices diagonalization
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
asked Dec 20 '17 at 4:37
BAYMAXBAYMAX
2,86621123
2,86621123
$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39
$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39
$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
$endgroup$
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
add a comment |
$begingroup$
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
$endgroup$
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
add a comment |
$begingroup$
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
$endgroup$
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
add a comment |
$begingroup$
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
$endgroup$
It isn't diagonalizable. Your matrix is of the form $I+N$ where $I$ is the identity matrix and $N$ is strictly upper triangular. You can check that being strictly upper triangular means that $N^n=0$, since if the $e_i$ are the basis vectors, and $E_m$ is the subspace spanned by $e_1,ldots,e_m$, with $E_0=0$ then you can check that $NE_m subseteq E_{m-1}$, so $N^nE_n = E_0=0$. However since $Ane I$, $Nne 0$. Now suppose $M$ diagonalizes $A$, so $MAM^{-1} = D$ for $D$ a diagonal matrix. Then $M(I+N)M^{-1}=MIM^{-1}+MNM^{-1} = D$, but $MIM^{-1}=I$, so we in fact have $MNM^{-1}=D-I$, so in fact $N$ must be diagonalizable. However we still have $(MNM^{-1})^n=0$, so if $D-I$ has elements $lambda_1,ldots,lambda_n$ on the diagonal, then $lambda_i^n=0$, which implies $lambda_i=0$, hence $D-I=0$. Thus we would have that $D=I$, and $N=0$, contradicting the assumption that $Ane I$, so $Nne 0$.
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
answered Dec 20 '17 at 4:46
jgonjgon
13.5k22041
13.5k22041
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
add a comment |
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
$N^{n} = 0$ for some $n$ means that it is Idempotent , then does this mean that we can give Upper triangular matrices as examples of Idempotent matrices?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:43
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Err, $N$ is nilpotent (nilpotence means that some power of something is 0). Idempotence is the property that $P^2=P$, where say $P$ is some matrix. Examples of idempotent operators are projections (say $(a,b)mapsto (a,0)$). But strictly upper triangular matrices are nilpotent, if that's the question you were asking, then yes.
$endgroup$
– jgon
Dec 20 '17 at 5:45
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the order of the matrix ? then is it possible that $N^{k} =0$ for $k < n$?
$endgroup$
– BAYMAX
Dec 20 '17 at 5:47
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
$begingroup$
Here $n$ is the dimension of the vector space. It is certainly possible that $N^k=0$ for $k<n$, but we can guarantee that it will become 0 by the time we reach $N^n$. An example with $N^k = 0$ for $k<n$ is $begin{pmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \ end{pmatrix}$, which squares to 0.
$endgroup$
– jgon
Dec 20 '17 at 5:48
add a comment |
$begingroup$
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
$endgroup$
add a comment |
$begingroup$
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
$endgroup$
add a comment |
$begingroup$
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
$endgroup$
Since the matrix is upper triangular with all diagonal entries $1$, its characteristic equation is $(1-lambda)^n=0$. This give an eigenvalue of $1$ with algebraic multiplicity $n$. Also, finding the corresponding eigenspace through the null space of $A-lambda I$ tells us that it is generated by a single vector $[1,0,...,0]^T$. Thus its geometric multiplicity is $1$, different from $n$ and hence it is not diagonalizable.
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
answered Nov 30 '18 at 15:47
RhaldrynRhaldryn
309414
309414
add a comment |
add a comment |
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$begingroup$
It depends. Refer to: en.wikipedia.org/wiki/Jordan_normal_form
$endgroup$
– Brian Cheung
Dec 20 '17 at 4:39