Mixing
Show that this matrix is not diagonalizable
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Say I have a matrix:
$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$
Is this matrix diagonalizable?
Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?
Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:
$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$
$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$
By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:
linear-algebra eigenvalues-eigenvectors diagonalization
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
Say I have a matrix:
$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$
Is this matrix diagonalizable?
Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?
Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:
$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$
$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$
By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:
linear-algebra eigenvalues-eigenvectors diagonalization
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Say I have a matrix:
$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$
Is this matrix diagonalizable?
Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?
Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:
$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$
$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$
By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:
linear-algebra eigenvalues-eigenvectors diagonalization
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Say I have a matrix:
$$A = beginbmatrix 2 & 0 \ -1 & 2 endbmatrix $$
Is this matrix diagonalizable?
Does a 2x2 matrix always have 2 eigenvalues (multipicity counts). Why is this? I know this matrix (because it's lower triangular) has the eigenvalue of 2 with multiplicity 2... but does a matrix of this size always have 2 eigenvalues. Why is this?
Is there any way to know if the eigenvalue of 2 has two eigenvectors or not quickly? Here's the way I know to find the eigenvector:
$$beginbmatrix 2 & 0 \ 0 & 2 endbmatrix - beginbmatrix 2 & 0 \ -1 & 2 endbmatrix = beginbmatrix 0 & 0 \ 1 & 0 endbmatrix $$
$$ eigenvector = beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 0 \ t endbmatrix = t * beginbmatrix 0 \ 1 endbmatrix$$
By this theorem, it is not diagonalizable because it only has 1 eigenvector right and the matrix has 2 rows and 2 columns:
linear-algebra eigenvalues-eigenvectors diagonalization
linear-algebra eigenvalues-eigenvectors diagonalization
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
edited 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
11.8k72143
asked 2 hours ago
Kitty Capital
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Kitty Capital
495
asked 2 hours ago
Kitty Capital
495
495
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kitty Capital is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago
add a comment |Â
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
answered 2 hours ago
John Hughes
60.1k23987
add a comment |Â
up vote
1
down vote
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
answered 2 hours ago
gimusi
75.9k73889
add a comment |Â
up vote
1
down vote
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
answered 2 hours ago
John Hughes
60.1k23987
add a comment |Â
up vote
4
down vote
accepted
No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
answered 2 hours ago
John Hughes
60.1k23987
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
answered 2 hours ago
John Hughes
60.1k23987
No, it's not diagonalizable. If the two eigenvalues of a $ 2 times 2 $ matrix were distinct, it would be; when they're the same, it might be (but in this case it's not).
The eigenvalues of an $n times n$ matrix turn out (as you'll probably learn soon) to be the roots of a degree-$n$ polynomial. Since every degree-$n$ polynomial has $n$ roots (when counted with multiplicity, and allowing for complex roots as well as real ones), this means that every $n times n$ matrix has $n$ eigenvalues (when counted with multiplicities).
By the way, it appears that you've done exactly the right thing to determine how many eigenvectors there are that correspond to a given evalue; in general, there's no obvious and simple way to do it except to look for the solution space of an associated system of equations, as you did.
answered 2 hours ago
John Hughes
60.1k23987
answered 2 hours ago
John Hughes
60.1k23987
answered 2 hours ago
John Hughes
60.1k23987
answered 2 hours ago
John Hughes
60.1k23987
60.1k23987
add a comment |Â
add a comment |Â
up vote
1
down vote
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
answered 2 hours ago
gimusi
75.9k73889
add a comment |Â
up vote
1
down vote
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
answered 2 hours ago
gimusi
75.9k73889
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
answered 2 hours ago
gimusi
75.9k73889
HINT
Recall what are the necessary and sufficient conditions for a matrix to be diagonalizable and note that here we have an eigenvalue $2$ with arithmetic multiplicity $2$ and geometric multiplicity $1$, that is an eigenspace with dimension $1$.
answered 2 hours ago
gimusi
75.9k73889
answered 2 hours ago
gimusi
75.9k73889
answered 2 hours ago
gimusi
75.9k73889
answered 2 hours ago
gimusi
75.9k73889
75.9k73889
add a comment |Â
add a comment |Â
up vote
1
down vote
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
add a comment |Â
up vote
1
down vote
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
add a comment |Â
up vote
1
down vote
up vote
1
down vote
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
- No. $A - 2I$ has only one linearly independent column. (The second column of $A-2I$ is zero.)
- To give a quick example, let's consider the 2D rotation matrix. $$beginpmatrixcos theta &-sin theta \sin theta &cos theta \endpmatrix$$ Eigenspace for a linear transformation is an example of invariant subspace. Since there's no proper invariant subspace in 2D rotation, the rotation matrix doesn't have real eigenvalues.
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
answered 2 hours ago
GNUSupporter 8964民主女神 地下教會
11.8k72143
11.8k72143
add a comment |Â
add a comment |Â
Kitty Capital is a new contributor. Be nice, and check out our Code of Conduct.
Kitty Capital is a new contributor. Be nice, and check out our Code of Conduct.
Kitty Capital is a new contributor. Be nice, and check out our Code of Conduct.
Kitty Capital is a new contributor. Be nice, and check out our Code of Conduct.
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%2f2934522%2fshow-that-this-matrix-is-not-diagonalizable%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
You've pointed out a reasonably quick way to see that the eigenvalue $2$ has a one-dimensional eigenspace, which is often summarized by saying the geometric multiplicity is strictly less than the algebraic multiplicity of eigenvalue $2$. Whenever this happens, the matrix is not diagonalizable (a diagonal matrix will have agreement between geometric and algebraic multiplicity).
– hardmath
2 hours ago