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Why does eigendecomposition is in the same form of $A=CBC^{-1}$ as using eigenvectors as basis for a...
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Let A be a linear transformation matrix f(x)=Ax, and if D is A in the eigenvector basis, $$D = Q^{-1}AQ$$, where Q is eigenvector matrix of A and a change of basis matrix changing A in standard basis to a basis of eigenvectors of A. It seems to be similar to eigendecomposition $$ A = QLambda Q^{-1} $$ $$ Lambda = Q^{-1}AQ$$ So the linear transformation matrix D with eigenvectors of A as basis is just the eigenvalues matrix of A. What are the connections here?
linear-algebra eigenvalues-eigenvectors change-of-basis
asked Jan 23 at 20:06
user14042
1087
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Let A be a linear transformation matrix f(x)=Ax, and if D is A in the eigenvector basis, $$D = Q^{-1}AQ$$, where Q is eigenvector matrix of A and a change of basis matrix changing A in standard basis to a basis of eigenvectors of A. It seems to be similar to eigendecomposition $$ A = QLambda Q^{-1} $$ $$ Lambda = Q^{-1}AQ$$ So the linear transformation matrix D with eigenvectors of A as basis is just the eigenvalues matrix of A. What are the connections here?
linear-algebra eigenvalues-eigenvectors change-of-basis
asked Jan 23 at 20:06
user14042
1087
Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let A be a linear transformation matrix f(x)=Ax, and if D is A in the eigenvector basis, $$D = Q^{-1}AQ$$, where Q is eigenvector matrix of A and a change of basis matrix changing A in standard basis to a basis of eigenvectors of A. It seems to be similar to eigendecomposition $$ A = QLambda Q^{-1} $$ $$ Lambda = Q^{-1}AQ$$ So the linear transformation matrix D with eigenvectors of A as basis is just the eigenvalues matrix of A. What are the connections here?
linear-algebra eigenvalues-eigenvectors change-of-basis
asked Jan 23 at 20:06
user14042
1087
Let A be a linear transformation matrix f(x)=Ax, and if D is A in the eigenvector basis, $$D = Q^{-1}AQ$$, where Q is eigenvector matrix of A and a change of basis matrix changing A in standard basis to a basis of eigenvectors of A. It seems to be similar to eigendecomposition $$ A = QLambda Q^{-1} $$ $$ Lambda = Q^{-1}AQ$$ So the linear transformation matrix D with eigenvectors of A as basis is just the eigenvalues matrix of A. What are the connections here?
linear-algebra eigenvalues-eigenvectors change-of-basis
linear-algebra eigenvalues-eigenvectors change-of-basis
asked Jan 23 at 20:06
user14042
1087
asked Jan 23 at 20:06
user14042
1087
edited Jan 23 at 20:14
asked Jan 23 at 20:06
user14042
1087
asked Jan 23 at 20:06
user14042
1087
asked Jan 23 at 20:06
user14042
1087
1087
Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17
add a comment |
Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17
Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17
Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17
add a comment |
1 Answer
1
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up vote
1
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accepted
You simply have different cases of the same result:
If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.
Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.
If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$
Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $Lambda$ that is given by the same formula $Lambda=Q^{-1}AQ$ (and $A=QLambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You simply have different cases of the same result:
If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.
Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.
If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$
Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $Lambda$ that is given by the same formula $Lambda=Q^{-1}AQ$ (and $A=QLambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
add a comment |
up vote
1
down vote
accepted
You simply have different cases of the same result:
If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.
Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.
If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$
Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $Lambda$ that is given by the same formula $Lambda=Q^{-1}AQ$ (and $A=QLambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You simply have different cases of the same result:
If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.
Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.
If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$
Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $Lambda$ that is given by the same formula $Lambda=Q^{-1}AQ$ (and $A=QLambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
You simply have different cases of the same result:
If $A$ is the matrix that represents a linear transformation in the standard basis and $Q$ is a matrix that has as columns the vectors of a new basis, than the same linear transformation is represented in this new basis by the matrix $B= Q^{-1}AQ $. This result is always true, for any matrix $A$.
Obviously we have also: $A=QBQ^{-1}$ and the eigendecomposition is a special case.
If $A$ is diagonalizable than it has a set of eigenvectors that forms a basis and, in this basis, the linear transformation is represented by a diagonal matrix $D$ that has as diagonal elements the eigenvalues. In this case the transformation matrix $Q$ has as columns the eigenvalues and, using the same formula, we have that $D=Q^{-1}AQ$
Also if the linear transformation is not diagonalizable than we can represent it in a canonical form (the Jordan canonical form) in a basis that is formed by the proper and the generalized eigenvectors. In this basis the transformation is represented by a matrix $Lambda$ that is given by the same formula $Lambda=Q^{-1}AQ$ (and $A=QLambda Q^{-1}$) where, now, $Q$ is the matrix that has as columns the proper and generalized eigenvectors of $A$.
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
edited yesterday
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
answered Jan 23 at 20:47
Emilio Novati
50.6k43372
50.6k43372
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
add a comment |
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
Thanks. If the basis is not eigenvectors, then is D still a diagonal matrix? (It is diagonal with eigenvalues when the eigenvectors are basis.)
– user14042
Jan 31 at 8:11
1
1
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
No. The transformation is represented by a diagonal matrix only in the eigenbasis.
– Emilio Novati
Jan 31 at 8:23
add a comment |
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Other than whether the diagonal matrix is called $D$ or $Lambda$ there seems to be no difference between your first and last equations.
– Henning Makholm
Jan 23 at 20:17