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







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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















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







share|cite|improve this question
























  • 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













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







share|cite|improve this question















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






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edited Jan 23 at 20:14

























asked Jan 23 at 20:06









user14042

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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


















  • 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










1 Answer
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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 $A$.






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  • 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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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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 $A$.






share|cite|improve this answer























  • 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















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 $A$.






share|cite|improve this answer























  • 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













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 $A$.






share|cite|improve this answer














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 $A$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























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


















  • 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


















 

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