Eigenvetor Property of a Matrix
If a matrix $A$ is complex orthogonally similar to an upper triangular matrix, that is, $A=QUQ^T, Q^TQ=I$ and $U$ is upper triangular matrix, then there exist at least one eigenvector $x$ of $A$ such that $x^Txneq 0.$
This is an exercise in Horn and Johnson. Don't know how to start. Any help or hint will be appreciated.
matrix-analysis
asked Jan 4 at 17:54
user602672user602672
253
add a comment |
If a matrix $A$ is complex orthogonally similar to an upper triangular matrix, that is, $A=QUQ^T, Q^TQ=I$ and $U$ is upper triangular matrix, then there exist at least one eigenvector $x$ of $A$ such that $x^Txneq 0.$
This is an exercise in Horn and Johnson. Don't know how to start. Any help or hint will be appreciated.
matrix-analysis
asked Jan 4 at 17:54
user602672user602672
253
First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00
add a comment |
If a matrix $A$ is complex orthogonally similar to an upper triangular matrix, that is, $A=QUQ^T, Q^TQ=I$ and $U$ is upper triangular matrix, then there exist at least one eigenvector $x$ of $A$ such that $x^Txneq 0.$
This is an exercise in Horn and Johnson. Don't know how to start. Any help or hint will be appreciated.
matrix-analysis
asked Jan 4 at 17:54
user602672user602672
253
If a matrix $A$ is complex orthogonally similar to an upper triangular matrix, that is, $A=QUQ^T, Q^TQ=I$ and $U$ is upper triangular matrix, then there exist at least one eigenvector $x$ of $A$ such that $x^Txneq 0.$
This is an exercise in Horn and Johnson. Don't know how to start. Any help or hint will be appreciated.
matrix-analysis
matrix-analysis
asked Jan 4 at 17:54
user602672user602672
253
asked Jan 4 at 17:54
user602672user602672
253
asked Jan 4 at 17:54
user602672user602672
253
asked Jan 4 at 17:54
user602672user602672
253
asked Jan 4 at 17:54
user602672user602672
253
253
First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00
add a comment |
First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00
First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00
First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00
add a comment |
1 Answer
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Hint: exhibit an eigenvector $y(neq 0)$ of $U$, and note that $x=Qy$ is then an eigenvector of $A$.
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
add a comment |
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Hint: exhibit an eigenvector $y(neq 0)$ of $U$, and note that $x=Qy$ is then an eigenvector of $A$.
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
add a comment |
Hint: exhibit an eigenvector $y(neq 0)$ of $U$, and note that $x=Qy$ is then an eigenvector of $A$.
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
add a comment |
Hint: exhibit an eigenvector $y(neq 0)$ of $U$, and note that $x=Qy$ is then an eigenvector of $A$.
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
Hint: exhibit an eigenvector $y(neq 0)$ of $U$, and note that $x=Qy$ is then an eigenvector of $A$.
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
answered Jan 4 at 18:02
metamorphymetamorphy
3,6821621
3,6821621
add a comment |
add a comment |
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First step: assume $A$ is upper triangular. Is it easier to prove there? How might your proof change if we throw in the $Q$s?
– user3482749
Jan 4 at 18:00