Prove that if A has eigenvalues then determinant of A is the product of eigenvalues [duplicate]
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Show that the determinant of $A$ is equal to the product of its eigenvalues
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How can I prove that if A has eigenvalues then determinant of A is the product of its eigenvalues?
linear-algebra
edited Nov 16 at 18:43
David G. Stork
9,07321232
asked Nov 16 at 18:34
alex
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marked as duplicate by José Carlos Santos, T. Bongers, Lord Shark the Unknown, Leucippus, Key Flex Nov 16 at 21:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
-2
down vote
favorite
This question already has an answer here:
Show that the determinant of $A$ is equal to the product of its eigenvalues
7 answers
How can I prove that if A has eigenvalues then determinant of A is the product of its eigenvalues?
linear-algebra
edited Nov 16 at 18:43
David G. Stork
9,07321232
asked Nov 16 at 18:34
alex
6
marked as duplicate by José Carlos Santos, T. Bongers, Lord Shark the Unknown, Leucippus, Key Flex Nov 16 at 21:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
This question already has an answer here:
Show that the determinant of $A$ is equal to the product of its eigenvalues
7 answers
How can I prove that if A has eigenvalues then determinant of A is the product of its eigenvalues?
linear-algebra
edited Nov 16 at 18:43
David G. Stork
9,07321232
asked Nov 16 at 18:34
alex
6
This question already has an answer here:
Show that the determinant of $A$ is equal to the product of its eigenvalues
7 answers
How can I prove that if A has eigenvalues then determinant of A is the product of its eigenvalues?
This question already has an answer here:
Show that the determinant of $A$ is equal to the product of its eigenvalues
7 answers
linear-algebra
linear-algebra
edited Nov 16 at 18:43
David G. Stork
9,07321232
asked Nov 16 at 18:34
alex
6
edited Nov 16 at 18:43
David G. Stork
9,07321232
asked Nov 16 at 18:34
alex
6
edited Nov 16 at 18:43
David G. Stork
9,07321232
edited Nov 16 at 18:43
David G. Stork
9,07321232
edited Nov 16 at 18:43
David G. Stork
9,07321232
9,07321232
asked Nov 16 at 18:34
alex
6
asked Nov 16 at 18:34
alex
6
asked Nov 16 at 18:34
alex
6
6
marked as duplicate by José Carlos Santos, T. Bongers, Lord Shark the Unknown, Leucippus, Key Flex Nov 16 at 21:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by José Carlos Santos, T. Bongers, Lord Shark the Unknown, Leucippus, Key Flex Nov 16 at 21:05
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42
add a comment |
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42
add a comment |
1 Answer
1
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votes
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Hints:
1) Jordan Canonical Form
2) $textrm{det}(AB) = textrm{det}(A)textrm{det}(B)$
answered Nov 16 at 18:38
BaronVT
11.5k11337
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hints:
1) Jordan Canonical Form
2) $textrm{det}(AB) = textrm{det}(A)textrm{det}(B)$
answered Nov 16 at 18:38
BaronVT
11.5k11337
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
add a comment |
up vote
0
down vote
Hints:
1) Jordan Canonical Form
2) $textrm{det}(AB) = textrm{det}(A)textrm{det}(B)$
answered Nov 16 at 18:38
BaronVT
11.5k11337
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
add a comment |
up vote
0
down vote
up vote
0
down vote
Hints:
1) Jordan Canonical Form
2) $textrm{det}(AB) = textrm{det}(A)textrm{det}(B)$
answered Nov 16 at 18:38
BaronVT
11.5k11337
Hints:
1) Jordan Canonical Form
2) $textrm{det}(AB) = textrm{det}(A)textrm{det}(B)$
answered Nov 16 at 18:38
BaronVT
11.5k11337
answered Nov 16 at 18:38
BaronVT
11.5k11337
answered Nov 16 at 18:38
BaronVT
11.5k11337
answered Nov 16 at 18:38
BaronVT
11.5k11337
11.5k11337
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
add a comment |
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
Can you write a full proof ? @BaronVT
– alex
Nov 16 at 18:46
add a comment |
Look at your the constant in the characteristic polynomial and use the relation between roots of equation and product of roots.
– Yadati Kiran
Nov 16 at 18:42