Mixing
If real matrices $A,B$ have the same ch. polynomial and rank, and rk$(A-lambda I)= text{rk}(B-lambda I)$ for...
$begingroup$
Some time ago I had a linear algebra private lesson, and rereading some notes of the professor, I've found
If two real matrices $A,B$ have the same characteristic polynomial and rank, and satisfy $ operatorname{rk}(A-lambda I)= operatorname{rk}(B-lambda I)$ for each eigenvalue, then they are similar
but with no proof. Is this true? And if so, how would one prove it? What I do know is that having the same characteristic polynomial and rank is not enough, but I'm not sure why the additional condition would fill the gap
linear-algebra matrices eigenvalues-eigenvectors determinant matrix-rank
edited Jan 19 at 9:38
Bernard
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
$endgroup$
add a comment |
$begingroup$
Some time ago I had a linear algebra private lesson, and rereading some notes of the professor, I've found
If two real matrices $A,B$ have the same characteristic polynomial and rank, and satisfy $ operatorname{rk}(A-lambda I)= operatorname{rk}(B-lambda I)$ for each eigenvalue, then they are similar
but with no proof. Is this true? And if so, how would one prove it? What I do know is that having the same characteristic polynomial and rank is not enough, but I'm not sure why the additional condition would fill the gap
linear-algebra matrices eigenvalues-eigenvectors determinant matrix-rank
edited Jan 19 at 9:38
Bernard
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
$endgroup$
add a comment |
$begingroup$
Some time ago I had a linear algebra private lesson, and rereading some notes of the professor, I've found
If two real matrices $A,B$ have the same characteristic polynomial and rank, and satisfy $ operatorname{rk}(A-lambda I)= operatorname{rk}(B-lambda I)$ for each eigenvalue, then they are similar
but with no proof. Is this true? And if so, how would one prove it? What I do know is that having the same characteristic polynomial and rank is not enough, but I'm not sure why the additional condition would fill the gap
linear-algebra matrices eigenvalues-eigenvectors determinant matrix-rank
edited Jan 19 at 9:38
Bernard
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
$endgroup$
Some time ago I had a linear algebra private lesson, and rereading some notes of the professor, I've found
If two real matrices $A,B$ have the same characteristic polynomial and rank, and satisfy $ operatorname{rk}(A-lambda I)= operatorname{rk}(B-lambda I)$ for each eigenvalue, then they are similar
but with no proof. Is this true? And if so, how would one prove it? What I do know is that having the same characteristic polynomial and rank is not enough, but I'm not sure why the additional condition would fill the gap
linear-algebra matrices eigenvalues-eigenvectors determinant matrix-rank
linear-algebra matrices eigenvalues-eigenvectors determinant matrix-rank
edited Jan 19 at 9:38
Bernard
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
edited Jan 19 at 9:38
Bernard
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
edited Jan 19 at 9:38
Bernard
121k740116
edited Jan 19 at 9:38
Bernard
121k740116
edited Jan 19 at 9:38
Bernard
121k740116
121k740116
asked Jan 19 at 8:18
LearnerLearner
17510
asked Jan 19 at 8:18
LearnerLearner
17510
asked Jan 19 at 8:18
LearnerLearner
17510
17510
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
How about
$$pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}$$
and
$$pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}?$$
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How about
$$pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}$$
and
$$pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}?$$
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
add a comment |
$begingroup$
How about
$$pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}$$
and
$$pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}?$$
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
add a comment |
$begingroup$
How about
$$pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}$$
and
$$pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}?$$
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
How about
$$pmatrix{0&1&0&0\0&0&0&0\0&0&0&1\0&0&0&0}$$
and
$$pmatrix{0&1&0&0\0&0&1&0\0&0&0&0\0&0&0&0}?$$
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Jan 19 at 8:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
add a comment |
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
$begingroup$
Yes then I saw it doesn't matter if the eigenvalues are non-zero haha, I just thought maybe it was "my fault", not the prof's mistake. Thanks!
$endgroup$
– Learner
Jan 19 at 8:26
add a comment |
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