Mixing
how does the characteristic polynomial of a matrix change with respect to the action of taking a principal...
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Let $A$ be an $ntimes n$ matrix of complex numbers. Let $B$ be the matrix $A$ with the last row and column removed. Is there a relation between the characteristic polynomials of $A$ and $B$? More generally, how does the characteristic polynomial of a matrix change with respect to the action of taking a principal minor?
We can ask the same questions for eigenvalues/eigenvectors, which are closely related with the char polynomial. Intuitively, I feel that the eigenvalues of $B$ are also the eigenvalues of $A$.
linear-algebra eigenvalues-eigenvectors
asked Jan 8 at 20:54
Tony BTony B
800418
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add a comment |
$begingroup$
Let $A$ be an $ntimes n$ matrix of complex numbers. Let $B$ be the matrix $A$ with the last row and column removed. Is there a relation between the characteristic polynomials of $A$ and $B$? More generally, how does the characteristic polynomial of a matrix change with respect to the action of taking a principal minor?
We can ask the same questions for eigenvalues/eigenvectors, which are closely related with the char polynomial. Intuitively, I feel that the eigenvalues of $B$ are also the eigenvalues of $A$.
linear-algebra eigenvalues-eigenvectors
asked Jan 8 at 20:54
Tony BTony B
800418
$endgroup$
2
$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01
add a comment |
$begingroup$
Let $A$ be an $ntimes n$ matrix of complex numbers. Let $B$ be the matrix $A$ with the last row and column removed. Is there a relation between the characteristic polynomials of $A$ and $B$? More generally, how does the characteristic polynomial of a matrix change with respect to the action of taking a principal minor?
We can ask the same questions for eigenvalues/eigenvectors, which are closely related with the char polynomial. Intuitively, I feel that the eigenvalues of $B$ are also the eigenvalues of $A$.
linear-algebra eigenvalues-eigenvectors
asked Jan 8 at 20:54
Tony BTony B
800418
$endgroup$
Let $A$ be an $ntimes n$ matrix of complex numbers. Let $B$ be the matrix $A$ with the last row and column removed. Is there a relation between the characteristic polynomials of $A$ and $B$? More generally, how does the characteristic polynomial of a matrix change with respect to the action of taking a principal minor?
We can ask the same questions for eigenvalues/eigenvectors, which are closely related with the char polynomial. Intuitively, I feel that the eigenvalues of $B$ are also the eigenvalues of $A$.
linear-algebra eigenvalues-eigenvectors
linear-algebra eigenvalues-eigenvectors
asked Jan 8 at 20:54
Tony BTony B
800418
asked Jan 8 at 20:54
Tony BTony B
800418
asked Jan 8 at 20:54
Tony BTony B
800418
asked Jan 8 at 20:54
Tony BTony B
800418
asked Jan 8 at 20:54
Tony BTony B
800418
800418
2
$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01
add a comment |
2
$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01
2
2
$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01
$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01
add a comment |
1 Answer
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Your intuition is completely wrong, as simple examples will show.
The best way I can relate the characteristic polynomials is this. Write
$$ A = pmatrix{B & ccr d & ecr} $$
where $B$ is $(n-1)times (n-1)$, $c$ is $(n-1) times 1$, $d = 1 times (n-1)$, and $e$ is a scalar. Then
$$ det(A-lambda I) = det(B-lambda I) (e - lambda) - d ; text{adj}(B-lambda I) c$$
where $text{adj}$ is the classical adjoint. In general the term $d ; text{adj}(B-lambda I) c$ will be a polynomial of degree $n-2$.
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
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add a comment |
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1 Answer
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$begingroup$
Your intuition is completely wrong, as simple examples will show.
The best way I can relate the characteristic polynomials is this. Write
$$ A = pmatrix{B & ccr d & ecr} $$
where $B$ is $(n-1)times (n-1)$, $c$ is $(n-1) times 1$, $d = 1 times (n-1)$, and $e$ is a scalar. Then
$$ det(A-lambda I) = det(B-lambda I) (e - lambda) - d ; text{adj}(B-lambda I) c$$
where $text{adj}$ is the classical adjoint. In general the term $d ; text{adj}(B-lambda I) c$ will be a polynomial of degree $n-2$.
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
Your intuition is completely wrong, as simple examples will show.
The best way I can relate the characteristic polynomials is this. Write
$$ A = pmatrix{B & ccr d & ecr} $$
where $B$ is $(n-1)times (n-1)$, $c$ is $(n-1) times 1$, $d = 1 times (n-1)$, and $e$ is a scalar. Then
$$ det(A-lambda I) = det(B-lambda I) (e - lambda) - d ; text{adj}(B-lambda I) c$$
where $text{adj}$ is the classical adjoint. In general the term $d ; text{adj}(B-lambda I) c$ will be a polynomial of degree $n-2$.
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
$endgroup$
add a comment |
$begingroup$
Your intuition is completely wrong, as simple examples will show.
The best way I can relate the characteristic polynomials is this. Write
$$ A = pmatrix{B & ccr d & ecr} $$
where $B$ is $(n-1)times (n-1)$, $c$ is $(n-1) times 1$, $d = 1 times (n-1)$, and $e$ is a scalar. Then
$$ det(A-lambda I) = det(B-lambda I) (e - lambda) - d ; text{adj}(B-lambda I) c$$
where $text{adj}$ is the classical adjoint. In general the term $d ; text{adj}(B-lambda I) c$ will be a polynomial of degree $n-2$.
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
$endgroup$
Your intuition is completely wrong, as simple examples will show.
The best way I can relate the characteristic polynomials is this. Write
$$ A = pmatrix{B & ccr d & ecr} $$
where $B$ is $(n-1)times (n-1)$, $c$ is $(n-1) times 1$, $d = 1 times (n-1)$, and $e$ is a scalar. Then
$$ det(A-lambda I) = det(B-lambda I) (e - lambda) - d ; text{adj}(B-lambda I) c$$
where $text{adj}$ is the classical adjoint. In general the term $d ; text{adj}(B-lambda I) c$ will be a polynomial of degree $n-2$.
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
edited Jan 8 at 21:42
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
answered Jan 8 at 21:35
Robert IsraelRobert Israel
321k23210463
321k23210463
add a comment |
add a comment |
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$begingroup$
How about for instance $begin{bmatrix}0 & 1 \ 1 & 1end{bmatrix}$?.
$endgroup$
– Mindlack
Jan 8 at 21:01