Mixing
Characterization of diagonal matrices reducing rank
$begingroup$
Given full rank matrix $A$. I am looking for a characterization of all diagonal matrices $D$ with
$$
det(D - A) = 0
$$
This set of rank-reducing matrices is then somehow a superset to the $lambda I$ used in the eigenvalue equation.
linear-algebra
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
$endgroup$
add a comment |
$begingroup$
Given full rank matrix $A$. I am looking for a characterization of all diagonal matrices $D$ with
$$
det(D - A) = 0
$$
This set of rank-reducing matrices is then somehow a superset to the $lambda I$ used in the eigenvalue equation.
linear-algebra
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
$endgroup$
add a comment |
$begingroup$
Given full rank matrix $A$. I am looking for a characterization of all diagonal matrices $D$ with
$$
det(D - A) = 0
$$
This set of rank-reducing matrices is then somehow a superset to the $lambda I$ used in the eigenvalue equation.
linear-algebra
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
$endgroup$
Given full rank matrix $A$. I am looking for a characterization of all diagonal matrices $D$ with
$$
det(D - A) = 0
$$
This set of rank-reducing matrices is then somehow a superset to the $lambda I$ used in the eigenvalue equation.
linear-algebra
linear-algebra
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
asked Jan 21 at 21:31
Sebastian SchlechtSebastian Schlecht
23018
23018
add a comment |
add a comment |
1 Answer
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$begingroup$
Actually there are a lot of such diagonal matrices. Notice that for any choice of the first $n-1$ elements of the diagonal, there always exists a unique $n$-th entry such that $det(A-D)=0$.
answered Jan 21 at 21:43
ExoddExodd
5,6821224
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Actually there are a lot of such diagonal matrices. Notice that for any choice of the first $n-1$ elements of the diagonal, there always exists a unique $n$-th entry such that $det(A-D)=0$.
answered Jan 21 at 21:43
ExoddExodd
5,6821224
$endgroup$
add a comment |
$begingroup$
Actually there are a lot of such diagonal matrices. Notice that for any choice of the first $n-1$ elements of the diagonal, there always exists a unique $n$-th entry such that $det(A-D)=0$.
answered Jan 21 at 21:43
ExoddExodd
5,6821224
$endgroup$
add a comment |
$begingroup$
Actually there are a lot of such diagonal matrices. Notice that for any choice of the first $n-1$ elements of the diagonal, there always exists a unique $n$-th entry such that $det(A-D)=0$.
answered Jan 21 at 21:43
ExoddExodd
5,6821224
$endgroup$
Actually there are a lot of such diagonal matrices. Notice that for any choice of the first $n-1$ elements of the diagonal, there always exists a unique $n$-th entry such that $det(A-D)=0$.
answered Jan 21 at 21:43
ExoddExodd
5,6821224
answered Jan 21 at 21:43
ExoddExodd
5,6821224
answered Jan 21 at 21:43
ExoddExodd
5,6821224
answered Jan 21 at 21:43
ExoddExodd
5,6821224
5,6821224
add a comment |
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