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Eigenvalues of $mathrmadj(A)$
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If A is a singular square matrix , what are the eigenvalue of $mathrmadj(A)$ ?
My attempt:
I know that if $mathrmrank(A) < n- 1$ then $mathrmadj(A) = 0$. I think that the eigenvalue of $mathrmadj(A)$ will $0$.
Is its correct ?
Any hints/solution will be aappreciated ..
linear-algebra eigenvalues-eigenvectors
edited Aug 1 at 17:51
Math Lover
12.2k21132
asked Aug 1 at 17:44
stupid
51218
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up vote
1
down vote
favorite
If A is a singular square matrix , what are the eigenvalue of $mathrmadj(A)$ ?
My attempt:
I know that if $mathrmrank(A) < n- 1$ then $mathrmadj(A) = 0$. I think that the eigenvalue of $mathrmadj(A)$ will $0$.
Is its correct ?
Any hints/solution will be aappreciated ..
linear-algebra eigenvalues-eigenvectors
edited Aug 1 at 17:51
Math Lover
12.2k21132
asked Aug 1 at 17:44
stupid
51218
1
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
2
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If A is a singular square matrix , what are the eigenvalue of $mathrmadj(A)$ ?
My attempt:
I know that if $mathrmrank(A) < n- 1$ then $mathrmadj(A) = 0$. I think that the eigenvalue of $mathrmadj(A)$ will $0$.
Is its correct ?
Any hints/solution will be aappreciated ..
linear-algebra eigenvalues-eigenvectors
edited Aug 1 at 17:51
Math Lover
12.2k21132
asked Aug 1 at 17:44
stupid
51218
If A is a singular square matrix , what are the eigenvalue of $mathrmadj(A)$ ?
My attempt:
I know that if $mathrmrank(A) < n- 1$ then $mathrmadj(A) = 0$. I think that the eigenvalue of $mathrmadj(A)$ will $0$.
Is its correct ?
Any hints/solution will be aappreciated ..
linear-algebra eigenvalues-eigenvectors
edited Aug 1 at 17:51
Math Lover
12.2k21132
asked Aug 1 at 17:44
stupid
51218
edited Aug 1 at 17:51
Math Lover
12.2k21132
edited Aug 1 at 17:51
Math Lover
12.2k21132
edited Aug 1 at 17:51
Math Lover
12.2k21132
12.2k21132
asked Aug 1 at 17:44
stupid
51218
asked Aug 1 at 17:44
stupid
51218
asked Aug 1 at 17:44
stupid
51218
51218
1
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
2
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47
add a comment |Â
1
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
2
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47
1
1
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
2
2
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
As this blog post points out, we have $operatornameadj(U^-1AU) = U^-1(operatornameadj(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^-1AU$, we have
$$ operatornameadj(B) = U^-1(operatornameadj(A)) U. $$
The diagonal elements of $B$ are $lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $prod_j neq i lambda_j$.
So the eigenvalues of $operatornameadj(A)$ are the same as those of $operatornameadj(B)$, which are the diagonal elements, i.e. $prod_j neq i lambda_j$. Of course, this means that if $dimkerA>1$, $operatornameadj(A) = 0$, while if $dimkerA = 1$, $operatornameadj(A)$ has exactly one nonzero eigenvalue, the product of the others.
answered Aug 1 at 18:44
Chappers
54.9k74190
add a comment |Â
up vote
1
down vote
The following is just a collection of direct properties:
If $mathrmrank(A)<n-1$, then the only eigenvalue of $mathrmadj(A)$ is $0$ as $mathrmadj(A)=mathbf0$.
If on the other hand $mathrmrank(A)=n-1$, then $0$ is again an eigenvalue of $mathrmadj(A)$ as $mathrmrank(mathrmadj(A))=1$, i.e. $mathrmdim(mathrmker(mathrmadj(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.
Note that this does not mean that there is not more to the $mathrmrank(A)=n-1$-case.
answered Aug 1 at 18:19
zzuussee
1,152419
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As this blog post points out, we have $operatornameadj(U^-1AU) = U^-1(operatornameadj(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^-1AU$, we have
$$ operatornameadj(B) = U^-1(operatornameadj(A)) U. $$
The diagonal elements of $B$ are $lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $prod_j neq i lambda_j$.
So the eigenvalues of $operatornameadj(A)$ are the same as those of $operatornameadj(B)$, which are the diagonal elements, i.e. $prod_j neq i lambda_j$. Of course, this means that if $dimkerA>1$, $operatornameadj(A) = 0$, while if $dimkerA = 1$, $operatornameadj(A)$ has exactly one nonzero eigenvalue, the product of the others.
answered Aug 1 at 18:44
Chappers
54.9k74190
add a comment |Â
up vote
3
down vote
accepted
As this blog post points out, we have $operatornameadj(U^-1AU) = U^-1(operatornameadj(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^-1AU$, we have
$$ operatornameadj(B) = U^-1(operatornameadj(A)) U. $$
The diagonal elements of $B$ are $lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $prod_j neq i lambda_j$.
So the eigenvalues of $operatornameadj(A)$ are the same as those of $operatornameadj(B)$, which are the diagonal elements, i.e. $prod_j neq i lambda_j$. Of course, this means that if $dimkerA>1$, $operatornameadj(A) = 0$, while if $dimkerA = 1$, $operatornameadj(A)$ has exactly one nonzero eigenvalue, the product of the others.
answered Aug 1 at 18:44
Chappers
54.9k74190
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As this blog post points out, we have $operatornameadj(U^-1AU) = U^-1(operatornameadj(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^-1AU$, we have
$$ operatornameadj(B) = U^-1(operatornameadj(A)) U. $$
The diagonal elements of $B$ are $lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $prod_j neq i lambda_j$.
So the eigenvalues of $operatornameadj(A)$ are the same as those of $operatornameadj(B)$, which are the diagonal elements, i.e. $prod_j neq i lambda_j$. Of course, this means that if $dimkerA>1$, $operatornameadj(A) = 0$, while if $dimkerA = 1$, $operatornameadj(A)$ has exactly one nonzero eigenvalue, the product of the others.
answered Aug 1 at 18:44
Chappers
54.9k74190
As this blog post points out, we have $operatornameadj(U^-1AU) = U^-1(operatornameadj(A)) U$ for any invertible $U$. Since any matrix $A$ can be reduced to a triangular one $B$ by such a transformation, $B=U^-1AU$, we have
$$ operatornameadj(B) = U^-1(operatornameadj(A)) U. $$
The diagonal elements of $B$ are $lambda_i$, the eigenvalues of $A$, and a calculation shows that the adjugate of an upper triangular matrix is also upper triangular, and the diagonal elements are $prod_j neq i lambda_j$.
So the eigenvalues of $operatornameadj(A)$ are the same as those of $operatornameadj(B)$, which are the diagonal elements, i.e. $prod_j neq i lambda_j$. Of course, this means that if $dimkerA>1$, $operatornameadj(A) = 0$, while if $dimkerA = 1$, $operatornameadj(A)$ has exactly one nonzero eigenvalue, the product of the others.
answered Aug 1 at 18:44
Chappers
54.9k74190
answered Aug 1 at 18:44
Chappers
54.9k74190
answered Aug 1 at 18:44
Chappers
54.9k74190
answered Aug 1 at 18:44
Chappers
54.9k74190
54.9k74190
add a comment |Â
add a comment |Â
up vote
1
down vote
The following is just a collection of direct properties:
If $mathrmrank(A)<n-1$, then the only eigenvalue of $mathrmadj(A)$ is $0$ as $mathrmadj(A)=mathbf0$.
If on the other hand $mathrmrank(A)=n-1$, then $0$ is again an eigenvalue of $mathrmadj(A)$ as $mathrmrank(mathrmadj(A))=1$, i.e. $mathrmdim(mathrmker(mathrmadj(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.
Note that this does not mean that there is not more to the $mathrmrank(A)=n-1$-case.
answered Aug 1 at 18:19
zzuussee
1,152419
add a comment |Â
up vote
1
down vote
The following is just a collection of direct properties:
If $mathrmrank(A)<n-1$, then the only eigenvalue of $mathrmadj(A)$ is $0$ as $mathrmadj(A)=mathbf0$.
If on the other hand $mathrmrank(A)=n-1$, then $0$ is again an eigenvalue of $mathrmadj(A)$ as $mathrmrank(mathrmadj(A))=1$, i.e. $mathrmdim(mathrmker(mathrmadj(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.
Note that this does not mean that there is not more to the $mathrmrank(A)=n-1$-case.
answered Aug 1 at 18:19
zzuussee
1,152419
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The following is just a collection of direct properties:
If $mathrmrank(A)<n-1$, then the only eigenvalue of $mathrmadj(A)$ is $0$ as $mathrmadj(A)=mathbf0$.
If on the other hand $mathrmrank(A)=n-1$, then $0$ is again an eigenvalue of $mathrmadj(A)$ as $mathrmrank(mathrmadj(A))=1$, i.e. $mathrmdim(mathrmker(mathrmadj(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.
Note that this does not mean that there is not more to the $mathrmrank(A)=n-1$-case.
answered Aug 1 at 18:19
zzuussee
1,152419
The following is just a collection of direct properties:
If $mathrmrank(A)<n-1$, then the only eigenvalue of $mathrmadj(A)$ is $0$ as $mathrmadj(A)=mathbf0$.
If on the other hand $mathrmrank(A)=n-1$, then $0$ is again an eigenvalue of $mathrmadj(A)$ as $mathrmrank(mathrmadj(A))=1$, i.e. $mathrmdim(mathrmker(mathrmadj(A)))=n-1$ by the rank-nullity theorem. Note, that $0$ thus has geometric multiplicity $n-1$ instead of $n$ as before.
Note that this does not mean that there is not more to the $mathrmrank(A)=n-1$-case.
answered Aug 1 at 18:19
zzuussee
1,152419
edited Aug 1 at 19:52
answered Aug 1 at 18:19
zzuussee
1,152419
answered Aug 1 at 18:19
zzuussee
1,152419
answered Aug 1 at 18:19
zzuussee
1,152419
1,152419
add a comment |Â
add a comment |Â
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1
If the rank of $A$ is $n-1$, then its adjugate has rank one.
– Lord Shark the Unknown
Aug 1 at 17:46
okks,@LordSharktheUnknown..let me thinks more
– stupid
Aug 1 at 17:49
2
Please not that before accepting my answer that it is just an observation and as @Chappers remarked there is more to the $n-1$-case. As his answer is more thorough, I encourage you to accept it instead.
– zzuussee
Aug 1 at 18:47