Prove that any eigenvectors of $A^-1$ must be in some subspace of $adj(A)$
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Prove that any eigenvectors of $A^-1$ must be in some subspace of $adj(A)$
Since $A^-1$=$frac1detAajdA$, so if $lambda$ is eigenvalue of $A^-1$ so it exist $xnot=0$ such that $A^-1x=lambda x$ then $frac1detAajdAx=lambda x$ since $frac1det$ is some scalar then $adjAx=detAlambda x$ and if I put $lambda´=detAlambda$ then $adjAx=lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:24
Marko Škorić
4008
add a comment |Â
up vote
3
down vote
favorite
Prove that any eigenvectors of $A^-1$ must be in some subspace of $adj(A)$
Since $A^-1$=$frac1detAajdA$, so if $lambda$ is eigenvalue of $A^-1$ so it exist $xnot=0$ such that $A^-1x=lambda x$ then $frac1detAajdAx=lambda x$ since $frac1det$ is some scalar then $adjAx=detAlambda x$ and if I put $lambda´=detAlambda$ then $adjAx=lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:24
Marko Škorić
4008
What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
1
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
Yes you are right
– Marko Å korić
Sep 8 at 9:17
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Prove that any eigenvectors of $A^-1$ must be in some subspace of $adj(A)$
Since $A^-1$=$frac1detAajdA$, so if $lambda$ is eigenvalue of $A^-1$ so it exist $xnot=0$ such that $A^-1x=lambda x$ then $frac1detAajdAx=lambda x$ since $frac1det$ is some scalar then $adjAx=detAlambda x$ and if I put $lambda´=detAlambda$ then $adjAx=lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:24
Marko Škorić
4008
Prove that any eigenvectors of $A^-1$ must be in some subspace of $adj(A)$
Since $A^-1$=$frac1detAajdA$, so if $lambda$ is eigenvalue of $A^-1$ so it exist $xnot=0$ such that $A^-1x=lambda x$ then $frac1detAajdAx=lambda x$ since $frac1det$ is some scalar then $adjAx=detAlambda x$ and if I put $lambda´=detAlambda$ then $adjAx=lambda^´ x$, so they have the same eigenvectors, I think that is easy way to show but what you think?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Sep 8 at 8:24
Marko Škorić
4008
asked Sep 8 at 8:24
Marko Škorić
4008
edited Sep 8 at 8:33
asked Sep 8 at 8:24
Marko Škorić
4008
asked Sep 8 at 8:24
Marko Škorić
4008
asked Sep 8 at 8:24
Marko Škorić
4008
4008
What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
1
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
Yes you are right
– Marko Å korić
Sep 8 at 9:17
add a comment |Â
What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
1
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
Yes you are right
– Marko Å korić
Sep 8 at 9:17
What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
1
1
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
Yes you are right
– Marko Å korić
Sep 8 at 9:17
Yes you are right
– Marko Å korić
Sep 8 at 9:17
add a comment |Â
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What's "some subspace of $adj(A)$ means?
– Vim
Sep 8 at 9:12
I think that I need to write eigensubspace
– Marko Å korić
Sep 8 at 9:14
1
As you have noted $A^-1$ is just a non-zero multiple of $textadj(A)$ so they trivially have the same eigenvectors don't they?
– Vim
Sep 8 at 9:15
Yes you are right
– Marko Å korić
Sep 8 at 9:17