Mixing
basis for eigenspace corresponding to eigenvalue that reduces to I?
$begingroup$
every single example I ever encountered for finding eignspace basis' always was a situation where the reduced matrix had a null space of 1 or more. but what if after plugging in the eignenvalue and reducing, there are no free variables, leading to a reduced matrix of I and no vectors in the basis? is it an empty basis? is this basis an empty set? I was hoping I could find a quick answer to this, I can't seem to find it anywhere online.
such as basis for the eigenspace corresponding to eigenvalue -1 for the matrix A =
$$ left[
begin{array}{cc}
1&4\
2&3
end{array}
right] $$
since after I plug in eigenvalue -1 to the characteristic eq. it reduces to I giving me no free variables, and no t parameters, how do I find the basis? is it an empty set basis?
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Dec 4 '14 at 7:28
J LJ L
629102653
$endgroup$
add a comment |
$begingroup$
every single example I ever encountered for finding eignspace basis' always was a situation where the reduced matrix had a null space of 1 or more. but what if after plugging in the eignenvalue and reducing, there are no free variables, leading to a reduced matrix of I and no vectors in the basis? is it an empty basis? is this basis an empty set? I was hoping I could find a quick answer to this, I can't seem to find it anywhere online.
such as basis for the eigenspace corresponding to eigenvalue -1 for the matrix A =
$$ left[
begin{array}{cc}
1&4\
2&3
end{array}
right] $$
since after I plug in eigenvalue -1 to the characteristic eq. it reduces to I giving me no free variables, and no t parameters, how do I find the basis? is it an empty set basis?
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Dec 4 '14 at 7:28
J LJ L
629102653
$endgroup$
$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
1
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16
add a comment |
$begingroup$
every single example I ever encountered for finding eignspace basis' always was a situation where the reduced matrix had a null space of 1 or more. but what if after plugging in the eignenvalue and reducing, there are no free variables, leading to a reduced matrix of I and no vectors in the basis? is it an empty basis? is this basis an empty set? I was hoping I could find a quick answer to this, I can't seem to find it anywhere online.
such as basis for the eigenspace corresponding to eigenvalue -1 for the matrix A =
$$ left[
begin{array}{cc}
1&4\
2&3
end{array}
right] $$
since after I plug in eigenvalue -1 to the characteristic eq. it reduces to I giving me no free variables, and no t parameters, how do I find the basis? is it an empty set basis?
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Dec 4 '14 at 7:28
J LJ L
629102653
$endgroup$
every single example I ever encountered for finding eignspace basis' always was a situation where the reduced matrix had a null space of 1 or more. but what if after plugging in the eignenvalue and reducing, there are no free variables, leading to a reduced matrix of I and no vectors in the basis? is it an empty basis? is this basis an empty set? I was hoping I could find a quick answer to this, I can't seem to find it anywhere online.
such as basis for the eigenspace corresponding to eigenvalue -1 for the matrix A =
$$ left[
begin{array}{cc}
1&4\
2&3
end{array}
right] $$
since after I plug in eigenvalue -1 to the characteristic eq. it reduces to I giving me no free variables, and no t parameters, how do I find the basis? is it an empty set basis?
linear-algebra vector-spaces eigenvalues-eigenvectors
linear-algebra vector-spaces eigenvalues-eigenvectors
asked Dec 4 '14 at 7:28
J LJ L
629102653
asked Dec 4 '14 at 7:28
J LJ L
629102653
asked Dec 4 '14 at 7:28
J LJ L
629102653
asked Dec 4 '14 at 7:28
J LJ L
629102653
asked Dec 4 '14 at 7:28
J LJ L
629102653
629102653
$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
1
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16
add a comment |
$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
1
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16
$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
1
1
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
if $lambda$ is an eigenvalue of $A,$ then it cannot happen the row reduced matrix $A - lambda I$ lacks at least one free variable.
the reason is if $lambda$ is an eigenvalue of $A,$ then $det(A - lambda I) = 0.$ this means $A - lambda I$ is not invertible and the $rank(A) le n-1.$ this in turn implies that null space of $A - lambda I$ is non trivial. this guarantees at least one free variable.
this is the reason we are able to find a nonzero vector(eigenvector)$u$ associated with the eigenvalue $lambda$ such that $A u = lambda u, u neq 0.$
p.s. you may not need this information but the dimension of $A - lambda A$ is called the geometric multiplicity of $lambda$ and tells you the number of jordan blocks associated with $lambda.$
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
$endgroup$
add a comment |
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1 Answer
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$begingroup$
if $lambda$ is an eigenvalue of $A,$ then it cannot happen the row reduced matrix $A - lambda I$ lacks at least one free variable.
the reason is if $lambda$ is an eigenvalue of $A,$ then $det(A - lambda I) = 0.$ this means $A - lambda I$ is not invertible and the $rank(A) le n-1.$ this in turn implies that null space of $A - lambda I$ is non trivial. this guarantees at least one free variable.
this is the reason we are able to find a nonzero vector(eigenvector)$u$ associated with the eigenvalue $lambda$ such that $A u = lambda u, u neq 0.$
p.s. you may not need this information but the dimension of $A - lambda A$ is called the geometric multiplicity of $lambda$ and tells you the number of jordan blocks associated with $lambda.$
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
$endgroup$
add a comment |
$begingroup$
if $lambda$ is an eigenvalue of $A,$ then it cannot happen the row reduced matrix $A - lambda I$ lacks at least one free variable.
the reason is if $lambda$ is an eigenvalue of $A,$ then $det(A - lambda I) = 0.$ this means $A - lambda I$ is not invertible and the $rank(A) le n-1.$ this in turn implies that null space of $A - lambda I$ is non trivial. this guarantees at least one free variable.
this is the reason we are able to find a nonzero vector(eigenvector)$u$ associated with the eigenvalue $lambda$ such that $A u = lambda u, u neq 0.$
p.s. you may not need this information but the dimension of $A - lambda A$ is called the geometric multiplicity of $lambda$ and tells you the number of jordan blocks associated with $lambda.$
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
$endgroup$
add a comment |
$begingroup$
if $lambda$ is an eigenvalue of $A,$ then it cannot happen the row reduced matrix $A - lambda I$ lacks at least one free variable.
the reason is if $lambda$ is an eigenvalue of $A,$ then $det(A - lambda I) = 0.$ this means $A - lambda I$ is not invertible and the $rank(A) le n-1.$ this in turn implies that null space of $A - lambda I$ is non trivial. this guarantees at least one free variable.
this is the reason we are able to find a nonzero vector(eigenvector)$u$ associated with the eigenvalue $lambda$ such that $A u = lambda u, u neq 0.$
p.s. you may not need this information but the dimension of $A - lambda A$ is called the geometric multiplicity of $lambda$ and tells you the number of jordan blocks associated with $lambda.$
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
$endgroup$
if $lambda$ is an eigenvalue of $A,$ then it cannot happen the row reduced matrix $A - lambda I$ lacks at least one free variable.
the reason is if $lambda$ is an eigenvalue of $A,$ then $det(A - lambda I) = 0.$ this means $A - lambda I$ is not invertible and the $rank(A) le n-1.$ this in turn implies that null space of $A - lambda I$ is non trivial. this guarantees at least one free variable.
this is the reason we are able to find a nonzero vector(eigenvector)$u$ associated with the eigenvalue $lambda$ such that $A u = lambda u, u neq 0.$
p.s. you may not need this information but the dimension of $A - lambda A$ is called the geometric multiplicity of $lambda$ and tells you the number of jordan blocks associated with $lambda.$
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
answered Dec 6 '14 at 0:09
abelabel
26.5k12048
26.5k12048
add a comment |
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$begingroup$
Whenever there is a non-trivial homogenous solution $AX=0$ or $(A-lambda I)X=0$, there is always a free parameter because any constant multiple of $X$ is also a solution.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 7:44
$begingroup$
so if there is a free parameter what is it? because it seems like the solution is x1= 0 and x2=0, theres no parameter t, so how are you supposed to express the basis? is it {(0,0)}?
$endgroup$
– J L
Dec 4 '14 at 8:02
1
$begingroup$
$left[begin{array}{cc}1 & 4\2 & 3end{array}right]-(-1)I = left[begin{array}{cc}2 & 4 \ 2 & 4end{array}right]$ has null space $tleft[begin{array}{c} -2 \ 1end{array}right]$ where $t$ can be anything.
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 8:55
$begingroup$
damnit, somehow i either reduced the matrix wrong or plugged in the values wrong.. there goes over an hour fixating on 1 mysteri0us error >.<
$endgroup$
– J L
Dec 4 '14 at 9:18
$begingroup$
I wish I could get all those hours back, too. :)
$endgroup$
– DisintegratingByParts
Dec 4 '14 at 10:16