characterising the square 2x2 matrices on types of eigenvalues
Multi tool use
If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-
- begin{bmatrix}
a & 0 \
0 & b
end{bmatrix}
begin{bmatrix}
a & 0 \
1 & a
end{bmatrix}
We know we get 2 eigenvalues since the matrix is over complex numbers.
If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?
linear-algebra matrices eigenvalues-eigenvectors
add a comment |
If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-
- begin{bmatrix}
a & 0 \
0 & b
end{bmatrix}
begin{bmatrix}
a & 0 \
1 & a
end{bmatrix}
We know we get 2 eigenvalues since the matrix is over complex numbers.
If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?
linear-algebra matrices eigenvalues-eigenvectors
add a comment |
If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-
- begin{bmatrix}
a & 0 \
0 & b
end{bmatrix}
begin{bmatrix}
a & 0 \
1 & a
end{bmatrix}
We know we get 2 eigenvalues since the matrix is over complex numbers.
If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?
linear-algebra matrices eigenvalues-eigenvectors
If A is a 2 X 2 matrix with complex entries, then A is similar over $C$ to a matrix of one of the two types :-
- begin{bmatrix}
a & 0 \
0 & b
end{bmatrix}
begin{bmatrix}
a & 0 \
1 & a
end{bmatrix}
We know we get 2 eigenvalues since the matrix is over complex numbers.
If the Eigenvalues are distinct, or if they are equal with geometric multiplicity 2 then it's similar to type 1. If they are equal but the geometric multiplicity is 1 then we get one eigenvector, how do I find out the other basis element?
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
edited Nov 24 at 12:10
José Carlos Santos
148k22117218
148k22117218
asked Nov 24 at 11:44
Rohan Rajagopal
5811
5811
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1 Answer
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If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
add a comment |
If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
add a comment |
If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.
If $vneq(0,0)$ is such that $A.v=av$, solve the equation $A.w=aw+v$. That will give you the other vector that you are looking for.
answered Nov 24 at 12:01
José Carlos Santos
148k22117218
148k22117218
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
add a comment |
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
I know, I got that. But how are we sure to get such a vector?
– Rohan Rajagopal
Nov 24 at 12:09
1
1
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Every square matrix is similar to a lower triangular matrix. In particular, if $A$ is a $2times2$ matrix whose only eigenvalue is $a$, then $A$ is similar to a matrix of the form $left(begin{smallmatrix}a&0\lambda&aend{smallmatrix}right),$with $lambdaneq0$. This means that there are vectors $v$ and $w$ such that $A.v=av$ and that $A.w=aw+lambda v$. It is now not hard to define a vector $w^star$ such that $A.w^star=aw^star+v$.
– José Carlos Santos
Nov 24 at 12:17
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
Thank you very much. Deeply appreciate it.
– Rohan Rajagopal
Nov 24 at 13:01
add a comment |
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