Trace in a generalized eigenspace
$begingroup$
I am reading a paper and there is a steep in a proof that I cannot understand. The proof
says:
Suppose $v_lambda$ is a generalized eigenvector corresponding to an eigenvalue $lambda$ of the map $T$, then there is a $n , in mathbb{N}$ such that
$(T-lambda,I)^n(v_lambda)=0$
and they arrive to:
$$Tr{(T-lambda,I)^n(v_lambda)}=(1-lambda)^n,Tr;v_lambda=0$$
Due to $T$ is trace-preserving, its trace is 1, but I am not able to see
why $v_lambda$ is treated as a matrix ( at least it is what I understand) and no as a vector. So I would like to know if someone has a reference to understand this or if someone can explain this to me. Thanks in advance.
linear-algebra linear-transformations generalized-eigenvector
asked Jan 17 at 13:22
morsmors
134
$endgroup$
add a comment |
$begingroup$
I am reading a paper and there is a steep in a proof that I cannot understand. The proof
says:
Suppose $v_lambda$ is a generalized eigenvector corresponding to an eigenvalue $lambda$ of the map $T$, then there is a $n , in mathbb{N}$ such that
$(T-lambda,I)^n(v_lambda)=0$
and they arrive to:
$$Tr{(T-lambda,I)^n(v_lambda)}=(1-lambda)^n,Tr;v_lambda=0$$
Due to $T$ is trace-preserving, its trace is 1, but I am not able to see
why $v_lambda$ is treated as a matrix ( at least it is what I understand) and no as a vector. So I would like to know if someone has a reference to understand this or if someone can explain this to me. Thanks in advance.
linear-algebra linear-transformations generalized-eigenvector
asked Jan 17 at 13:22
morsmors
134
$endgroup$
1
$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01
add a comment |
$begingroup$
I am reading a paper and there is a steep in a proof that I cannot understand. The proof
says:
Suppose $v_lambda$ is a generalized eigenvector corresponding to an eigenvalue $lambda$ of the map $T$, then there is a $n , in mathbb{N}$ such that
$(T-lambda,I)^n(v_lambda)=0$
and they arrive to:
$$Tr{(T-lambda,I)^n(v_lambda)}=(1-lambda)^n,Tr;v_lambda=0$$
Due to $T$ is trace-preserving, its trace is 1, but I am not able to see
why $v_lambda$ is treated as a matrix ( at least it is what I understand) and no as a vector. So I would like to know if someone has a reference to understand this or if someone can explain this to me. Thanks in advance.
linear-algebra linear-transformations generalized-eigenvector
asked Jan 17 at 13:22
morsmors
134
$endgroup$
I am reading a paper and there is a steep in a proof that I cannot understand. The proof
says:
Suppose $v_lambda$ is a generalized eigenvector corresponding to an eigenvalue $lambda$ of the map $T$, then there is a $n , in mathbb{N}$ such that
$(T-lambda,I)^n(v_lambda)=0$
and they arrive to:
$$Tr{(T-lambda,I)^n(v_lambda)}=(1-lambda)^n,Tr;v_lambda=0$$
Due to $T$ is trace-preserving, its trace is 1, but I am not able to see
why $v_lambda$ is treated as a matrix ( at least it is what I understand) and no as a vector. So I would like to know if someone has a reference to understand this or if someone can explain this to me. Thanks in advance.
linear-algebra linear-transformations generalized-eigenvector
linear-algebra linear-transformations generalized-eigenvector
asked Jan 17 at 13:22
morsmors
134
asked Jan 17 at 13:22
morsmors
134
asked Jan 17 at 13:22
morsmors
134
asked Jan 17 at 13:22
morsmors
134
asked Jan 17 at 13:22
morsmors
134
134
1
$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01
add a comment |
1
$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01
1
1
$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01
$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01
add a comment |
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$begingroup$
Could you give a reference to the paper?
$endgroup$
– Zeekless
Jan 17 at 15:01