Decomposing a Vector Space into a direct sum of Generalized Eigenspaces
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I've seen a proof that doesn't involve any induction, but now I am trying to do this inductively since it is bothering me.
If $T$ is a linear operator over a finite-dimensional vector space $V$ and the characteristic polynomial $phi(t)$ of $T$ factors into
begin{gather}
phi(t) = prod^n_{i = 1}(t - lambda_i)^{m_i}
end{gather}
where each $lambda_i$ is an eigenvalue of $T.$ Then for each generalized eigenspace $K_{lambda_i}$ we have that
begin{gather}
V = bigoplus^n_{i = 1}K_{lambda_i}.
end{gather}
My idea was to use a base case of two eigenvalues. Our base case will show that when $phi(t) = (t - lambda_1)^{m_1}(t - lambda_2)^{m_2}$ we have that. If we consider $f(t) = (t - lambda_1)^{m_1}, g(t) = (t - lambda_2)^{m_2},$ we see that the following comes from the fact that $f,g$ are both co-prime. We have that
begin{gather}
V = N(f(t)) oplus R(f(t)).
end{gather}
Additionally, we note that $N(f(t)) = K_{lambda_1}$ and we have that $R(f(t)) = N(g(t)) = K_{lambda_2}.$ So we have that
begin{gather}
V = K_{lambda_1} oplus K_{lambda_2}.
end{gather}
So, this is where I am not sure where to continue. I don't know what to suppose as my inductive hypothesis. My ultimate goal is to express the final polynomial
begin{gather}
phi(t) = (t - lambda_1)^{m_1} cdots (t - lambda_n)^{m_n}.
end{gather}
we can express one polynomial as $f(t) = (t - lambda_1)^{m_1}$ and $g(t) = (t - lambda_2)^{m_2} cdots (t - lambda_n)^{m_n}.$ Then with the discussion above argue that
begin{gather}
R(f(t)) = bigoplus^n_{i = 2} K_{lambda_i}.
end{gather}
and then we will have that $V$ is a direct sum of these generalized eigenspaces. But I am not sure what the inductive step here is.
linear-algebra
asked Nov 13 at 22:11
Josabanks
434
add a comment |
up vote
2
down vote
favorite
I've seen a proof that doesn't involve any induction, but now I am trying to do this inductively since it is bothering me.
If $T$ is a linear operator over a finite-dimensional vector space $V$ and the characteristic polynomial $phi(t)$ of $T$ factors into
begin{gather}
phi(t) = prod^n_{i = 1}(t - lambda_i)^{m_i}
end{gather}
where each $lambda_i$ is an eigenvalue of $T.$ Then for each generalized eigenspace $K_{lambda_i}$ we have that
begin{gather}
V = bigoplus^n_{i = 1}K_{lambda_i}.
end{gather}
My idea was to use a base case of two eigenvalues. Our base case will show that when $phi(t) = (t - lambda_1)^{m_1}(t - lambda_2)^{m_2}$ we have that. If we consider $f(t) = (t - lambda_1)^{m_1}, g(t) = (t - lambda_2)^{m_2},$ we see that the following comes from the fact that $f,g$ are both co-prime. We have that
begin{gather}
V = N(f(t)) oplus R(f(t)).
end{gather}
Additionally, we note that $N(f(t)) = K_{lambda_1}$ and we have that $R(f(t)) = N(g(t)) = K_{lambda_2}.$ So we have that
begin{gather}
V = K_{lambda_1} oplus K_{lambda_2}.
end{gather}
So, this is where I am not sure where to continue. I don't know what to suppose as my inductive hypothesis. My ultimate goal is to express the final polynomial
begin{gather}
phi(t) = (t - lambda_1)^{m_1} cdots (t - lambda_n)^{m_n}.
end{gather}
we can express one polynomial as $f(t) = (t - lambda_1)^{m_1}$ and $g(t) = (t - lambda_2)^{m_2} cdots (t - lambda_n)^{m_n}.$ Then with the discussion above argue that
begin{gather}
R(f(t)) = bigoplus^n_{i = 2} K_{lambda_i}.
end{gather}
and then we will have that $V$ is a direct sum of these generalized eigenspaces. But I am not sure what the inductive step here is.
linear-algebra
asked Nov 13 at 22:11
Josabanks
434
My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I've seen a proof that doesn't involve any induction, but now I am trying to do this inductively since it is bothering me.
If $T$ is a linear operator over a finite-dimensional vector space $V$ and the characteristic polynomial $phi(t)$ of $T$ factors into
begin{gather}
phi(t) = prod^n_{i = 1}(t - lambda_i)^{m_i}
end{gather}
where each $lambda_i$ is an eigenvalue of $T.$ Then for each generalized eigenspace $K_{lambda_i}$ we have that
begin{gather}
V = bigoplus^n_{i = 1}K_{lambda_i}.
end{gather}
My idea was to use a base case of two eigenvalues. Our base case will show that when $phi(t) = (t - lambda_1)^{m_1}(t - lambda_2)^{m_2}$ we have that. If we consider $f(t) = (t - lambda_1)^{m_1}, g(t) = (t - lambda_2)^{m_2},$ we see that the following comes from the fact that $f,g$ are both co-prime. We have that
begin{gather}
V = N(f(t)) oplus R(f(t)).
end{gather}
Additionally, we note that $N(f(t)) = K_{lambda_1}$ and we have that $R(f(t)) = N(g(t)) = K_{lambda_2}.$ So we have that
begin{gather}
V = K_{lambda_1} oplus K_{lambda_2}.
end{gather}
So, this is where I am not sure where to continue. I don't know what to suppose as my inductive hypothesis. My ultimate goal is to express the final polynomial
begin{gather}
phi(t) = (t - lambda_1)^{m_1} cdots (t - lambda_n)^{m_n}.
end{gather}
we can express one polynomial as $f(t) = (t - lambda_1)^{m_1}$ and $g(t) = (t - lambda_2)^{m_2} cdots (t - lambda_n)^{m_n}.$ Then with the discussion above argue that
begin{gather}
R(f(t)) = bigoplus^n_{i = 2} K_{lambda_i}.
end{gather}
and then we will have that $V$ is a direct sum of these generalized eigenspaces. But I am not sure what the inductive step here is.
linear-algebra
asked Nov 13 at 22:11
Josabanks
434
I've seen a proof that doesn't involve any induction, but now I am trying to do this inductively since it is bothering me.
If $T$ is a linear operator over a finite-dimensional vector space $V$ and the characteristic polynomial $phi(t)$ of $T$ factors into
begin{gather}
phi(t) = prod^n_{i = 1}(t - lambda_i)^{m_i}
end{gather}
where each $lambda_i$ is an eigenvalue of $T.$ Then for each generalized eigenspace $K_{lambda_i}$ we have that
begin{gather}
V = bigoplus^n_{i = 1}K_{lambda_i}.
end{gather}
My idea was to use a base case of two eigenvalues. Our base case will show that when $phi(t) = (t - lambda_1)^{m_1}(t - lambda_2)^{m_2}$ we have that. If we consider $f(t) = (t - lambda_1)^{m_1}, g(t) = (t - lambda_2)^{m_2},$ we see that the following comes from the fact that $f,g$ are both co-prime. We have that
begin{gather}
V = N(f(t)) oplus R(f(t)).
end{gather}
Additionally, we note that $N(f(t)) = K_{lambda_1}$ and we have that $R(f(t)) = N(g(t)) = K_{lambda_2}.$ So we have that
begin{gather}
V = K_{lambda_1} oplus K_{lambda_2}.
end{gather}
So, this is where I am not sure where to continue. I don't know what to suppose as my inductive hypothesis. My ultimate goal is to express the final polynomial
begin{gather}
phi(t) = (t - lambda_1)^{m_1} cdots (t - lambda_n)^{m_n}.
end{gather}
we can express one polynomial as $f(t) = (t - lambda_1)^{m_1}$ and $g(t) = (t - lambda_2)^{m_2} cdots (t - lambda_n)^{m_n}.$ Then with the discussion above argue that
begin{gather}
R(f(t)) = bigoplus^n_{i = 2} K_{lambda_i}.
end{gather}
and then we will have that $V$ is a direct sum of these generalized eigenspaces. But I am not sure what the inductive step here is.
linear-algebra
linear-algebra
asked Nov 13 at 22:11
Josabanks
434
asked Nov 13 at 22:11
Josabanks
434
asked Nov 13 at 22:11
Josabanks
434
asked Nov 13 at 22:11
Josabanks
434
asked Nov 13 at 22:11
Josabanks
434
434
My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27
add a comment |
My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27
My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27
My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27
add a comment |
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My feeling here is that you would start with/show $V=K_{lambda_{1}}oplus S$, for some $S$ and then proceed to decompose $S$ .
– AnyAD
Nov 13 at 22:27