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Why is $W = R((T-lambda I)^m)$ $T$-invariant?
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$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.
$T_W$ is the $T$-invariant subspace.
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.
So, I'm having some difficulty interpreting the second paragraph.
Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant.
Is $W$ $T$-invariant because $(T-lambda I)^m(v) = 0$ for some $v in V$?
Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
I'm unable to see why $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
linear-algebra
asked 4 hours ago
K.M
449312
add a comment |Â
up vote
1
down vote
favorite
$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.
$T_W$ is the $T$-invariant subspace.
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.
So, I'm having some difficulty interpreting the second paragraph.
Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant.
Is $W$ $T$-invariant because $(T-lambda I)^m(v) = 0$ for some $v in V$?
Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
I'm unable to see why $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
linear-algebra
asked 4 hours ago
K.M
449312
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.
$T_W$ is the $T$-invariant subspace.
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.
So, I'm having some difficulty interpreting the second paragraph.
Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant.
Is $W$ $T$-invariant because $(T-lambda I)^m(v) = 0$ for some $v in V$?
Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
I'm unable to see why $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
linear-algebra
asked 4 hours ago
K.M
449312
$K_lambda_i$ = $v in V: (T-lambda_i I)^p(v) = 0$ for some positive integer $p$.
$T_W$ is the $T$-invariant subspace.
Theorem 7.3. Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial of $T$ splits, and lett $lambda_1,lambda_2,...,lambda_k$ be the distinct eigenvalues of $T$. Then, for every $xin V$, there exist vectors $v_i in K_lambda_i$, $1le i le k$, such that
$x = v_1 + v_2 + cdots +v_k$.
Proof. The proof is by mathematical induction on the number $k$ of distinct eigenvalues of $T$. First suppose that $k = 1$, and let $m$ be the multiplicity of $lambda_1$. Then $(lambda_1 - t)^m$is the characteristic polynomial of $T$, and hence $(lambda_1 I - T)^m = T_0$ by the Cayley-Hamilton theorem(p.317). Thus $V=K_lambda_i$, and the result follows.
Now suppose that for some integer $k>1$, the result is established whenever $T$ has fewer than $k$ distinct eigenvalues, and suppose that $T$ has $k$ distinct eigenvalues. Let $m$ be the multiplicity of $lambda_k$, and let $f(t)$ be the characteristic polynomial of $T$. Then $f(t) = (t - lambda_k)^mg(t)$ from some polynomial $g(t)$ not divisible by $(t-lambda_k)$. Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant. Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$. For suppose that $i<k$. Since $(T - lambda I)^m$ maps $K_lambda_i$ into itself and $lambda_k ne lambda_i$, the restriction of $T-lambda_k I$ to $K_lambda_i$ is one-to-one and hence is onto. One consequence of this is that for $i<k$, $K_lambda_i$ is contained in $W$, and hence $lambda_i$ is an eigenvalue of $T_W$ with corresponding generalized eigenspace $K_lambda_i$.
So, I'm having some difficulty interpreting the second paragraph.
Let $W = R((T-lambda I)^m)$. Clearly $W$ is $T$-invariant.
Is $W$ $T$-invariant because $(T-lambda I)^m(v) = 0$ for some $v in V$?
Observe that $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
I'm unable to see why $(T-lambda_k I)^m$ maps $K_lambda_i$ onto itself for $i<k$
linear-algebra
asked 4 hours ago
K.M
449312
edited 4 hours ago
asked 4 hours ago
K.M
449312
asked 4 hours ago
K.M
449312
asked 4 hours ago
K.M
449312
449312
add a comment |Â
add a comment |Â
1 Answer
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These two facts are both true because the relevant operators commute.
First let's recall that we say that $W$ is $T$-invariant if $T(W) subseteq W$.
If $x in W$, then by definition of $W$, $x = (T-lambda I)^m y$ for some $y in V$. As a result, $Tx = T(T-lambda I)^m y = (T-lambda I)^m Ty in R((T-lambda I)^m) = W$. This shows that $T(W) subseteq W$ as desired.
For the second part, by the justification given in the proof, it is enough to see that $(T-lambda_k)^m$ maps $K_lambda_i$ into itself.
So suppose $x in K_lambda_i$. This means that there is a $p$ such that $(T-lambda_i)^p x = 0$. Then we have
$$(T-lambda_i)^p (T-lambda_k) x = (T- lambda_k) (T-lambda_i)^p x = (T- lambda_k) 0 = 0$$
and so $(T-lambda_k)x in K_lambda_i$ which is what we wanted to show.
answered 4 hours ago
Rhys Steele
5,4151727
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
These two facts are both true because the relevant operators commute.
First let's recall that we say that $W$ is $T$-invariant if $T(W) subseteq W$.
If $x in W$, then by definition of $W$, $x = (T-lambda I)^m y$ for some $y in V$. As a result, $Tx = T(T-lambda I)^m y = (T-lambda I)^m Ty in R((T-lambda I)^m) = W$. This shows that $T(W) subseteq W$ as desired.
For the second part, by the justification given in the proof, it is enough to see that $(T-lambda_k)^m$ maps $K_lambda_i$ into itself.
So suppose $x in K_lambda_i$. This means that there is a $p$ such that $(T-lambda_i)^p x = 0$. Then we have
$$(T-lambda_i)^p (T-lambda_k) x = (T- lambda_k) (T-lambda_i)^p x = (T- lambda_k) 0 = 0$$
and so $(T-lambda_k)x in K_lambda_i$ which is what we wanted to show.
answered 4 hours ago
Rhys Steele
5,4151727
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
These two facts are both true because the relevant operators commute.
First let's recall that we say that $W$ is $T$-invariant if $T(W) subseteq W$.
If $x in W$, then by definition of $W$, $x = (T-lambda I)^m y$ for some $y in V$. As a result, $Tx = T(T-lambda I)^m y = (T-lambda I)^m Ty in R((T-lambda I)^m) = W$. This shows that $T(W) subseteq W$ as desired.
For the second part, by the justification given in the proof, it is enough to see that $(T-lambda_k)^m$ maps $K_lambda_i$ into itself.
So suppose $x in K_lambda_i$. This means that there is a $p$ such that $(T-lambda_i)^p x = 0$. Then we have
$$(T-lambda_i)^p (T-lambda_k) x = (T- lambda_k) (T-lambda_i)^p x = (T- lambda_k) 0 = 0$$
and so $(T-lambda_k)x in K_lambda_i$ which is what we wanted to show.
answered 4 hours ago
Rhys Steele
5,4151727
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
These two facts are both true because the relevant operators commute.
First let's recall that we say that $W$ is $T$-invariant if $T(W) subseteq W$.
If $x in W$, then by definition of $W$, $x = (T-lambda I)^m y$ for some $y in V$. As a result, $Tx = T(T-lambda I)^m y = (T-lambda I)^m Ty in R((T-lambda I)^m) = W$. This shows that $T(W) subseteq W$ as desired.
For the second part, by the justification given in the proof, it is enough to see that $(T-lambda_k)^m$ maps $K_lambda_i$ into itself.
So suppose $x in K_lambda_i$. This means that there is a $p$ such that $(T-lambda_i)^p x = 0$. Then we have
$$(T-lambda_i)^p (T-lambda_k) x = (T- lambda_k) (T-lambda_i)^p x = (T- lambda_k) 0 = 0$$
and so $(T-lambda_k)x in K_lambda_i$ which is what we wanted to show.
answered 4 hours ago
Rhys Steele
5,4151727
These two facts are both true because the relevant operators commute.
First let's recall that we say that $W$ is $T$-invariant if $T(W) subseteq W$.
If $x in W$, then by definition of $W$, $x = (T-lambda I)^m y$ for some $y in V$. As a result, $Tx = T(T-lambda I)^m y = (T-lambda I)^m Ty in R((T-lambda I)^m) = W$. This shows that $T(W) subseteq W$ as desired.
For the second part, by the justification given in the proof, it is enough to see that $(T-lambda_k)^m$ maps $K_lambda_i$ into itself.
So suppose $x in K_lambda_i$. This means that there is a $p$ such that $(T-lambda_i)^p x = 0$. Then we have
$$(T-lambda_i)^p (T-lambda_k) x = (T- lambda_k) (T-lambda_i)^p x = (T- lambda_k) 0 = 0$$
and so $(T-lambda_k)x in K_lambda_i$ which is what we wanted to show.
answered 4 hours ago
Rhys Steele
5,4151727
answered 4 hours ago
Rhys Steele
5,4151727
answered 4 hours ago
Rhys Steele
5,4151727
answered 4 hours ago
Rhys Steele
5,4151727
5,4151727
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
add a comment |Â
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
For the second part, I thought we were trying to show that $(T-lambda_k)^m$ maps $K_lambda$ into itself, so since $0 in K_lambda_i$ , $(T-lambda_k I)^m0=0 in K_lambda_i$? I don't know if I'm misunderstanding your reasoning.
– K.M
3 hours ago
1
1
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
$(T-lambda_k)^m$ maps $K_lambda_i$ into itself means that for any $x in K_lambda_i$ we have that $(T-lambda_k)^m x in K_lambda_i$. It is not enough to check that this is true for some special $x in K_lambda_i$ like $x = 0$ since we want it to be true for all such $x$. My argument deals with an arbitrary $x in K_lambda_i$.
– Rhys Steele
2 hours ago
add a comment |Â
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