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Prove that every invariant spaces of $varphi'$ is invariant spaces of $varphi$
Problem: Let $E$ be a invariant space of linear operator $varphi$, $varphi'$ is a restriction of $varphi$ on $E'$. Prove that every invariant spaces of $varphi'$ is invariant spaces of $varphi$.
My attempt: Follows the hypothesis, we have $E' le E$. $E$ be a invariant space of linear operator $varphi$ and $varphi'$ is a restriction of $varphi$ on $E'$, so $varphi'(E') = varphi(E') subset E Rightarrow varphi$ is a linear operator on $E'$. To prove the problem, we just prove $E'$ is an invariant space of $E$. Suppose $S = {x_1, dots, x_n}$ is a base of $E$ and $R = {x_1, dots, x_r}$ is a base of $E'$. Since $E$ is an invariant space with operator $varphi$ implies $varphi(R) = displaystylesum_{i=1}^{n} a_ix_i in E$. Q.E.D
Could you check my work? Thank all!
linear-algebra
asked Nov 20 '18 at 4:37
Minh
1788
add a comment |
Problem: Let $E$ be a invariant space of linear operator $varphi$, $varphi'$ is a restriction of $varphi$ on $E'$. Prove that every invariant spaces of $varphi'$ is invariant spaces of $varphi$.
My attempt: Follows the hypothesis, we have $E' le E$. $E$ be a invariant space of linear operator $varphi$ and $varphi'$ is a restriction of $varphi$ on $E'$, so $varphi'(E') = varphi(E') subset E Rightarrow varphi$ is a linear operator on $E'$. To prove the problem, we just prove $E'$ is an invariant space of $E$. Suppose $S = {x_1, dots, x_n}$ is a base of $E$ and $R = {x_1, dots, x_r}$ is a base of $E'$. Since $E$ is an invariant space with operator $varphi$ implies $varphi(R) = displaystylesum_{i=1}^{n} a_ix_i in E$. Q.E.D
Could you check my work? Thank all!
linear-algebra
asked Nov 20 '18 at 4:37
Minh
1788
"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35
add a comment |
Problem: Let $E$ be a invariant space of linear operator $varphi$, $varphi'$ is a restriction of $varphi$ on $E'$. Prove that every invariant spaces of $varphi'$ is invariant spaces of $varphi$.
My attempt: Follows the hypothesis, we have $E' le E$. $E$ be a invariant space of linear operator $varphi$ and $varphi'$ is a restriction of $varphi$ on $E'$, so $varphi'(E') = varphi(E') subset E Rightarrow varphi$ is a linear operator on $E'$. To prove the problem, we just prove $E'$ is an invariant space of $E$. Suppose $S = {x_1, dots, x_n}$ is a base of $E$ and $R = {x_1, dots, x_r}$ is a base of $E'$. Since $E$ is an invariant space with operator $varphi$ implies $varphi(R) = displaystylesum_{i=1}^{n} a_ix_i in E$. Q.E.D
Could you check my work? Thank all!
linear-algebra
asked Nov 20 '18 at 4:37
Minh
1788
Problem: Let $E$ be a invariant space of linear operator $varphi$, $varphi'$ is a restriction of $varphi$ on $E'$. Prove that every invariant spaces of $varphi'$ is invariant spaces of $varphi$.
My attempt: Follows the hypothesis, we have $E' le E$. $E$ be a invariant space of linear operator $varphi$ and $varphi'$ is a restriction of $varphi$ on $E'$, so $varphi'(E') = varphi(E') subset E Rightarrow varphi$ is a linear operator on $E'$. To prove the problem, we just prove $E'$ is an invariant space of $E$. Suppose $S = {x_1, dots, x_n}$ is a base of $E$ and $R = {x_1, dots, x_r}$ is a base of $E'$. Since $E$ is an invariant space with operator $varphi$ implies $varphi(R) = displaystylesum_{i=1}^{n} a_ix_i in E$. Q.E.D
Could you check my work? Thank all!
linear-algebra
linear-algebra
asked Nov 20 '18 at 4:37
Minh
1788
asked Nov 20 '18 at 4:37
Minh
1788
edited Nov 20 '18 at 5:15
asked Nov 20 '18 at 4:37
Minh
1788
asked Nov 20 '18 at 4:37
Minh
1788
asked Nov 20 '18 at 4:37
Minh
1788
1788
"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35
add a comment |
"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35
"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35
add a comment |
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"Let $E$ be an invariant space of linear operator $varphi$". Do you mean $E'$ instead of $E$ here?
– Theo Bendit
Nov 20 '18 at 4:48
No, $E$ is an invariant space with operator $varphi$ and from the hypothesis I think that $E'$ is a subspace of $E$.
– Minh
Nov 20 '18 at 4:53
I was thinking that $E$ might be the domain of $varphi$, and $E'$ is the invariant subspace. It's not true that any subspace of an invariant subspace is invariant. Also, simply introducing a new symbol $E'$ without any specification is poor practice when posing a problem.
– Theo Bendit
Nov 20 '18 at 5:29
@Theo_Bendit You right. But the problem's request here that is: every invariant subspaces of $varphi'$ also be invariant subspaces of $varphi$.
– Minh
Nov 20 '18 at 5:35