N(T) and R(T) are T-invariant: application to proofs.
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Am I correct in assuming that since $N(T)$ and $R(T)$ are $T$-invariant,
$N(T^k)= N(T)$ and $R(T^k)=R(T)$,
$dim(N(T^k))= dim(N(T))$ and $dim(R(T^k))=dim(R(T))$
for all $k$?
If true, can someone give me the intuition behind this?
linear-algebra
asked Nov 19 at 2:08
alwaysiamcaesar
505
add a comment |
up vote
0
down vote
favorite
Am I correct in assuming that since $N(T)$ and $R(T)$ are $T$-invariant,
$N(T^k)= N(T)$ and $R(T^k)=R(T)$,
$dim(N(T^k))= dim(N(T))$ and $dim(R(T^k))=dim(R(T))$
for all $k$?
If true, can someone give me the intuition behind this?
linear-algebra
asked Nov 19 at 2:08
alwaysiamcaesar
505
I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
2
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Am I correct in assuming that since $N(T)$ and $R(T)$ are $T$-invariant,
$N(T^k)= N(T)$ and $R(T^k)=R(T)$,
$dim(N(T^k))= dim(N(T))$ and $dim(R(T^k))=dim(R(T))$
for all $k$?
If true, can someone give me the intuition behind this?
linear-algebra
asked Nov 19 at 2:08
alwaysiamcaesar
505
Am I correct in assuming that since $N(T)$ and $R(T)$ are $T$-invariant,
$N(T^k)= N(T)$ and $R(T^k)=R(T)$,
$dim(N(T^k))= dim(N(T))$ and $dim(R(T^k))=dim(R(T))$
for all $k$?
If true, can someone give me the intuition behind this?
linear-algebra
linear-algebra
asked Nov 19 at 2:08
alwaysiamcaesar
505
asked Nov 19 at 2:08
alwaysiamcaesar
505
asked Nov 19 at 2:08
alwaysiamcaesar
505
asked Nov 19 at 2:08
alwaysiamcaesar
505
asked Nov 19 at 2:08
alwaysiamcaesar
505
505
I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
2
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24
add a comment |
I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
2
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24
I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
2
2
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24
add a comment |
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I can only assume that $N(T)$ is the null space and $R(T)$ is the row space. The notation is not standard. Both equations are false.
– Matt Samuel
Nov 19 at 2:19
2
Invariant is a bit of a misnomer. We say a subspace $A$ is $T$-invariant if $T(A) subseteq A$. Equality is not required.
– Matt Samuel
Nov 19 at 2:24