Linear operator image subspace chain
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How to prove the proposition: $A: V to V$ is a linear operator on a finite-dimensional vector space $V $, if $Im(A^p)=Im(A^{p+1})$,then $Im(A^{p+1})=Im(A^{p+2})$
The "Kernel" version is simple but I am stuck at the Image version of this proposition.
linear-algebra operator-theory
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
$endgroup$
add a comment |
$begingroup$
How to prove the proposition: $A: V to V$ is a linear operator on a finite-dimensional vector space $V $, if $Im(A^p)=Im(A^{p+1})$,then $Im(A^{p+1})=Im(A^{p+2})$
The "Kernel" version is simple but I am stuck at the Image version of this proposition.
linear-algebra operator-theory
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
$endgroup$
$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15
add a comment |
$begingroup$
How to prove the proposition: $A: V to V$ is a linear operator on a finite-dimensional vector space $V $, if $Im(A^p)=Im(A^{p+1})$,then $Im(A^{p+1})=Im(A^{p+2})$
The "Kernel" version is simple but I am stuck at the Image version of this proposition.
linear-algebra operator-theory
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
$endgroup$
How to prove the proposition: $A: V to V$ is a linear operator on a finite-dimensional vector space $V $, if $Im(A^p)=Im(A^{p+1})$,then $Im(A^{p+1})=Im(A^{p+2})$
The "Kernel" version is simple but I am stuck at the Image version of this proposition.
linear-algebra operator-theory
linear-algebra operator-theory
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
asked Jan 22 at 14:13
Hyz YuzhouHyz Yuzhou
31
31
$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15
add a comment |
$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15
$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15
$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First direction:
$A^{p+2} subseteq A^{p+1}$:
If $v = A^{p+2}(w)$ for some $w in V$, then for some $w' in V$
$$ v = A(A^{p+1}(w)) = A(A^p(w')) = A^{p+1}(w') in Im(A^{p+1})$$
Other direction:
$A^{p+1} subseteq A^{p+2}$:
If $ v = A^{p+1}(w)$ for some $w in V$, then for some $w' in V$
$$v = A(A^p(w)) = A(A^{p+1}(w')) = A^{p+2}(w') in Im(A^{p+2}) $$
Shorter version
The above is instructive, but there's a one-liner to this, using the hint that states $Im(Acirc B) = A(Im(B))$:
$$ Im(A^{p+2}) = Im(A circ A^{p+1}) = A(Im(A^{p+1})) = A(Im(A^p)) = Im(A^{p+1}) $$
answered Jan 22 at 23:48
MetricMetric
1,23649
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First direction:
$A^{p+2} subseteq A^{p+1}$:
If $v = A^{p+2}(w)$ for some $w in V$, then for some $w' in V$
$$ v = A(A^{p+1}(w)) = A(A^p(w')) = A^{p+1}(w') in Im(A^{p+1})$$
Other direction:
$A^{p+1} subseteq A^{p+2}$:
If $ v = A^{p+1}(w)$ for some $w in V$, then for some $w' in V$
$$v = A(A^p(w)) = A(A^{p+1}(w')) = A^{p+2}(w') in Im(A^{p+2}) $$
Shorter version
The above is instructive, but there's a one-liner to this, using the hint that states $Im(Acirc B) = A(Im(B))$:
$$ Im(A^{p+2}) = Im(A circ A^{p+1}) = A(Im(A^{p+1})) = A(Im(A^p)) = Im(A^{p+1}) $$
answered Jan 22 at 23:48
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
First direction:
$A^{p+2} subseteq A^{p+1}$:
If $v = A^{p+2}(w)$ for some $w in V$, then for some $w' in V$
$$ v = A(A^{p+1}(w)) = A(A^p(w')) = A^{p+1}(w') in Im(A^{p+1})$$
Other direction:
$A^{p+1} subseteq A^{p+2}$:
If $ v = A^{p+1}(w)$ for some $w in V$, then for some $w' in V$
$$v = A(A^p(w)) = A(A^{p+1}(w')) = A^{p+2}(w') in Im(A^{p+2}) $$
Shorter version
The above is instructive, but there's a one-liner to this, using the hint that states $Im(Acirc B) = A(Im(B))$:
$$ Im(A^{p+2}) = Im(A circ A^{p+1}) = A(Im(A^{p+1})) = A(Im(A^p)) = Im(A^{p+1}) $$
answered Jan 22 at 23:48
MetricMetric
1,23649
$endgroup$
add a comment |
$begingroup$
First direction:
$A^{p+2} subseteq A^{p+1}$:
If $v = A^{p+2}(w)$ for some $w in V$, then for some $w' in V$
$$ v = A(A^{p+1}(w)) = A(A^p(w')) = A^{p+1}(w') in Im(A^{p+1})$$
Other direction:
$A^{p+1} subseteq A^{p+2}$:
If $ v = A^{p+1}(w)$ for some $w in V$, then for some $w' in V$
$$v = A(A^p(w)) = A(A^{p+1}(w')) = A^{p+2}(w') in Im(A^{p+2}) $$
Shorter version
The above is instructive, but there's a one-liner to this, using the hint that states $Im(Acirc B) = A(Im(B))$:
$$ Im(A^{p+2}) = Im(A circ A^{p+1}) = A(Im(A^{p+1})) = A(Im(A^p)) = Im(A^{p+1}) $$
answered Jan 22 at 23:48
MetricMetric
1,23649
$endgroup$
First direction:
$A^{p+2} subseteq A^{p+1}$:
If $v = A^{p+2}(w)$ for some $w in V$, then for some $w' in V$
$$ v = A(A^{p+1}(w)) = A(A^p(w')) = A^{p+1}(w') in Im(A^{p+1})$$
Other direction:
$A^{p+1} subseteq A^{p+2}$:
If $ v = A^{p+1}(w)$ for some $w in V$, then for some $w' in V$
$$v = A(A^p(w)) = A(A^{p+1}(w')) = A^{p+2}(w') in Im(A^{p+2}) $$
Shorter version
The above is instructive, but there's a one-liner to this, using the hint that states $Im(Acirc B) = A(Im(B))$:
$$ Im(A^{p+2}) = Im(A circ A^{p+1}) = A(Im(A^{p+1})) = A(Im(A^p)) = Im(A^{p+1}) $$
answered Jan 22 at 23:48
MetricMetric
1,23649
answered Jan 22 at 23:48
MetricMetric
1,23649
answered Jan 22 at 23:48
MetricMetric
1,23649
answered Jan 22 at 23:48
MetricMetric
1,23649
1,23649
add a comment |
add a comment |
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$begingroup$
Hint: $mbox{Im}(AB)=A(mbox{Im}(B))$.
$endgroup$
– SmileyCraft
Jan 22 at 14:15