Prove that $text{rank } T = operatorname{rank} T^2 iff operatorname{Im}T cap ker T = { vec 0}$
$begingroup$
$newcommand{r}{ operatorname{rank} } $
Let $T: Vto V$ be a linear transformation with $dim V< infty$. Prove that:
$$ r T = r T^2 iff operatorname{Im} T cap ker T = { vec 0 }.$$
$"Rightarrow"$ Let $r T = r T^2$. Then, by rank - nullity theorem we have that $$dim ker T =dim ker T^2 tag 1.$$ But it is always true that:
$ker T subseteq ker T^2 .tag 2$
By $(1),(2)$ we have that $ker T = ker T^2.$ So, instead of $r T = r T^2$ we can say that $ker T = ker T^2$ and we need to prove that $operatorname{Im} T cap ker T = { vec 0}$.
Proof:
Suppose that there is a $z in operatorname{Im}T cap ker T$ with $z neq 0$. Since $z in ker T implies T(z) = 0$. Also, since $z in operatorname{Im}T implies exists yin V$ such that $T(y) = z implies T^2(y) = T(z) = 0.$ But this implies that $y in ker T^2 $ and by our hypothesis we have that $y in ker T implies T(y) = 0 = z, $ which is absurd, because we assumed that $z neq 0$.
$"Leftarrow"$
We need to prove that $ker T = ker T^2$ or $ker T^2 subseteq ker T.$
Proof:
Let $x in ker T^2$, which implies $T^2(x) = Tleft(T(x)right) = 0$. It is implied $T(x) in ker T,$ but also $T(x) in operatorname{Im}T.$ Thus, $T(x) in operatorname{Im}T cap ker T = {0}$. Thus, $T(x) = 0 implies x in ker T.$
I would like to know if my reasoning is correct and if all the points are clear. Also, I would like to know if there is any shorter proof.
linear-algebra proof-verification linear-transformations
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
$endgroup$
add a comment |
$begingroup$
$newcommand{r}{ operatorname{rank} } $
Let $T: Vto V$ be a linear transformation with $dim V< infty$. Prove that:
$$ r T = r T^2 iff operatorname{Im} T cap ker T = { vec 0 }.$$
$"Rightarrow"$ Let $r T = r T^2$. Then, by rank - nullity theorem we have that $$dim ker T =dim ker T^2 tag 1.$$ But it is always true that:
$ker T subseteq ker T^2 .tag 2$
By $(1),(2)$ we have that $ker T = ker T^2.$ So, instead of $r T = r T^2$ we can say that $ker T = ker T^2$ and we need to prove that $operatorname{Im} T cap ker T = { vec 0}$.
Proof:
Suppose that there is a $z in operatorname{Im}T cap ker T$ with $z neq 0$. Since $z in ker T implies T(z) = 0$. Also, since $z in operatorname{Im}T implies exists yin V$ such that $T(y) = z implies T^2(y) = T(z) = 0.$ But this implies that $y in ker T^2 $ and by our hypothesis we have that $y in ker T implies T(y) = 0 = z, $ which is absurd, because we assumed that $z neq 0$.
$"Leftarrow"$
We need to prove that $ker T = ker T^2$ or $ker T^2 subseteq ker T.$
Proof:
Let $x in ker T^2$, which implies $T^2(x) = Tleft(T(x)right) = 0$. It is implied $T(x) in ker T,$ but also $T(x) in operatorname{Im}T.$ Thus, $T(x) in operatorname{Im}T cap ker T = {0}$. Thus, $T(x) = 0 implies x in ker T.$
I would like to know if my reasoning is correct and if all the points are clear. Also, I would like to know if there is any shorter proof.
linear-algebra proof-verification linear-transformations
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
$endgroup$
1
$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11
add a comment |
$begingroup$
$newcommand{r}{ operatorname{rank} } $
Let $T: Vto V$ be a linear transformation with $dim V< infty$. Prove that:
$$ r T = r T^2 iff operatorname{Im} T cap ker T = { vec 0 }.$$
$"Rightarrow"$ Let $r T = r T^2$. Then, by rank - nullity theorem we have that $$dim ker T =dim ker T^2 tag 1.$$ But it is always true that:
$ker T subseteq ker T^2 .tag 2$
By $(1),(2)$ we have that $ker T = ker T^2.$ So, instead of $r T = r T^2$ we can say that $ker T = ker T^2$ and we need to prove that $operatorname{Im} T cap ker T = { vec 0}$.
Proof:
Suppose that there is a $z in operatorname{Im}T cap ker T$ with $z neq 0$. Since $z in ker T implies T(z) = 0$. Also, since $z in operatorname{Im}T implies exists yin V$ such that $T(y) = z implies T^2(y) = T(z) = 0.$ But this implies that $y in ker T^2 $ and by our hypothesis we have that $y in ker T implies T(y) = 0 = z, $ which is absurd, because we assumed that $z neq 0$.
$"Leftarrow"$
We need to prove that $ker T = ker T^2$ or $ker T^2 subseteq ker T.$
Proof:
Let $x in ker T^2$, which implies $T^2(x) = Tleft(T(x)right) = 0$. It is implied $T(x) in ker T,$ but also $T(x) in operatorname{Im}T.$ Thus, $T(x) in operatorname{Im}T cap ker T = {0}$. Thus, $T(x) = 0 implies x in ker T.$
I would like to know if my reasoning is correct and if all the points are clear. Also, I would like to know if there is any shorter proof.
linear-algebra proof-verification linear-transformations
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
$endgroup$
$newcommand{r}{ operatorname{rank} } $
Let $T: Vto V$ be a linear transformation with $dim V< infty$. Prove that:
$$ r T = r T^2 iff operatorname{Im} T cap ker T = { vec 0 }.$$
$"Rightarrow"$ Let $r T = r T^2$. Then, by rank - nullity theorem we have that $$dim ker T =dim ker T^2 tag 1.$$ But it is always true that:
$ker T subseteq ker T^2 .tag 2$
By $(1),(2)$ we have that $ker T = ker T^2.$ So, instead of $r T = r T^2$ we can say that $ker T = ker T^2$ and we need to prove that $operatorname{Im} T cap ker T = { vec 0}$.
Proof:
Suppose that there is a $z in operatorname{Im}T cap ker T$ with $z neq 0$. Since $z in ker T implies T(z) = 0$. Also, since $z in operatorname{Im}T implies exists yin V$ such that $T(y) = z implies T^2(y) = T(z) = 0.$ But this implies that $y in ker T^2 $ and by our hypothesis we have that $y in ker T implies T(y) = 0 = z, $ which is absurd, because we assumed that $z neq 0$.
$"Leftarrow"$
We need to prove that $ker T = ker T^2$ or $ker T^2 subseteq ker T.$
Proof:
Let $x in ker T^2$, which implies $T^2(x) = Tleft(T(x)right) = 0$. It is implied $T(x) in ker T,$ but also $T(x) in operatorname{Im}T.$ Thus, $T(x) in operatorname{Im}T cap ker T = {0}$. Thus, $T(x) = 0 implies x in ker T.$
I would like to know if my reasoning is correct and if all the points are clear. Also, I would like to know if there is any shorter proof.
linear-algebra proof-verification linear-transformations
linear-algebra proof-verification linear-transformations
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
edited Jan 25 at 11:36
thanasissdr
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
asked May 26 '16 at 23:53
thanasissdrthanasissdr
5,56111325
5,56111325
1
$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11
add a comment |
1
$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11
1
1
$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11
add a comment |
1 Answer
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$begingroup$
Let $ W = textrm{Im}, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ ker T_W = W cap ker T $.
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
$endgroup$
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
add a comment |
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$begingroup$
Let $ W = textrm{Im}, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ ker T_W = W cap ker T $.
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
$endgroup$
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
add a comment |
$begingroup$
Let $ W = textrm{Im}, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ ker T_W = W cap ker T $.
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
$endgroup$
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
add a comment |
$begingroup$
Let $ W = textrm{Im}, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ ker T_W = W cap ker T $.
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
$endgroup$
Let $ W = textrm{Im}, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ ker T_W = W cap ker T $.
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
edited May 27 '16 at 0:18
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
answered May 27 '16 at 0:12
StarfallStarfall
13.3k11139
13.3k11139
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
add a comment |
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
$begingroup$
Yes, didn't see that!
$endgroup$
– thanasissdr
May 27 '16 at 0:46
add a comment |
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$begingroup$
Your proof is very good. I can't see a faster way, at least.
$endgroup$
– Ivo Terek
May 27 '16 at 0:10
$begingroup$
You wrote textrm{rank }, with a blank space after the "k". If you write operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write Aoperatorname{rank}B you see $Aoperatorname{rank}B$ and if you write Aoperatorname{rank}(B) you see $Aoperatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like sin and max and log and det, etc. That's the right way to do that. $qquad$
$endgroup$
– Michael Hardy
May 27 '16 at 0:11