Rank of a linear operator.
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Let $V$ be a vector space of dimension 3 and $T $ be a linear operator on $V $ such that $T^3=0 $ and $T^2 neq 0$. The question is to show the rank of $T=2$.
My attempt : As I had no other idea, I tried using proof by contradiction. I succeeded in showing that a rank of 0 or 3 will lead to a contradiction.
Now suppose rank($T$) = 1. Let ${v_1,v_2,v_3} $ be a basis of $V $ and WLOG assume ${v_1,v_2}$ spans the kernel of $T$ and $Tv_3$ spans the range of $T$. Then I can show that Range($T^2$) is spanned by $T^2(v_3)$ and rank($T^2$)= 1. I'm completely stuck after this. I'm not even sure this result has any importance. Any hint is appreciated. Thanks.
linear-algebra linear-transformations
asked Nov 18 at 14:29
Thomas Shelby
1,063116
add a comment |
up vote
0
down vote
favorite
Let $V$ be a vector space of dimension 3 and $T $ be a linear operator on $V $ such that $T^3=0 $ and $T^2 neq 0$. The question is to show the rank of $T=2$.
My attempt : As I had no other idea, I tried using proof by contradiction. I succeeded in showing that a rank of 0 or 3 will lead to a contradiction.
Now suppose rank($T$) = 1. Let ${v_1,v_2,v_3} $ be a basis of $V $ and WLOG assume ${v_1,v_2}$ spans the kernel of $T$ and $Tv_3$ spans the range of $T$. Then I can show that Range($T^2$) is spanned by $T^2(v_3)$ and rank($T^2$)= 1. I'm completely stuck after this. I'm not even sure this result has any importance. Any hint is appreciated. Thanks.
linear-algebra linear-transformations
asked Nov 18 at 14:29
Thomas Shelby
1,063116
Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
1
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $V$ be a vector space of dimension 3 and $T $ be a linear operator on $V $ such that $T^3=0 $ and $T^2 neq 0$. The question is to show the rank of $T=2$.
My attempt : As I had no other idea, I tried using proof by contradiction. I succeeded in showing that a rank of 0 or 3 will lead to a contradiction.
Now suppose rank($T$) = 1. Let ${v_1,v_2,v_3} $ be a basis of $V $ and WLOG assume ${v_1,v_2}$ spans the kernel of $T$ and $Tv_3$ spans the range of $T$. Then I can show that Range($T^2$) is spanned by $T^2(v_3)$ and rank($T^2$)= 1. I'm completely stuck after this. I'm not even sure this result has any importance. Any hint is appreciated. Thanks.
linear-algebra linear-transformations
asked Nov 18 at 14:29
Thomas Shelby
1,063116
Let $V$ be a vector space of dimension 3 and $T $ be a linear operator on $V $ such that $T^3=0 $ and $T^2 neq 0$. The question is to show the rank of $T=2$.
My attempt : As I had no other idea, I tried using proof by contradiction. I succeeded in showing that a rank of 0 or 3 will lead to a contradiction.
Now suppose rank($T$) = 1. Let ${v_1,v_2,v_3} $ be a basis of $V $ and WLOG assume ${v_1,v_2}$ spans the kernel of $T$ and $Tv_3$ spans the range of $T$. Then I can show that Range($T^2$) is spanned by $T^2(v_3)$ and rank($T^2$)= 1. I'm completely stuck after this. I'm not even sure this result has any importance. Any hint is appreciated. Thanks.
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Nov 18 at 14:29
Thomas Shelby
1,063116
asked Nov 18 at 14:29
Thomas Shelby
1,063116
edited Nov 18 at 15:38
asked Nov 18 at 14:29
Thomas Shelby
1,063116
asked Nov 18 at 14:29
Thomas Shelby
1,063116
asked Nov 18 at 14:29
Thomas Shelby
1,063116
1,063116
Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
1
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00
add a comment |
Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
1
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00
Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
1
1
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00
add a comment |
1 Answer
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1
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Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$. Since $T^2not=0$, we know that $operatorname{rank}(T^2)geq 1$. Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $operatorname{rank}(T^2)leqoperatorname{rank}(T)$. Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2not=0$, there is some vector $v$ so that $T^2(v)not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$. This implies that the kernel of $T$ is at least $1$-dimensional. This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that). If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$. Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image. But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero. But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$. To prove that this is even possible, an example would be
$$
begin{bmatrix}0&1&0\0&0&1\0&0&0end{bmatrix}.
$$
answered Nov 18 at 17:02
Michael Burr
26.4k23262
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$. Since $T^2not=0$, we know that $operatorname{rank}(T^2)geq 1$. Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $operatorname{rank}(T^2)leqoperatorname{rank}(T)$. Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2not=0$, there is some vector $v$ so that $T^2(v)not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$. This implies that the kernel of $T$ is at least $1$-dimensional. This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that). If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$. Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image. But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero. But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$. To prove that this is even possible, an example would be
$$
begin{bmatrix}0&1&0\0&0&1\0&0&0end{bmatrix}.
$$
answered Nov 18 at 17:02
Michael Burr
26.4k23262
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
add a comment |
up vote
1
down vote
accepted
Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$. Since $T^2not=0$, we know that $operatorname{rank}(T^2)geq 1$. Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $operatorname{rank}(T^2)leqoperatorname{rank}(T)$. Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2not=0$, there is some vector $v$ so that $T^2(v)not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$. This implies that the kernel of $T$ is at least $1$-dimensional. This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that). If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$. Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image. But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero. But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$. To prove that this is even possible, an example would be
$$
begin{bmatrix}0&1&0\0&0&1\0&0&0end{bmatrix}.
$$
answered Nov 18 at 17:02
Michael Burr
26.4k23262
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$. Since $T^2not=0$, we know that $operatorname{rank}(T^2)geq 1$. Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $operatorname{rank}(T^2)leqoperatorname{rank}(T)$. Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2not=0$, there is some vector $v$ so that $T^2(v)not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$. This implies that the kernel of $T$ is at least $1$-dimensional. This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that). If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$. Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image. But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero. But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$. To prove that this is even possible, an example would be
$$
begin{bmatrix}0&1&0\0&0&1\0&0&0end{bmatrix}.
$$
answered Nov 18 at 17:02
Michael Burr
26.4k23262
Since we know that $T^3=0$, we know that the null space of $T^3$ contains the image of $T^2$. Since $T^2not=0$, we know that $operatorname{rank}(T^2)geq 1$. Since $T^2(v)=T(T(v))$, i.e., $T$ applied to some vector $T(v)$, we know that $operatorname{rank}(T^2)leqoperatorname{rank}(T)$. Therefore, the rank of $T$ is at least one (but you already knew that).
On the other hand, since $T^2not=0$, there is some vector $v$ so that $T^2(v)not=0$, but since $T^3(v)=T(T^2(v))=0$, it follows that there is some nonzero vector, i.e., $T^2(v)$ which is mapped to zero by $T$. This implies that the kernel of $T$ is at least $1$-dimensional. This implies, by the rank-nullity theorem that the rank of $T$ is at most $2$ (but you already knew that). If you don't know the rank-nullity theorem, keep a count of free and pivot variables and you'll get the same conclusions.
Now, suppose that the rank of $T$ is $1$. Then, since $T^2$ has rank $1$ as well, this implies that $T^2$ and $T$ have the same image. But, since $T$ applied to the image of $T^2$ is zero, then $T$ applied to the image of $T$ is also zero. But then, $T^2$ is zero, but this contradicts the assumptions.
This shows that the rank of $T$ must be $2$. To prove that this is even possible, an example would be
$$
begin{bmatrix}0&1&0\0&0&1\0&0&0end{bmatrix}.
$$
answered Nov 18 at 17:02
Michael Burr
26.4k23262
answered Nov 18 at 17:02
Michael Burr
26.4k23262
answered Nov 18 at 17:02
Michael Burr
26.4k23262
answered Nov 18 at 17:02
Michael Burr
26.4k23262
26.4k23262
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
add a comment |
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
Thank you for your wonderful answer. But in the first paragraph, how did you conclude rank($T^2$) $leq$ rank($T$) ?
– Thomas Shelby
Nov 18 at 17:11
1
1
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
The rank is the dimension of the image. The image of $T^2$ is a subset of the image of $T$. This implies that the rank of $T^2$ is less than the rank of $T$.
– Michael Burr
Nov 18 at 18:49
add a comment |
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Do you know the Jordan form? If so, what blocks are possible?
– Michael Burr
Nov 18 at 14:37
I haven't learnt about it. Sorry.
– Thomas Shelby
Nov 18 at 15:06
1
In your basis $v_i$, the matrix for $T$ is all zeros except a $1$ in the bottom-right corner. This matrix has the property that $T^2=T$, so you contradict that $T^3=0$, since we would also have $T^3=T^2=T$.
– Nick
Nov 18 at 15:52
@Nick Thank you. I didn't consider the matrix at all.
– Thomas Shelby
Nov 18 at 17:00