Find a example of $A$ be $4 times 4$ matrix such that $A$ has rank $2$ but $A^2 =0 $?
0
Find a example of $A$ be $4 times 4$ matrix such that $A$ has rank $2$ but $A^2 =0 $?
My attempt :
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 1 & 0\0 &0 &0 &0 \0 &0 &0 &0 \end{bmatrix}$$
Is it correct ??
Any hints/solution will be apprecaited
thanks u
linear-algebra matrices matrix-rank nilpotence
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
|
show 1 more comment
0
Find a example of $A$ be $4 times 4$ matrix such that $A$ has rank $2$ but $A^2 =0 $?
My attempt :
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 1 & 0\0 &0 &0 &0 \0 &0 &0 &0 \end{bmatrix}$$
Is it correct ??
Any hints/solution will be apprecaited
thanks u
linear-algebra matrices matrix-rank nilpotence
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
3
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
1
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
1
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
1
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42
|
show 1 more comment
0
0
0
Find a example of $A$ be $4 times 4$ matrix such that $A$ has rank $2$ but $A^2 =0 $?
My attempt :
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 1 & 0\0 &0 &0 &0 \0 &0 &0 &0 \end{bmatrix}$$
Is it correct ??
Any hints/solution will be apprecaited
thanks u
linear-algebra matrices matrix-rank nilpotence
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
Find a example of $A$ be $4 times 4$ matrix such that $A$ has rank $2$ but $A^2 =0 $?
My attempt :
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 1 & 0\0 &0 &0 &0 \0 &0 &0 &0 \end{bmatrix}$$
Is it correct ??
Any hints/solution will be apprecaited
thanks u
linear-algebra matrices matrix-rank nilpotence
linear-algebra matrices matrix-rank nilpotence
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
edited Dec 8 at 9:45
Rodrigo de Azevedo
12.8k41854
12.8k41854
asked Dec 8 at 9:22
jasmine
1,520416
asked Dec 8 at 9:22
jasmine
1,520416
asked Dec 8 at 9:22
jasmine
1,520416
1,520416
3
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
1
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
1
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
1
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42
|
show 1 more comment
3
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
1
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
1
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
1
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42
3
3
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
1
1
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
1
1
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
1
1
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42
|
show 1 more comment
1 Answer
1
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3
Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$
answered Dec 8 at 9:29
J. Doe
936
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
add a comment |
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1 Answer
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1 Answer
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3
Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$
answered Dec 8 at 9:29
J. Doe
936
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
add a comment |
3
Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$
answered Dec 8 at 9:29
J. Doe
936
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
add a comment |
3
3
3
Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$
answered Dec 8 at 9:29
J. Doe
936
Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$
answered Dec 8 at 9:29
J. Doe
936
answered Dec 8 at 9:29
J. Doe
936
answered Dec 8 at 9:29
J. Doe
936
answered Dec 8 at 9:29
J. Doe
936
936
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
add a comment |
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
thanks @J. Doe this matrix didn't came in my mind
– jasmine
Dec 8 at 9:30
add a comment |
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3
Your example has rank $1$. You might be better off considering the Jordan canonical form.
– EuYu
Dec 8 at 9:28
1
Along the right lines, but looks like rank $1$ to me ...
– Mark Bennet
Dec 8 at 9:28
Ya ,that is my misunderstanding @ Mark and @EuYu
– jasmine
Dec 8 at 9:29
1
Your question is unclear. Do you mean to find an example (rather than a counterexample) of a 4-by-4 matrix $A$ such that $A$ has rank 2 but $A^2=0$?
– user1551
Dec 8 at 9:38
1
One way of thinking about the rank is as the dimension of the image space of a linear map. The image here will be a subspace $W$ of four dimensional space $V$ - because you want $A^2=0$ you end up applying $A$ to $W$ and getting zero. (I'm using terms loosely here). Rank $2$ means that the rank-nullity theorem tells you that $W$ has dimension $2$. Pick a convenient two dimensional subspace $W$ to go to zero, and then make sure you map everything in $V$ to $W$ and that both basis vectors in $W$ appear in the image of $V$. It is this last bit which didn't work for your first attempt.
– Mark Bennet
Dec 8 at 9:42