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







share|cite|improve this question




















  • 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


















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







share|cite|improve this question




















  • 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
















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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 at 9:45









Rodrigo de Azevedo

12.8k41854




12.8k41854










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
















  • 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












1 Answer
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3














Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$






share|cite|improve this answer





















  • thanks @J. Doe this matrix didn't came in my mind
    – jasmine
    Dec 8 at 9:30











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1 Answer
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1 Answer
1






active

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active

oldest

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active

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3














Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$






share|cite|improve this answer





















  • thanks @J. Doe this matrix didn't came in my mind
    – jasmine
    Dec 8 at 9:30
















3














Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$






share|cite|improve this answer





















  • thanks @J. Doe this matrix didn't came in my mind
    – jasmine
    Dec 8 at 9:30














3












3








3






Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$






share|cite|improve this answer












Your first 2 lines are not independent thus the rank is 1 and not 2.
Look at
$$A=begin{bmatrix} 0 & 0 & 1 &0\0 & 0 & 0 & 0\0 &0 &0 &0 \0 &1 &0 &0 \end{bmatrix}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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