rank and orthogonal projection [on hold]

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If $A$ is an orthogonal projection matrix onto a subspace $W$ of dimension $2$ in $mathbbR^4$,



how can I prove that
$$rm rank(A) = 2$$
$$rm rank(A^2) = 2$$
$$rm rank(A^2) = 1$$
$$rm rank(A^2) = 4$$




linear-algebra linear-transformations matrix-rank dimension-theory




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put on hold as off-topic by Lord Shark the Unknown, John Ma, Adrian Keister, Shailesh, Xander Henderson Aug 4 at 2:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Shailesh, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $A^2=A$ surely? What do you want the rank of $A^2$ to be?
    – Lord Shark the Unknown
    Aug 3 at 14:39










  • rank(A) is 2, 1 and 4?
    – Jonas
    Aug 3 at 14:39










  • @Jonas yes that is correct
    – KhanMan
    Aug 3 at 14:41










  • @LordSharktheUnknown where do you get A^2 = A?
    – KhanMan
    Aug 3 at 14:43






  • 3




    But how would rank(A^2) be three different values at the same time for a fixed A?
    – Jonas
    Aug 3 at 14:45














up vote
-1
down vote

favorite












If $A$ is an orthogonal projection matrix onto a subspace $W$ of dimension $2$ in $mathbbR^4$,



how can I prove that
$$rm rank(A) = 2$$
$$rm rank(A^2) = 2$$
$$rm rank(A^2) = 1$$
$$rm rank(A^2) = 4$$




linear-algebra linear-transformations matrix-rank dimension-theory




share|cite|improve this question













put on hold as off-topic by Lord Shark the Unknown, John Ma, Adrian Keister, Shailesh, Xander Henderson Aug 4 at 2:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Shailesh, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $A^2=A$ surely? What do you want the rank of $A^2$ to be?
    – Lord Shark the Unknown
    Aug 3 at 14:39










  • rank(A) is 2, 1 and 4?
    – Jonas
    Aug 3 at 14:39










  • @Jonas yes that is correct
    – KhanMan
    Aug 3 at 14:41










  • @LordSharktheUnknown where do you get A^2 = A?
    – KhanMan
    Aug 3 at 14:43






  • 3




    But how would rank(A^2) be three different values at the same time for a fixed A?
    – Jonas
    Aug 3 at 14:45












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











If $A$ is an orthogonal projection matrix onto a subspace $W$ of dimension $2$ in $mathbbR^4$,



how can I prove that
$$rm rank(A) = 2$$
$$rm rank(A^2) = 2$$
$$rm rank(A^2) = 1$$
$$rm rank(A^2) = 4$$




linear-algebra linear-transformations matrix-rank dimension-theory




share|cite|improve this question













If $A$ is an orthogonal projection matrix onto a subspace $W$ of dimension $2$ in $mathbbR^4$,



how can I prove that
$$rm rank(A) = 2$$
$$rm rank(A^2) = 2$$
$$rm rank(A^2) = 1$$
$$rm rank(A^2) = 4$$





linear-algebra linear-transformations matrix-rank dimension-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 14:40









Jonas

259210




259210









asked Aug 3 at 14:34









KhanMan

387




387




put on hold as off-topic by Lord Shark the Unknown, John Ma, Adrian Keister, Shailesh, Xander Henderson Aug 4 at 2:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Shailesh, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Lord Shark the Unknown, John Ma, Adrian Keister, Shailesh, Xander Henderson Aug 4 at 2:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Shailesh, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    $A^2=A$ surely? What do you want the rank of $A^2$ to be?
    – Lord Shark the Unknown
    Aug 3 at 14:39










  • rank(A) is 2, 1 and 4?
    – Jonas
    Aug 3 at 14:39










  • @Jonas yes that is correct
    – KhanMan
    Aug 3 at 14:41










  • @LordSharktheUnknown where do you get A^2 = A?
    – KhanMan
    Aug 3 at 14:43






  • 3




    But how would rank(A^2) be three different values at the same time for a fixed A?
    – Jonas
    Aug 3 at 14:45












  • 1




    $A^2=A$ surely? What do you want the rank of $A^2$ to be?
    – Lord Shark the Unknown
    Aug 3 at 14:39










  • rank(A) is 2, 1 and 4?
    – Jonas
    Aug 3 at 14:39










  • @Jonas yes that is correct
    – KhanMan
    Aug 3 at 14:41










  • @LordSharktheUnknown where do you get A^2 = A?
    – KhanMan
    Aug 3 at 14:43






  • 3




    But how would rank(A^2) be three different values at the same time for a fixed A?
    – Jonas
    Aug 3 at 14:45







1




1




$A^2=A$ surely? What do you want the rank of $A^2$ to be?
– Lord Shark the Unknown
Aug 3 at 14:39




$A^2=A$ surely? What do you want the rank of $A^2$ to be?
– Lord Shark the Unknown
Aug 3 at 14:39












rank(A) is 2, 1 and 4?
– Jonas
Aug 3 at 14:39




rank(A) is 2, 1 and 4?
– Jonas
Aug 3 at 14:39












@Jonas yes that is correct
– KhanMan
Aug 3 at 14:41




@Jonas yes that is correct
– KhanMan
Aug 3 at 14:41












@LordSharktheUnknown where do you get A^2 = A?
– KhanMan
Aug 3 at 14:43




@LordSharktheUnknown where do you get A^2 = A?
– KhanMan
Aug 3 at 14:43




3




3




But how would rank(A^2) be three different values at the same time for a fixed A?
– Jonas
Aug 3 at 14:45




But how would rank(A^2) be three different values at the same time for a fixed A?
– Jonas
Aug 3 at 14:45










1 Answer
1






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oldest

votes

















up vote
0
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accepted










Let $x$ be any vector in $mathbbR^4$. Then $Ax in W$ by definition of projection, so since $W$ is two-dimensional, it follows that $textrank(A) = 2$. Also, by the definition of projection, $A^2 = A$. So for the same reasons as before, $textrank(A^2) = 2$. Orthogonality of $A$ is not needed.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    Let $x$ be any vector in $mathbbR^4$. Then $Ax in W$ by definition of projection, so since $W$ is two-dimensional, it follows that $textrank(A) = 2$. Also, by the definition of projection, $A^2 = A$. So for the same reasons as before, $textrank(A^2) = 2$. Orthogonality of $A$ is not needed.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      Let $x$ be any vector in $mathbbR^4$. Then $Ax in W$ by definition of projection, so since $W$ is two-dimensional, it follows that $textrank(A) = 2$. Also, by the definition of projection, $A^2 = A$. So for the same reasons as before, $textrank(A^2) = 2$. Orthogonality of $A$ is not needed.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        Let $x$ be any vector in $mathbbR^4$. Then $Ax in W$ by definition of projection, so since $W$ is two-dimensional, it follows that $textrank(A) = 2$. Also, by the definition of projection, $A^2 = A$. So for the same reasons as before, $textrank(A^2) = 2$. Orthogonality of $A$ is not needed.






        share|cite|improve this answer













        Let $x$ be any vector in $mathbbR^4$. Then $Ax in W$ by definition of projection, so since $W$ is two-dimensional, it follows that $textrank(A) = 2$. Also, by the definition of projection, $A^2 = A$. So for the same reasons as before, $textrank(A^2) = 2$. Orthogonality of $A$ is not needed.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 3 at 15:43









        amarney

        978215




        978215












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