Prove that if matrix $C = I - 2M $ and $M = M^2, $ then $ C^3 = C.$Matrix properties after multiplication.If...

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Prove that if matrix $C = I - 2M $ and $M = M^2, $ then $ C^3 = C.$


Matrix properties after multiplication.If AB is a projection then BA is a projectionProve that Det(A-E)=0 if and only if AC=CSum of identity and idempotent (projection) matrixThe Zero Matrixquestion about idempotent matrixAngle from rotation matrixMutual coherence for identity matrix and DFT matrixWhy if add a multiple of $I$ to a symmetric matrix then we have positive semidefinite matrixFinding the idempotent matrix













0












$begingroup$


I'm faced with a problem for which I haven't been given a correction, so I expected you could tell me if I am right or not, and in the later case give me the appropriate answer.



Let $M$ be an idempotent matrix such that $M=M^2$, and let $C = I - 2M$, with $I$ being the identity matrix.



Prove that $C^3 = C$



My answer:



$$C=I-2M$$
$$Rightarrow C^2=(I-2M)^2$$
$$=I^2-4IM+4M^2$$
$$=I^2-4M^2+4M^2$$
$$=I$$
$$Rightarrow C^3 = IC=C$$



Thank you.










share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
    $endgroup$
    – mfl
    Mar 11 at 20:10












  • $begingroup$
    I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
    $endgroup$
    – J. W. Tanner
    Mar 11 at 20:14










  • $begingroup$
    I changed the implication according to your correction. Thank you.
    $endgroup$
    – torito verdejo
    Mar 11 at 20:17






  • 2




    $begingroup$
    Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
    $endgroup$
    – Wolfgang Kais
    Mar 11 at 20:36
















0












$begingroup$


I'm faced with a problem for which I haven't been given a correction, so I expected you could tell me if I am right or not, and in the later case give me the appropriate answer.



Let $M$ be an idempotent matrix such that $M=M^2$, and let $C = I - 2M$, with $I$ being the identity matrix.



Prove that $C^3 = C$



My answer:



$$C=I-2M$$
$$Rightarrow C^2=(I-2M)^2$$
$$=I^2-4IM+4M^2$$
$$=I^2-4M^2+4M^2$$
$$=I$$
$$Rightarrow C^3 = IC=C$$



Thank you.










share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
    $endgroup$
    – mfl
    Mar 11 at 20:10












  • $begingroup$
    I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
    $endgroup$
    – J. W. Tanner
    Mar 11 at 20:14










  • $begingroup$
    I changed the implication according to your correction. Thank you.
    $endgroup$
    – torito verdejo
    Mar 11 at 20:17






  • 2




    $begingroup$
    Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
    $endgroup$
    – Wolfgang Kais
    Mar 11 at 20:36














0












0








0


1



$begingroup$


I'm faced with a problem for which I haven't been given a correction, so I expected you could tell me if I am right or not, and in the later case give me the appropriate answer.



Let $M$ be an idempotent matrix such that $M=M^2$, and let $C = I - 2M$, with $I$ being the identity matrix.



Prove that $C^3 = C$



My answer:



$$C=I-2M$$
$$Rightarrow C^2=(I-2M)^2$$
$$=I^2-4IM+4M^2$$
$$=I^2-4M^2+4M^2$$
$$=I$$
$$Rightarrow C^3 = IC=C$$



Thank you.










share|cite|improve this question











$endgroup$




I'm faced with a problem for which I haven't been given a correction, so I expected you could tell me if I am right or not, and in the later case give me the appropriate answer.



Let $M$ be an idempotent matrix such that $M=M^2$, and let $C = I - 2M$, with $I$ being the identity matrix.



Prove that $C^3 = C$



My answer:



$$C=I-2M$$
$$Rightarrow C^2=(I-2M)^2$$
$$=I^2-4IM+4M^2$$
$$=I^2-4M^2+4M^2$$
$$=I$$
$$Rightarrow C^3 = IC=C$$



Thank you.







linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 11 at 20:15









J. W. Tanner

3,3351320




3,3351320










asked Mar 11 at 20:05









torito verdejotorito verdejo

745




745








  • 6




    $begingroup$
    In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
    $endgroup$
    – mfl
    Mar 11 at 20:10












  • $begingroup$
    I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
    $endgroup$
    – J. W. Tanner
    Mar 11 at 20:14










  • $begingroup$
    I changed the implication according to your correction. Thank you.
    $endgroup$
    – torito verdejo
    Mar 11 at 20:17






  • 2




    $begingroup$
    Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
    $endgroup$
    – Wolfgang Kais
    Mar 11 at 20:36














  • 6




    $begingroup$
    In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
    $endgroup$
    – mfl
    Mar 11 at 20:10












  • $begingroup$
    I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
    $endgroup$
    – J. W. Tanner
    Mar 11 at 20:14










  • $begingroup$
    I changed the implication according to your correction. Thank you.
    $endgroup$
    – torito verdejo
    Mar 11 at 20:17






  • 2




    $begingroup$
    Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
    $endgroup$
    – Wolfgang Kais
    Mar 11 at 20:36








6




6




$begingroup$
In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
$endgroup$
– mfl
Mar 11 at 20:10






$begingroup$
In my opinion the idea of your proof is correct. However, you can't use the symbol $iff$. Note that $C=I-2M$ is not equivalent to $C^2=I$. If you use $implies$ it is correct. The same happens in $C^2=Iiff C^3=C.$
$endgroup$
– mfl
Mar 11 at 20:10














$begingroup$
I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
$endgroup$
– J. W. Tanner
Mar 11 at 20:14




$begingroup$
I agree with the above. Also, to be clear, it might be helpful to write $C^3=C^2C=IC=C$
$endgroup$
– J. W. Tanner
Mar 11 at 20:14












$begingroup$
I changed the implication according to your correction. Thank you.
$endgroup$
– torito verdejo
Mar 11 at 20:17




$begingroup$
I changed the implication according to your correction. Thank you.
$endgroup$
– torito verdejo
Mar 11 at 20:17




2




2




$begingroup$
Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
$endgroup$
– Wolfgang Kais
Mar 11 at 20:36




$begingroup$
Maybe it's worth noting that $(I-2M)^2 = I^2-2IM-2MI+4M^2$ and $IM=MI=M$.
$endgroup$
– Wolfgang Kais
Mar 11 at 20:36










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