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
$begingroup$
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.
linear-algebra matrices
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra matrices
$endgroup$
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
add a comment |
$begingroup$
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.
linear-algebra matrices
$endgroup$
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
linear-algebra matrices
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
add a comment |
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
add a comment |
0
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$begingroup$
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