If a diagonalizable matrix is equal to its cube, then its rank is equal to the trace of its square
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Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
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add a comment |
$begingroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
$endgroup$
add a comment |
$begingroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
$endgroup$
Suppose $A in Bbb R ^{n times n}$ is such that $A^3=A$, and $A$ is diagonalizable. Prove that $operatorname{trace}(A^2) = operatorname{rank}(A)$.
I know that the possible eigenvalues of $A$ are $0,1,-1$. But how does that help to prove the required relation between trace and rank?
linear-algebra matrices eigenvalues-eigenvectors diagonalization
linear-algebra matrices eigenvalues-eigenvectors diagonalization
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
edited Oct 11 '15 at 4:55
user147263
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
asked Oct 6 '15 at 7:53
StabiloStabilo
734512
734512
add a comment |
add a comment |
1 Answer
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$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
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Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
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– Learnmore
Oct 6 '15 at 10:13
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@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
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– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
$endgroup$
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
$endgroup$
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
$endgroup$
$Tr(A^2)=sum_{i}lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
answered Oct 6 '15 at 7:58
Samrat MukhopadhyaySamrat Mukhopadhyay
13.7k2047
13.7k2047
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
Two questions 1.Where is the fact that $A$ is diagonalizable used?2.how is rank of A=no. of non-zero eigen values
$endgroup$
– Learnmore
Oct 6 '15 at 10:13
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
@learnmore A is similar to diagonal matrix D that its diagonal elements are A's eigenvalues, so tr(A)=tr(D) which is multiplication of diagonal elements = multiplication of eigenvalues. Also rank(A)=rank(D) = no. of non-zero eigenvalues.
$endgroup$
– Stabilo
Oct 6 '15 at 10:40
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
$begingroup$
$tr(D)$ =sum of diagonal elements ;anyway got your point
$endgroup$
– Learnmore
Oct 6 '15 at 10:44
add a comment |
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