Prove that $r(A)=operatorname{tr}(A^2)$












0












$begingroup$


Let $Ain M_n(mathbb{C})$. Show that if $A^3=A$, then $r(A)=operatorname{tr}(A^2)$.




Since $A^3=A$, the possible eigenvalues are $0,1,-1$. I don't know from here how to compute the rank of $A$.



Edited



Since eigenvalues of $A$ are $0,1,-1$ SO eigenvalues of $A^2=0,1$ So $r(A^2)=tr(A^2)$ Now we have to show $r(A^2)=r(A)$ where rank of $A^2$ is the number of non-zero eigenvalues .






linear-algebra matrix-rank trace







share|cite|improve this question











$endgroup$












  • $begingroup$
    See also this question. The arguments are the same.
    $endgroup$
    – Dietrich Burde
    Dec 7 '18 at 19:39
















0












$begingroup$


Let $Ain M_n(mathbb{C})$. Show that if $A^3=A$, then $r(A)=operatorname{tr}(A^2)$.




Since $A^3=A$, the possible eigenvalues are $0,1,-1$. I don't know from here how to compute the rank of $A$.



Edited



Since eigenvalues of $A$ are $0,1,-1$ SO eigenvalues of $A^2=0,1$ So $r(A^2)=tr(A^2)$ Now we have to show $r(A^2)=r(A)$ where rank of $A^2$ is the number of non-zero eigenvalues .






linear-algebra matrix-rank trace







share|cite|improve this question











$endgroup$












  • $begingroup$
    See also this question. The arguments are the same.
    $endgroup$
    – Dietrich Burde
    Dec 7 '18 at 19:39














0












0








0


1



$begingroup$


Let $Ain M_n(mathbb{C})$. Show that if $A^3=A$, then $r(A)=operatorname{tr}(A^2)$.




Since $A^3=A$, the possible eigenvalues are $0,1,-1$. I don't know from here how to compute the rank of $A$.



Edited



Since eigenvalues of $A$ are $0,1,-1$ SO eigenvalues of $A^2=0,1$ So $r(A^2)=tr(A^2)$ Now we have to show $r(A^2)=r(A)$ where rank of $A^2$ is the number of non-zero eigenvalues .






linear-algebra matrix-rank trace







share|cite|improve this question











$endgroup$




Let $Ain M_n(mathbb{C})$. Show that if $A^3=A$, then $r(A)=operatorname{tr}(A^2)$.




Since $A^3=A$, the possible eigenvalues are $0,1,-1$. I don't know from here how to compute the rank of $A$.



Edited



Since eigenvalues of $A$ are $0,1,-1$ SO eigenvalues of $A^2=0,1$ So $r(A^2)=tr(A^2)$ Now we have to show $r(A^2)=r(A)$ where rank of $A^2$ is the number of non-zero eigenvalues .






linear-algebra matrix-rank trace




linear-algebra matrix-rank trace






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 20:00







RAM_3R

















asked Dec 7 '18 at 19:26









RAM_3RRAM_3R

561214




561214












  • $begingroup$
    See also this question. The arguments are the same.
    $endgroup$
    – Dietrich Burde
    Dec 7 '18 at 19:39


















  • $begingroup$
    See also this question. The arguments are the same.
    $endgroup$
    – Dietrich Burde
    Dec 7 '18 at 19:39
















$begingroup$
See also this question. The arguments are the same.
$endgroup$
– Dietrich Burde
Dec 7 '18 at 19:39




$begingroup$
See also this question. The arguments are the same.
$endgroup$
– Dietrich Burde
Dec 7 '18 at 19:39










1 Answer
1






active

oldest

votes


















3












$begingroup$

The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but why both equal ??
    $endgroup$
    – RAM_3R
    Dec 7 '18 at 19:34










  • $begingroup$
    @RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
    $endgroup$
    – Zach Langley
    Dec 7 '18 at 19:38










  • $begingroup$
    @RAM_3R Because the trace is the sum of the eigenvalues
    $endgroup$
    – Alan
    Dec 7 '18 at 19:39






  • 1




    $begingroup$
    This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
    $endgroup$
    – user1551
    Dec 7 '18 at 19:44










  • $begingroup$
    It's the number of non-zero eigenvalues counted with multiplicity.
    $endgroup$
    – Bernard
    Dec 7 '18 at 19:46











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but why both equal ??
    $endgroup$
    – RAM_3R
    Dec 7 '18 at 19:34










  • $begingroup$
    @RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
    $endgroup$
    – Zach Langley
    Dec 7 '18 at 19:38










  • $begingroup$
    @RAM_3R Because the trace is the sum of the eigenvalues
    $endgroup$
    – Alan
    Dec 7 '18 at 19:39






  • 1




    $begingroup$
    This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
    $endgroup$
    – user1551
    Dec 7 '18 at 19:44










  • $begingroup$
    It's the number of non-zero eigenvalues counted with multiplicity.
    $endgroup$
    – Bernard
    Dec 7 '18 at 19:46
















3












$begingroup$

The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    but why both equal ??
    $endgroup$
    – RAM_3R
    Dec 7 '18 at 19:34










  • $begingroup$
    @RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
    $endgroup$
    – Zach Langley
    Dec 7 '18 at 19:38










  • $begingroup$
    @RAM_3R Because the trace is the sum of the eigenvalues
    $endgroup$
    – Alan
    Dec 7 '18 at 19:39






  • 1




    $begingroup$
    This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
    $endgroup$
    – user1551
    Dec 7 '18 at 19:44










  • $begingroup$
    It's the number of non-zero eigenvalues counted with multiplicity.
    $endgroup$
    – Bernard
    Dec 7 '18 at 19:46














3












3








3





$begingroup$

The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.






share|cite|improve this answer









$endgroup$



The rank is the number of nonzero eigenvalues and as their squares are 0 or 1, this number is just the same as the sum of the squares.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 19:29









Richard MartinRichard Martin

1,61318




1,61318












  • $begingroup$
    but why both equal ??
    $endgroup$
    – RAM_3R
    Dec 7 '18 at 19:34










  • $begingroup$
    @RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
    $endgroup$
    – Zach Langley
    Dec 7 '18 at 19:38










  • $begingroup$
    @RAM_3R Because the trace is the sum of the eigenvalues
    $endgroup$
    – Alan
    Dec 7 '18 at 19:39






  • 1




    $begingroup$
    This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
    $endgroup$
    – user1551
    Dec 7 '18 at 19:44










  • $begingroup$
    It's the number of non-zero eigenvalues counted with multiplicity.
    $endgroup$
    – Bernard
    Dec 7 '18 at 19:46


















  • $begingroup$
    but why both equal ??
    $endgroup$
    – RAM_3R
    Dec 7 '18 at 19:34










  • $begingroup$
    @RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
    $endgroup$
    – Zach Langley
    Dec 7 '18 at 19:38










  • $begingroup$
    @RAM_3R Because the trace is the sum of the eigenvalues
    $endgroup$
    – Alan
    Dec 7 '18 at 19:39






  • 1




    $begingroup$
    This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
    $endgroup$
    – user1551
    Dec 7 '18 at 19:44










  • $begingroup$
    It's the number of non-zero eigenvalues counted with multiplicity.
    $endgroup$
    – Bernard
    Dec 7 '18 at 19:46
















$begingroup$
but why both equal ??
$endgroup$
– RAM_3R
Dec 7 '18 at 19:34




$begingroup$
but why both equal ??
$endgroup$
– RAM_3R
Dec 7 '18 at 19:34












$begingroup$
@RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
$endgroup$
– Zach Langley
Dec 7 '18 at 19:38




$begingroup$
@RAM_3R Since every non-zero eigenvalue of $A^2$ is 1, the trace counts exactly the number of non-zero eigenvalues, which is the same as the rank.
$endgroup$
– Zach Langley
Dec 7 '18 at 19:38












$begingroup$
@RAM_3R Because the trace is the sum of the eigenvalues
$endgroup$
– Alan
Dec 7 '18 at 19:39




$begingroup$
@RAM_3R Because the trace is the sum of the eigenvalues
$endgroup$
– Alan
Dec 7 '18 at 19:39




1




1




$begingroup$
This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
$endgroup$
– user1551
Dec 7 '18 at 19:44




$begingroup$
This is not correct. The rank of a matrix in general is not equal to the number of nonzero eigenvalues (consider e.g. any nontrivial nilpotent block). You need to show that $A$ is diagonalisable.
$endgroup$
– user1551
Dec 7 '18 at 19:44












$begingroup$
It's the number of non-zero eigenvalues counted with multiplicity.
$endgroup$
– Bernard
Dec 7 '18 at 19:46




$begingroup$
It's the number of non-zero eigenvalues counted with multiplicity.
$endgroup$
– Bernard
Dec 7 '18 at 19:46


















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