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
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
$endgroup$
add a comment |
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
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
$endgroup$
$begingroup$
See also this question. The arguments are the same.
$endgroup$
– Dietrich Burde
Dec 7 '18 at 19:39
add a comment |
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
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
$endgroup$
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
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
edited Dec 7 '18 at 20:00
RAM_3R
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
asked Dec 7 '18 at 19:26
RAM_3RRAM_3R
561214
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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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.
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
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
|
show 2 more comments
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1 Answer
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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.
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
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
|
show 2 more comments
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.
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
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
|
show 2 more comments
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.
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
$endgroup$
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.
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
answered Dec 7 '18 at 19:29
Richard MartinRichard Martin
1,61318
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
|
show 2 more comments
$begingroup$
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
|
show 2 more comments
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$begingroup$
See also this question. The arguments are the same.
$endgroup$
– Dietrich Burde
Dec 7 '18 at 19:39