$A$ is a symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent

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Let $A$ be a real symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent.

I have tried using eigenvalues and only inferred that the eigen values may be $0,1$. But I cannot proceed with this.

linear-algebra matrices matrix-equations symmetric-matrices idempotents

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edited Jul 25 at 11:38

Rodrigo de Azevedo

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asked Jul 25 at 3:57

Legend Killer

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up vote
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down vote

favorite

2

Let $A$ be a real symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent.

I have tried using eigenvalues and only inferred that the eigen values may be $0,1$. But I cannot proceed with this.

linear-algebra matrices matrix-equations symmetric-matrices idempotents

share|cite|improve this question

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 3:57

Legend Killer

1,500523

add a comment | 

up vote
4
down vote

favorite

2

up vote
4
down vote

favorite

2

2

Let $A$ be a real symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent.

I have tried using eigenvalues and only inferred that the eigen values may be $0,1$. But I cannot proceed with this.

linear-algebra matrices matrix-equations symmetric-matrices idempotents

share|cite|improve this question

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 3:57

Legend Killer

1,500523

Let $A$ be a real symmetric matrix such that $A^4=A$. Prove that $A$ is idempotent.

I have tried using eigenvalues and only inferred that the eigen values may be $0,1$. But I cannot proceed with this.

linear-algebra matrices matrix-equations symmetric-matrices idempotents

share|cite|improve this question

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

asked Jul 25 at 3:57

Legend Killer

1,500523

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

edited Jul 25 at 11:38

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 25 at 3:57

Legend Killer

1,500523

asked Jul 25 at 3:57

Legend Killer

1,500523

asked Jul 25 at 3:57

Legend Killer

1,500523

1,500523

add a comment | 

add a comment | 

5 Answers
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up vote
4
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Let $A=PDP^-1$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^-1PDP^-1=PD^2P^-1=PDP^-1=A$$

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edited Jul 25 at 4:03

P Vanchinathan

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answered Jul 25 at 4:00

Parcly Taxel

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up vote
2
down vote

Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:

We have $A^4 - A = 0$. Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = left(A + frac 12 Iright)^Tleft(A + frac 12 Iright) + frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible). Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 implies A^2 - A = 0 implies A^2 = A
$$
Thus, $A$ is idempotent.

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answered Jul 25 at 11:45

Omnomnomnom

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up vote
0
down vote

If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
  – amd
  Jul 25 at 7:40
  

  
  
  
  

add a comment | 

up vote
0
down vote

I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:

if

$A^4 = A, tag 1$

then

$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; tag 2$

since

$A^T = A, tag 3$

we have for any vector $x$

$langle x, Ax rangle = langle x, A^4 x rangle = langle (A^T)^2 x, A^2 x rangle = langle A^2 x, A^2 x rangle ge 0; tag 4$

$langle x, A^2 x rangle = langle A^T x, Ax rangle = langle Ax, Ax rangle ge 0; tag 5$

thus

$langle x, (A^2 + A + I)x rangle = langle x, A^2 x rangle + langle x, Ax rangle + langle x, x rangle > 0, tag 6$

by (4), (5) and the fact that $langle x, x rangle > 0$; now if any matrix $B$ satisfies

$langle x, Bx rangle ne 0, forall x, tag 7$

then $B$ is invertible; if not,

$exists y ne 0, ; By = 0 Longrightarrow langle y, By rangle = 0, tag 8$

which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that

$A^2 - A = A(A - I) = 0, tag 9$

or

$A^2 = A. tag10$

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answered Jul 25 at 4:45

Robert Lewis

36.9k22155

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up vote
0
down vote

You are given that
$$
0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
lambda^2+lambda +1 = (lambda+1/2)^2+3/4 ge 3/4.
$$

So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.

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answered Jul 25 at 23:47

DisintegratingByParts

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Let $A=PDP^-1$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^-1PDP^-1=PD^2P^-1=PDP^-1=A$$

share|cite|improve this answer

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

add a comment | 

up vote
4
down vote

Let $A=PDP^-1$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^-1PDP^-1=PD^2P^-1=PDP^-1=A$$

share|cite|improve this answer

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

add a comment | 

up vote
4
down vote

up vote
4
down vote

Let $A=PDP^-1$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^-1PDP^-1=PD^2P^-1=PDP^-1=A$$

share|cite|improve this answer

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

Let $A=PDP^-1$ be the eigendecomposition of the given matrix. Then, since the diagonal matrix of eigenvalues $D$ contains only 0s and 1s, $D^2=D$ and
$$A^2=PDP^-1PDP^-1=PD^2P^-1=PDP^-1=A$$

share|cite|improve this answer

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

share|cite|improve this answer

share|cite|improve this answer

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

edited Jul 25 at 4:03

P Vanchinathan

13.9k12035

13.9k12035

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

answered Jul 25 at 4:00

Parcly Taxel

33.5k136588

33.5k136588

add a comment | 

add a comment | 

up vote
2
down vote

Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:

We have $A^4 - A = 0$. Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = left(A + frac 12 Iright)^Tleft(A + frac 12 Iright) + frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible). Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 implies A^2 - A = 0 implies A^2 = A
$$
Thus, $A$ is idempotent.

share|cite|improve this answer

answered Jul 25 at 11:45

Omnomnomnom

121k784170

add a comment | 

up vote
2
down vote

Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:

We have $A^4 - A = 0$. Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = left(A + frac 12 Iright)^Tleft(A + frac 12 Iright) + frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible). Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 implies A^2 - A = 0 implies A^2 = A
$$
Thus, $A$ is idempotent.

share|cite|improve this answer

answered Jul 25 at 11:45

Omnomnomnom

121k784170

add a comment | 

up vote
2
down vote

up vote
2
down vote

Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:

We have $A^4 - A = 0$. Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = left(A + frac 12 Iright)^Tleft(A + frac 12 Iright) + frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible). Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 implies A^2 - A = 0 implies A^2 = A
$$
Thus, $A$ is idempotent.

share|cite|improve this answer

answered Jul 25 at 11:45

Omnomnomnom

121k784170

Here's a proof which doesn't invoke Cayley Hamilton or eigenstructures:

We have $A^4 - A = 0$. Factoring this yields
$$
(A^2 + A + I)(A^2 - A) = 0
$$
However, we may write
$$
A^2 + A + I = left(A + frac 12 Iright)^Tleft(A + frac 12 Iright) + frac 34 I
$$
Since $A^2 + A + I$ can be written as the sum of a positive semidefinite and positive definite matrix, it must be positive definite (and hence invertible). Thus, we have
$$
(A^2 + A + I)(A^2 - A) = 0 implies A^2 - A = 0 implies A^2 = A
$$
Thus, $A$ is idempotent.

share|cite|improve this answer

answered Jul 25 at 11:45

Omnomnomnom

121k784170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 11:45

Omnomnomnom

121k784170

answered Jul 25 at 11:45

Omnomnomnom

121k784170

answered Jul 25 at 11:45

Omnomnomnom

121k784170

121k784170

add a comment | 

add a comment | 

up vote
0
down vote

If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
  – amd
  Jul 25 at 7:40
  

  
  
  
  

add a comment | 

up vote
0
down vote

If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
  – amd
  Jul 25 at 7:40
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

If A=0 or A=$I_n$ ok, you can use the Spectral theorem or eingvalues, $A^4=A$ implies that $p(x)=x^4-x$ is a multiple of Characteristic polynomial of A note that $p(x)=x(x-1)(x^2+x+1)$ every ortogonal operator have real eingvalue, thus the characteristic polynomial is x or x-1 or x(x-1), but by Cayley Theorem the characteristic polynomial(A)=0 thus x(x-1) again Cayley Theorem $A^2-A=0$.

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 4:28

Olecram

1364

answered Jul 25 at 4:28

Olecram

1364

answered Jul 25 at 4:28

Olecram

1364

1364

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
  – amd
  Jul 25 at 7:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
  – amd
  Jul 25 at 7:40
  

  
  
  
  

If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
– amd
Jul 25 at 7:40

If $A$ is of order 5 or larger, $x^4-x$ cannot be a multiple of its characteristic polynomial. It is, however, a multiple of the minimal polynomial.
– amd
Jul 25 at 7:40

add a comment | 

up vote
0
down vote

I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:

if

$A^4 = A, tag 1$

then

$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; tag 2$

since

$A^T = A, tag 3$

we have for any vector $x$

$langle x, Ax rangle = langle x, A^4 x rangle = langle (A^T)^2 x, A^2 x rangle = langle A^2 x, A^2 x rangle ge 0; tag 4$

$langle x, A^2 x rangle = langle A^T x, Ax rangle = langle Ax, Ax rangle ge 0; tag 5$

thus

$langle x, (A^2 + A + I)x rangle = langle x, A^2 x rangle + langle x, Ax rangle + langle x, x rangle > 0, tag 6$

by (4), (5) and the fact that $langle x, x rangle > 0$; now if any matrix $B$ satisfies

$langle x, Bx rangle ne 0, forall x, tag 7$

then $B$ is invertible; if not,

$exists y ne 0, ; By = 0 Longrightarrow langle y, By rangle = 0, tag 8$

which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that

$A^2 - A = A(A - I) = 0, tag 9$

or

$A^2 = A. tag10$

share|cite|improve this answer

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

add a comment | 

up vote
0
down vote

I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:

if

$A^4 = A, tag 1$

then

$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; tag 2$

since

$A^T = A, tag 3$

we have for any vector $x$

$langle x, Ax rangle = langle x, A^4 x rangle = langle (A^T)^2 x, A^2 x rangle = langle A^2 x, A^2 x rangle ge 0; tag 4$

$langle x, A^2 x rangle = langle A^T x, Ax rangle = langle Ax, Ax rangle ge 0; tag 5$

thus

$langle x, (A^2 + A + I)x rangle = langle x, A^2 x rangle + langle x, Ax rangle + langle x, x rangle > 0, tag 6$

by (4), (5) and the fact that $langle x, x rangle > 0$; now if any matrix $B$ satisfies

$langle x, Bx rangle ne 0, forall x, tag 7$

then $B$ is invertible; if not,

$exists y ne 0, ; By = 0 Longrightarrow langle y, By rangle = 0, tag 8$

which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that

$A^2 - A = A(A - I) = 0, tag 9$

or

$A^2 = A. tag10$

share|cite|improve this answer

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

add a comment | 

up vote
0
down vote

up vote
0
down vote

I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:

if

$A^4 = A, tag 1$

then

$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; tag 2$

since

$A^T = A, tag 3$

we have for any vector $x$

$langle x, Ax rangle = langle x, A^4 x rangle = langle (A^T)^2 x, A^2 x rangle = langle A^2 x, A^2 x rangle ge 0; tag 4$

$langle x, A^2 x rangle = langle A^T x, Ax rangle = langle Ax, Ax rangle ge 0; tag 5$

thus

$langle x, (A^2 + A + I)x rangle = langle x, A^2 x rangle + langle x, Ax rangle + langle x, x rangle > 0, tag 6$

by (4), (5) and the fact that $langle x, x rangle > 0$; now if any matrix $B$ satisfies

$langle x, Bx rangle ne 0, forall x, tag 7$

then $B$ is invertible; if not,

$exists y ne 0, ; By = 0 Longrightarrow langle y, By rangle = 0, tag 8$

which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that

$A^2 - A = A(A - I) = 0, tag 9$

or

$A^2 = A. tag10$

share|cite|improve this answer

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

I tried to cook up a basic proof which doesn't directly invoke the eigenstructure of $A$ or Cayley-Hamilton, etc.; here's what I've got:

if

$A^4 = A, tag 1$

then

$A(A - I)(A^2 + A + I) = A(A^3 - I) = A^4 - A = 0; tag 2$

since

$A^T = A, tag 3$

we have for any vector $x$

$langle x, Ax rangle = langle x, A^4 x rangle = langle (A^T)^2 x, A^2 x rangle = langle A^2 x, A^2 x rangle ge 0; tag 4$

$langle x, A^2 x rangle = langle A^T x, Ax rangle = langle Ax, Ax rangle ge 0; tag 5$

thus

$langle x, (A^2 + A + I)x rangle = langle x, A^2 x rangle + langle x, Ax rangle + langle x, x rangle > 0, tag 6$

by (4), (5) and the fact that $langle x, x rangle > 0$; now if any matrix $B$ satisfies

$langle x, Bx rangle ne 0, forall x, tag 7$

then $B$ is invertible; if not,

$exists y ne 0, ; By = 0 Longrightarrow langle y, By rangle = 0, tag 8$

which contradicts (7); applying this conclusion to $A^2 + A + I$ we see that it must be invertible; then by (2) we see that

$A^2 - A = A(A - I) = 0, tag 9$

or

$A^2 = A. tag10$

share|cite|improve this answer

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

answered Jul 25 at 4:45

Robert Lewis

36.9k22155

36.9k22155

add a comment | 

add a comment | 

up vote
0
down vote

You are given that
$$
0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
lambda^2+lambda +1 = (lambda+1/2)^2+3/4 ge 3/4.
$$

So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.

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answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

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0
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You are given that
$$
0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
lambda^2+lambda +1 = (lambda+1/2)^2+3/4 ge 3/4.
$$

So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.

share|cite|improve this answer

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

add a comment | 

up vote
0
down vote

up vote
0
down vote

You are given that
$$
0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
lambda^2+lambda +1 = (lambda+1/2)^2+3/4 ge 3/4.
$$

So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.

share|cite|improve this answer

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

You are given that
$$
0 = A^4-A=(A^2-A)(A^2+A+I)
$$
Because $A$ is assumed to be real and symmetric, then the matrix $A^2+A+I$ is real and symmetric, and its eigenvalues have the form
$$
lambda^2+lambda +1 = (lambda+1/2)^2+3/4 ge 3/4.
$$

So $A^2+A+I$ is invertible, which forces $A^2-A=0$, or $A^2=A$.

share|cite|improve this answer

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

answered Jul 25 at 23:47

DisintegratingByParts

55.6k42273

55.6k42273

add a comment | 

add a comment | 

 

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