If $A^2 = I$ (Identity Matrix) then $A = pm I$
Clash Royale CLAN TAG#URR8PPP
up vote
14
down vote
favorite
So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this:
If $A^2 = I$ (Identity Matrix) Then $A = pm I$ ?
I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks.
linear-algebra matrices examples-counterexamples
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
add a comment |Â
up vote
14
down vote
favorite
So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this:
If $A^2 = I$ (Identity Matrix) Then $A = pm I$ ?
I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks.
linear-algebra matrices examples-counterexamples
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
23
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
1
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43
add a comment |Â
up vote
14
down vote
favorite
up vote
14
down vote
favorite
So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this:
If $A^2 = I$ (Identity Matrix) Then $A = pm I$ ?
I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks.
linear-algebra matrices examples-counterexamples
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the question is this:
If $A^2 = I$ (Identity Matrix) Then $A = pm I$ ?
I'm pretty sure it is true but the answer say it's false. How can this be false (maybe its a typography error on the book)?. Thanks.
linear-algebra matrices examples-counterexamples
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
edited Mar 16 '16 at 5:55
Martin Sleziak
43.5k6113260
43.5k6113260
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3751519
3751519
23
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
1
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43
add a comment |Â
23
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
1
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43
23
23
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
1
1
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
33
down vote
accepted
A simple counterexample is $$A = beginbmatrix 1 & 0 \ 0 & -1 endbmatrix $$ We have $A neq pm I$, but $A^2 = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
add a comment |Â
up vote
19
down vote
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,212915
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
add a comment |Â
up vote
12
down vote
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1beginpmatrix
0&1\
1&0
endpmatrix+c_2beginpmatrix
1&0\
0&-1
endpmatrixpmsqrtc_1^2+c_2^2pm1beginpmatrix
0&-1\
1&0
endpmatrix,$$
see Pauli Matrices $sigma_i$.
These are all such matrices and can be written as $A=vec e· vec sigma$, where $vec e^2=pm1$.
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
add a comment |Â
up vote
7
down vote
The following matrix is a conterexample $
A =
left( beginarraycc
-1 & 0 \
0 & 1 \
endarray right)
$
answered Feb 5 '12 at 20:20
azarel
11k22331
add a comment |Â
up vote
6
down vote
"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
33
down vote
accepted
A simple counterexample is $$A = beginbmatrix 1 & 0 \ 0 & -1 endbmatrix $$ We have $A neq pm I$, but $A^2 = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
add a comment |Â
up vote
33
down vote
accepted
A simple counterexample is $$A = beginbmatrix 1 & 0 \ 0 & -1 endbmatrix $$ We have $A neq pm I$, but $A^2 = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
add a comment |Â
up vote
33
down vote
accepted
up vote
33
down vote
accepted
A simple counterexample is $$A = beginbmatrix 1 & 0 \ 0 & -1 endbmatrix $$ We have $A neq pm I$, but $A^2 = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
A simple counterexample is $$A = beginbmatrix 1 & 0 \ 0 & -1 endbmatrix $$ We have $A neq pm I$, but $A^2 = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5321215
2,5321215
add a comment |Â
add a comment |Â
up vote
19
down vote
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,212915
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
add a comment |Â
up vote
19
down vote
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,212915
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
add a comment |Â
up vote
19
down vote
up vote
19
down vote
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,212915
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,212915
answered Feb 5 '12 at 20:16
Blah
4,212915
answered Feb 5 '12 at 20:16
Blah
4,212915
answered Feb 5 '12 at 20:16
Blah
4,212915
4,212915
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
add a comment |Â
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
If you want to exchange the (standard) basis vectors $e_i$ and $e_j$ ($1 leq i,j leq n$), then use the matrix $A = [m_ij]$ with $m_kk = 1, kneq i,j$, $m_ij = m_ji = 1$ and $m_kl = 0$ for all other values of $k$ and $l$. For example, if you want $e_2$ and $e_3$ exhanged in $mathbbR^3$, take $$A = beginbmatrix 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 endbmatrix$$ It is clear that such a matrix always satisfies $A^2 = I$, since applying it twice always gets you back to where you started.
– Martin Wanvik
Feb 5 '12 at 21:01
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
Thank you @Martin Wanvik, pretty clear explanation.
– Randolf Rincón Fadul
Feb 5 '12 at 21:52
add a comment |Â
up vote
12
down vote
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1beginpmatrix
0&1\
1&0
endpmatrix+c_2beginpmatrix
1&0\
0&-1
endpmatrixpmsqrtc_1^2+c_2^2pm1beginpmatrix
0&-1\
1&0
endpmatrix,$$
see Pauli Matrices $sigma_i$.
These are all such matrices and can be written as $A=vec e· vec sigma$, where $vec e^2=pm1$.
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
add a comment |Â
up vote
12
down vote
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1beginpmatrix
0&1\
1&0
endpmatrix+c_2beginpmatrix
1&0\
0&-1
endpmatrixpmsqrtc_1^2+c_2^2pm1beginpmatrix
0&-1\
1&0
endpmatrix,$$
see Pauli Matrices $sigma_i$.
These are all such matrices and can be written as $A=vec e· vec sigma$, where $vec e^2=pm1$.
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
add a comment |Â
up vote
12
down vote
up vote
12
down vote
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1beginpmatrix
0&1\
1&0
endpmatrix+c_2beginpmatrix
1&0\
0&-1
endpmatrixpmsqrtc_1^2+c_2^2pm1beginpmatrix
0&-1\
1&0
endpmatrix,$$
see Pauli Matrices $sigma_i$.
These are all such matrices and can be written as $A=vec e· vec sigma$, where $vec e^2=pm1$.
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1beginpmatrix
0&1\
1&0
endpmatrix+c_2beginpmatrix
1&0\
0&-1
endpmatrixpmsqrtc_1^2+c_2^2pm1beginpmatrix
0&-1\
1&0
endpmatrix,$$
see Pauli Matrices $sigma_i$.
These are all such matrices and can be written as $A=vec e· vec sigma$, where $vec e^2=pm1$.
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
edited Feb 6 '12 at 18:40
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
answered Feb 5 '12 at 20:57
Nikolaj-K
5,73222967
5,73222967
add a comment |Â
add a comment |Â
up vote
7
down vote
The following matrix is a conterexample $
A =
left( beginarraycc
-1 & 0 \
0 & 1 \
endarray right)
$
answered Feb 5 '12 at 20:20
azarel
11k22331
add a comment |Â
up vote
7
down vote
The following matrix is a conterexample $
A =
left( beginarraycc
-1 & 0 \
0 & 1 \
endarray right)
$
answered Feb 5 '12 at 20:20
azarel
11k22331
add a comment |Â
up vote
7
down vote
up vote
7
down vote
The following matrix is a conterexample $
A =
left( beginarraycc
-1 & 0 \
0 & 1 \
endarray right)
$
answered Feb 5 '12 at 20:20
azarel
11k22331
The following matrix is a conterexample $
A =
left( beginarraycc
-1 & 0 \
0 & 1 \
endarray right)
$
answered Feb 5 '12 at 20:20
azarel
11k22331
answered Feb 5 '12 at 20:20
azarel
11k22331
answered Feb 5 '12 at 20:20
azarel
11k22331
answered Feb 5 '12 at 20:20
azarel
11k22331
11k22331
add a comment |Â
add a comment |Â
up vote
6
down vote
"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
add a comment |Â
up vote
6
down vote
"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
add a comment |Â
up vote
6
down vote
up vote
6
down vote
"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
"Most" (read: diagonalizable) matrices can be viewed simply as a list of numbers -- its eigenvalues -- in the right basis. When doing arithmetic with just this matrix (or with other matrices that diagonalize in the same basis), you just do arithmetic on the eigenvalues.
So, to find diagonalizable solutions to $A^2 = I$, we just need to write down a matrix whose eigenvalues satisfy $lambda^2 = 1$ -- and any such matrix will do.
When thinking about matrices in this way -- as a list of independent numbers -- it makes it easy to think your way through problems like this.
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
answered Feb 6 '12 at 4:56
Hurkyl
108k9113254
108k9113254
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
add a comment |Â
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
1
1
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
Every matrix satisfying $A^2=I$ is diagonalizable, because either it is $pm I$ or its minimal polynomial is $(x-1)(x+1)$. The general solution is obtained by taking all diagonal matrices with entries $pm 1$ on the diagonal and conjugating by invertible matrices.
– Jonas Meyer
Feb 6 '12 at 5:03
2
2
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
Jonas Meyer, this is only true if $char F ne 2$. Otherwise, there are such matrices which are not diagonalizable,
– the L
Feb 6 '12 at 8:18
1
1
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
@Jonas: That's a good point to mention as an appendix, but dealing properly with non-diagonalizable matrices in this fashion is somewhat more sophisticated. The only reason I mentioned the word was so that I didn't mislead Randolf into thinking this method works (unmodified) for all matrices; e.g. that the argument I gave isn't sufficient to tell us that this (or any!) equation has only diagonalizable solutions.
– Hurkyl
Feb 6 '12 at 9:53
1
1
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
@anonymous: Good point, e.g. $beginbmatrix1&1\ 0&1endbmatrix$. @@Hurkyl: I agree, it is best as an appendix. I appreciate your caution, but wanted to point out that your method does lead to the general solution (in the characteristic $0$ case that the OP is probably working in).
– Jonas Meyer
Feb 6 '12 at 15:49
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f106070%2fif-a2-i-identity-matrix-then-a-pm-i%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
23
Try $$ A = beginpmatrix 0 & 1 \ 1 & 0 endpmatrix. $$
– Dylan Moreland
Feb 5 '12 at 20:13
1
I'd point out that it is true if you're working with $1$-by-$1$ matrices (over $mathbb C$, or any other integral domain). But for $n geq 2$ the ring of $n$-by-$n$ matrices over any non-trivial ring is not an integral domain: this means that $(A+I)(A-I) = 0$ doesn't necessarily imply that $A + I = 0 $ or $A - I = 0$.
– Matt
Feb 5 '12 at 20:30
possible duplicate of Finding number of matrices whose square is the identity matrix
– Jonas Meyer
Feb 5 '12 at 20:56
There's an entire family of so-called involutory matrices. Look up Householder reflectors, for instance.
– J. M. is not a mathematician
Feb 6 '12 at 5:11
What book is that exercise from?
– Rhaldryn
Jan 22 '17 at 17:43