If $A^2 = I$ (Identity Matrix) then $A = pm I$
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)?
linear-algebra matrices examples-counterexamples
edited Nov 30 at 15:19
Abcd
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
add a comment |
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)?
linear-algebra matrices examples-counterexamples
edited Nov 30 at 15:19
Abcd
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
24
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– 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 |
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)?
linear-algebra matrices examples-counterexamples
edited Nov 30 at 15:19
Abcd
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
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)?
linear-algebra matrices examples-counterexamples
linear-algebra matrices examples-counterexamples
edited Nov 30 at 15:19
Abcd
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
edited Nov 30 at 15:19
Abcd
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
edited Nov 30 at 15:19
Abcd
3,00121132
edited Nov 30 at 15:19
Abcd
3,00121132
edited Nov 30 at 15:19
Abcd
3,00121132
3,00121132
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
asked Feb 5 '12 at 20:11
Randolf Rincón Fadul
3901519
3901519
24
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– 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 |
24
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– 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
24
24
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– Dylan Moreland
Feb 5 '12 at 20:13
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– 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 |
5 Answers
5
active
oldest
votes
A simple counterexample is $$A = begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix} $$ We have $A neq pm I$, but $A^{2} = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
add a comment |
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,242915
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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 |
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1begin{pmatrix}
0&1\
1&0
end{pmatrix}+c_2begin{pmatrix}
1&0\
0&-1
end{pmatrix}pmsqrt{c_1^2+c_2^2pm1}begin{pmatrix}
0&-1\
1&0
end{pmatrix},$$
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
5,76223068
add a comment |
The following matrix is a conterexample $
A =
left( {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right)
$
answered Feb 5 '12 at 20:20
azarel
11.1k22431
add a comment |
"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
111k9117259
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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
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
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
A simple counterexample is $$A = begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix} $$ We have $A neq pm I$, but $A^{2} = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
add a comment |
A simple counterexample is $$A = begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix} $$ We have $A neq pm I$, but $A^{2} = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
add a comment |
A simple counterexample is $$A = begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix} $$ We have $A neq pm I$, but $A^{2} = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
A simple counterexample is $$A = begin{bmatrix} 1 & 0 \ 0 & -1 end{bmatrix} $$ We have $A neq pm I$, but $A^{2} = I$.
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
answered Feb 5 '12 at 20:15
Martin Wanvik
2,5621215
2,5621215
add a comment |
add a comment |
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,242915
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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 |
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,242915
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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 |
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,242915
In dimension $geq 2$ take the matrix that exchanges two basis vectors ("a transposition")
answered Feb 5 '12 at 20:16
Blah
4,242915
answered Feb 5 '12 at 20:16
Blah
4,242915
answered Feb 5 '12 at 20:16
Blah
4,242915
answered Feb 5 '12 at 20:16
Blah
4,242915
4,242915
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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 |
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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 $mathbb{R}^{3}$, take $$A = begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ 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 |
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1begin{pmatrix}
0&1\
1&0
end{pmatrix}+c_2begin{pmatrix}
1&0\
0&-1
end{pmatrix}pmsqrt{c_1^2+c_2^2pm1}begin{pmatrix}
0&-1\
1&0
end{pmatrix},$$
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
5,76223068
add a comment |
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1begin{pmatrix}
0&1\
1&0
end{pmatrix}+c_2begin{pmatrix}
1&0\
0&-1
end{pmatrix}pmsqrt{c_1^2+c_2^2pm1}begin{pmatrix}
0&-1\
1&0
end{pmatrix},$$
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
5,76223068
add a comment |
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1begin{pmatrix}
0&1\
1&0
end{pmatrix}+c_2begin{pmatrix}
1&0\
0&-1
end{pmatrix}pmsqrt{c_1^2+c_2^2pm1}begin{pmatrix}
0&-1\
1&0
end{pmatrix},$$
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
5,76223068
I know $2·mathbb C^2$ many counterexamples, namely
$$A=c_1begin{pmatrix}
0&1\
1&0
end{pmatrix}+c_2begin{pmatrix}
1&0\
0&-1
end{pmatrix}pmsqrt{c_1^2+c_2^2pm1}begin{pmatrix}
0&-1\
1&0
end{pmatrix},$$
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
5,76223068
edited Feb 6 '12 at 18:40
answered Feb 5 '12 at 20:57
Nikolaj-K
5,76223068
answered Feb 5 '12 at 20:57
Nikolaj-K
5,76223068
answered Feb 5 '12 at 20:57
Nikolaj-K
5,76223068
5,76223068
add a comment |
add a comment |
The following matrix is a conterexample $
A =
left( {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right)
$
answered Feb 5 '12 at 20:20
azarel
11.1k22431
add a comment |
The following matrix is a conterexample $
A =
left( {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right)
$
answered Feb 5 '12 at 20:20
azarel
11.1k22431
add a comment |
The following matrix is a conterexample $
A =
left( {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right)
$
answered Feb 5 '12 at 20:20
azarel
11.1k22431
The following matrix is a conterexample $
A =
left( {begin{array}{cc}
-1 & 0 \
0 & 1 \
end{array} } right)
$
answered Feb 5 '12 at 20:20
azarel
11.1k22431
answered Feb 5 '12 at 20:20
azarel
11.1k22431
answered Feb 5 '12 at 20:20
azarel
11.1k22431
answered Feb 5 '12 at 20:20
azarel
11.1k22431
11.1k22431
add a comment |
add a comment |
"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
111k9117259
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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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 |
"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
111k9117259
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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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 |
"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
111k9117259
"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
111k9117259
answered Feb 6 '12 at 4:56
Hurkyl
111k9117259
answered Feb 6 '12 at 4:56
Hurkyl
111k9117259
answered Feb 6 '12 at 4:56
Hurkyl
111k9117259
111k9117259
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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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 |
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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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. $begin{bmatrix}1&1\ 0&1end{bmatrix}$. @@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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
24
Try $$ A = begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}. $$
– 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