Why is this matrix not a reflection in a plane matrix?
$begingroup$
$$
begin{pmatrix}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1 \
end{pmatrix}
$$
This matrix clearly has det -1 and is orthogonal. Why is it not reflection in a plane matrix?
linear-algebra matrices linear-transformations reflection
asked Dec 19 '18 at 15:09
TitterTitter
41
$endgroup$
add a comment |
$begingroup$
$$
begin{pmatrix}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1 \
end{pmatrix}
$$
This matrix clearly has det -1 and is orthogonal. Why is it not reflection in a plane matrix?
linear-algebra matrices linear-transformations reflection
asked Dec 19 '18 at 15:09
TitterTitter
41
$endgroup$
$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
1
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15
add a comment |
$begingroup$
$$
begin{pmatrix}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1 \
end{pmatrix}
$$
This matrix clearly has det -1 and is orthogonal. Why is it not reflection in a plane matrix?
linear-algebra matrices linear-transformations reflection
asked Dec 19 '18 at 15:09
TitterTitter
41
$endgroup$
$$
begin{pmatrix}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1 \
end{pmatrix}
$$
This matrix clearly has det -1 and is orthogonal. Why is it not reflection in a plane matrix?
linear-algebra matrices linear-transformations reflection
linear-algebra matrices linear-transformations reflection
asked Dec 19 '18 at 15:09
TitterTitter
41
asked Dec 19 '18 at 15:09
TitterTitter
41
asked Dec 19 '18 at 15:09
TitterTitter
41
asked Dec 19 '18 at 15:09
TitterTitter
41
asked Dec 19 '18 at 15:09
TitterTitter
41
41
$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
1
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15
add a comment |
$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
1
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15
$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
1
1
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In your definition of a reflection of $mathbb{R}^3$, you have to replace your condition $det = -1$ by a condition about the dimensions of the eigenspaces $E_1$ and $E_{-1}$ associated to the eigenvalues $1$ and $-1$, namely that $dim E_1=2$ and $dim E_{-1}=1$ (note that this condition automatically implies that your determinant is $-1$). Here we clearly have that $dim E_{-1}=3$.
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
It is easy to see that this matrix corresponds to the reflection through the origin $(0,0,0)$ hence the origin is the only invariant point of this tranformation.
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your definition of a reflection of $mathbb{R}^3$, you have to replace your condition $det = -1$ by a condition about the dimensions of the eigenspaces $E_1$ and $E_{-1}$ associated to the eigenvalues $1$ and $-1$, namely that $dim E_1=2$ and $dim E_{-1}=1$ (note that this condition automatically implies that your determinant is $-1$). Here we clearly have that $dim E_{-1}=3$.
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
In your definition of a reflection of $mathbb{R}^3$, you have to replace your condition $det = -1$ by a condition about the dimensions of the eigenspaces $E_1$ and $E_{-1}$ associated to the eigenvalues $1$ and $-1$, namely that $dim E_1=2$ and $dim E_{-1}=1$ (note that this condition automatically implies that your determinant is $-1$). Here we clearly have that $dim E_{-1}=3$.
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
$endgroup$
add a comment |
$begingroup$
In your definition of a reflection of $mathbb{R}^3$, you have to replace your condition $det = -1$ by a condition about the dimensions of the eigenspaces $E_1$ and $E_{-1}$ associated to the eigenvalues $1$ and $-1$, namely that $dim E_1=2$ and $dim E_{-1}=1$ (note that this condition automatically implies that your determinant is $-1$). Here we clearly have that $dim E_{-1}=3$.
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
$endgroup$
In your definition of a reflection of $mathbb{R}^3$, you have to replace your condition $det = -1$ by a condition about the dimensions of the eigenspaces $E_1$ and $E_{-1}$ associated to the eigenvalues $1$ and $-1$, namely that $dim E_1=2$ and $dim E_{-1}=1$ (note that this condition automatically implies that your determinant is $-1$). Here we clearly have that $dim E_{-1}=3$.
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
answered Dec 19 '18 at 15:18
BalloonBalloon
4,625822
4,625822
add a comment |
add a comment |
$begingroup$
It is easy to see that this matrix corresponds to the reflection through the origin $(0,0,0)$ hence the origin is the only invariant point of this tranformation.
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
$endgroup$
add a comment |
$begingroup$
It is easy to see that this matrix corresponds to the reflection through the origin $(0,0,0)$ hence the origin is the only invariant point of this tranformation.
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
$endgroup$
add a comment |
$begingroup$
It is easy to see that this matrix corresponds to the reflection through the origin $(0,0,0)$ hence the origin is the only invariant point of this tranformation.
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
$endgroup$
It is easy to see that this matrix corresponds to the reflection through the origin $(0,0,0)$ hence the origin is the only invariant point of this tranformation.
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
answered Dec 19 '18 at 15:17
giannispapavgiannispapav
1,632324
1,632324
add a comment |
add a comment |
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$begingroup$
Welcome to Math.SE. You should provide more information. For instance, what troubles you in the question ? What have you tried (examples, counter-examples) to find the answer yourself and that didn't work ? Main principle in the community : show your work.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:13
1
$begingroup$
Remember that the points in a plane are invariant under the reflection induced by it.
$endgroup$
– Tom-Tom
Dec 19 '18 at 15:15