Is the maximum rotation by multiplying a positive definite matrix is less than 90 degrees?
$begingroup$
To have intuition, I think of rotation of a vector $v$ in a two dimensional space. A vector can be rotated by multiplying the rotation matrix as
$$
R=begin{bmatrix}
costheta & -sintheta \
sintheta & costheta \
end{bmatrix}
$$
However, the rotation matrix is positive definite only or $-frac{pi}{2}leq theta leq frac{pi}{2}$.
So, if we impose the positive definiteness onto the rotation matrix, it just can rotate the vector at most 90 degrees (clockwise or counterclockwise).
Now, since every matrix multiplication has its rotational effect, can we conclude that the maximum rotation by multiplying a positive definite matrix is less than 90 degrees?
linear-algebra rotations
edited Jan 4 at 4:10
Diglett
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
$endgroup$
add a comment |
$begingroup$
To have intuition, I think of rotation of a vector $v$ in a two dimensional space. A vector can be rotated by multiplying the rotation matrix as
$$
R=begin{bmatrix}
costheta & -sintheta \
sintheta & costheta \
end{bmatrix}
$$
However, the rotation matrix is positive definite only or $-frac{pi}{2}leq theta leq frac{pi}{2}$.
So, if we impose the positive definiteness onto the rotation matrix, it just can rotate the vector at most 90 degrees (clockwise or counterclockwise).
Now, since every matrix multiplication has its rotational effect, can we conclude that the maximum rotation by multiplying a positive definite matrix is less than 90 degrees?
linear-algebra rotations
edited Jan 4 at 4:10
Diglett
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
$endgroup$
2
$begingroup$
Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25
add a comment |
$begingroup$
To have intuition, I think of rotation of a vector $v$ in a two dimensional space. A vector can be rotated by multiplying the rotation matrix as
$$
R=begin{bmatrix}
costheta & -sintheta \
sintheta & costheta \
end{bmatrix}
$$
However, the rotation matrix is positive definite only or $-frac{pi}{2}leq theta leq frac{pi}{2}$.
So, if we impose the positive definiteness onto the rotation matrix, it just can rotate the vector at most 90 degrees (clockwise or counterclockwise).
Now, since every matrix multiplication has its rotational effect, can we conclude that the maximum rotation by multiplying a positive definite matrix is less than 90 degrees?
linear-algebra rotations
edited Jan 4 at 4:10
Diglett
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
$endgroup$
To have intuition, I think of rotation of a vector $v$ in a two dimensional space. A vector can be rotated by multiplying the rotation matrix as
$$
R=begin{bmatrix}
costheta & -sintheta \
sintheta & costheta \
end{bmatrix}
$$
However, the rotation matrix is positive definite only or $-frac{pi}{2}leq theta leq frac{pi}{2}$.
So, if we impose the positive definiteness onto the rotation matrix, it just can rotate the vector at most 90 degrees (clockwise or counterclockwise).
Now, since every matrix multiplication has its rotational effect, can we conclude that the maximum rotation by multiplying a positive definite matrix is less than 90 degrees?
linear-algebra rotations
linear-algebra rotations
edited Jan 4 at 4:10
Diglett
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
edited Jan 4 at 4:10
Diglett
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
edited Jan 4 at 4:10
Diglett
9821521
edited Jan 4 at 4:10
Diglett
9821521
edited Jan 4 at 4:10
Diglett
9821521
9821521
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
asked May 15 '18 at 17:10
SaeedSaeed
1,105310
1,105310
2
$begingroup$
Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25
add a comment |
2
$begingroup$
Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25
2
2
$begingroup$
Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25
add a comment |
1 Answer
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Yes. Suppose A is a 2x2 positive definite matrix. Then for any non-zero x, $x' A x > 0$. But $x' A x = |x| |Ax| cos(theta)$, for $theta$ the angle between x and Ax, from which we see that $cos(theta) > 0$ and thus $theta$ must be between $-pi/2$ and $pi/2$.
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
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add a comment |
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$begingroup$
Yes. Suppose A is a 2x2 positive definite matrix. Then for any non-zero x, $x' A x > 0$. But $x' A x = |x| |Ax| cos(theta)$, for $theta$ the angle between x and Ax, from which we see that $cos(theta) > 0$ and thus $theta$ must be between $-pi/2$ and $pi/2$.
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
$endgroup$
add a comment |
$begingroup$
Yes. Suppose A is a 2x2 positive definite matrix. Then for any non-zero x, $x' A x > 0$. But $x' A x = |x| |Ax| cos(theta)$, for $theta$ the angle between x and Ax, from which we see that $cos(theta) > 0$ and thus $theta$ must be between $-pi/2$ and $pi/2$.
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
$endgroup$
add a comment |
$begingroup$
Yes. Suppose A is a 2x2 positive definite matrix. Then for any non-zero x, $x' A x > 0$. But $x' A x = |x| |Ax| cos(theta)$, for $theta$ the angle between x and Ax, from which we see that $cos(theta) > 0$ and thus $theta$ must be between $-pi/2$ and $pi/2$.
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
$endgroup$
Yes. Suppose A is a 2x2 positive definite matrix. Then for any non-zero x, $x' A x > 0$. But $x' A x = |x| |Ax| cos(theta)$, for $theta$ the angle between x and Ax, from which we see that $cos(theta) > 0$ and thus $theta$ must be between $-pi/2$ and $pi/2$.
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
answered Jan 4 at 3:32
Jonathan RothJonathan Roth
1
1
add a comment |
add a comment |
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Aren't positive definite matrices usually assumed to be symmetric? So in general a rotation matrix would not qualify. In any case, one key property of a positive definite matrix is that it has real, positive eigenvalues. Rotation matrices don't have that property.
$endgroup$
– Bungo
May 15 '18 at 17:36
$begingroup$
Thank you. However, can we compare the maximum rotation of a vector which is multiplied by a positive definite matrix?
$endgroup$
– Saeed
May 15 '18 at 18:52
$begingroup$
@Bungo If the field is complex then positive definite implies symmetric; but in case of the real field, positive definite need not mean symmetric.
$endgroup$
– zimbra314
Jan 4 at 4:25