How to represent a vector in terms of another vector? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Dot product of two vectors without a common originMatrix Becomes a vector spaceProject vector onto another vector a maximum angle $theta$Intuitive understanding of vector / matrix calculcationUnderstanding components of a vectorHow to prove Vector Angles?Which of the following sets attached with an associated vector addition and scalar multiplication with real numbers describe a vector space?How to calculate an unknown vector from a known vector and an angleVector equation: $left<cos(a), sin(a)right> cdot left<cos(a),-sin(a) right> = -cos(a)$what is the result type 3D vector, point and scalar arithmetic operations?
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How to represent a vector in terms of another vector?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Dot product of two vectors without a common originMatrix Becomes a vector spaceProject vector onto another vector a maximum angle $theta$Intuitive understanding of vector / matrix calculcationUnderstanding components of a vectorHow to prove Vector Angles?Which of the following sets attached with an associated vector addition and scalar multiplication with real numbers describe a vector space?How to calculate an unknown vector from a known vector and an angleVector equation: $left<cos(a), sin(a)right> cdot left<cos(a),-sin(a) right> = -cos(a)$what is the result type 3D vector, point and scalar arithmetic operations?
$begingroup$
If $u, v$ be two unit vectors. $textbfThen how to represent $u$ in terms of $v$? $
I can find a matrix, $R$ such that $u=Rv$, from trial and error. How to derive the matrix analytically?
Unit vectors essentially represent direction. So $cos(x)= u.v $ should be giving the angle between $u$ and $v$.
P.S: I am looking for the relationship between the eigenvectors of two different matrices.
I know that a scalar multiplication of one vector will not give another vector.
matrices vector-spaces eigenvalues-eigenvectors vectors matrix-equations
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
$endgroup$
add a comment |
$begingroup$
If $u, v$ be two unit vectors. $textbfThen how to represent $u$ in terms of $v$? $
I can find a matrix, $R$ such that $u=Rv$, from trial and error. How to derive the matrix analytically?
Unit vectors essentially represent direction. So $cos(x)= u.v $ should be giving the angle between $u$ and $v$.
P.S: I am looking for the relationship between the eigenvectors of two different matrices.
I know that a scalar multiplication of one vector will not give another vector.
matrices vector-spaces eigenvalues-eigenvectors vectors matrix-equations
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
$endgroup$
1
$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41
add a comment |
$begingroup$
If $u, v$ be two unit vectors. $textbfThen how to represent $u$ in terms of $v$? $
I can find a matrix, $R$ such that $u=Rv$, from trial and error. How to derive the matrix analytically?
Unit vectors essentially represent direction. So $cos(x)= u.v $ should be giving the angle between $u$ and $v$.
P.S: I am looking for the relationship between the eigenvectors of two different matrices.
I know that a scalar multiplication of one vector will not give another vector.
matrices vector-spaces eigenvalues-eigenvectors vectors matrix-equations
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
$endgroup$
If $u, v$ be two unit vectors. $textbfThen how to represent $u$ in terms of $v$? $
I can find a matrix, $R$ such that $u=Rv$, from trial and error. How to derive the matrix analytically?
Unit vectors essentially represent direction. So $cos(x)= u.v $ should be giving the angle between $u$ and $v$.
P.S: I am looking for the relationship between the eigenvectors of two different matrices.
I know that a scalar multiplication of one vector will not give another vector.
matrices vector-spaces eigenvalues-eigenvectors vectors matrix-equations
matrices vector-spaces eigenvalues-eigenvectors vectors matrix-equations
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
edited 2 days ago
Abhiram V P
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
asked Apr 1 at 18:26
Abhiram V PAbhiram V P
166
166
1
$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41
add a comment |
1
$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41
1
1
$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $u,v$ are unit vectors, the matrix $R=uv^T$ does the trick:
$$
Rv = uv^Tv = u.
$$
answered Apr 3 at 6:56
dawdaw
25.2k1745
$endgroup$
add a comment |
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$begingroup$
If $u,v$ are unit vectors, the matrix $R=uv^T$ does the trick:
$$
Rv = uv^Tv = u.
$$
answered Apr 3 at 6:56
dawdaw
25.2k1745
$endgroup$
add a comment |
$begingroup$
If $u,v$ are unit vectors, the matrix $R=uv^T$ does the trick:
$$
Rv = uv^Tv = u.
$$
answered Apr 3 at 6:56
dawdaw
25.2k1745
$endgroup$
add a comment |
$begingroup$
If $u,v$ are unit vectors, the matrix $R=uv^T$ does the trick:
$$
Rv = uv^Tv = u.
$$
answered Apr 3 at 6:56
dawdaw
25.2k1745
$endgroup$
If $u,v$ are unit vectors, the matrix $R=uv^T$ does the trick:
$$
Rv = uv^Tv = u.
$$
answered Apr 3 at 6:56
dawdaw
25.2k1745
answered Apr 3 at 6:56
dawdaw
25.2k1745
answered Apr 3 at 6:56
dawdaw
25.2k1745
answered Apr 3 at 6:56
dawdaw
25.2k1745
25.2k1745
add a comment |
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$begingroup$
Could you post everything you know about the two vectors? If the vectors are in 2D, then you could describe the vector $v$ as the value along $u$ and the value along the perpendicular of $u$. However, this does not work in higher dimensions. Eg: consider the vectors $(1, 0, 0)$ and $(0, 1, 2)$. There is no way to express $(0, 1, 2)$ in terms of just $(1, 0, 0)
$endgroup$
– Siddharth Bhat
Apr 1 at 18:32
$begingroup$
@SiddharthBhat vectors are of any dimensions. I was thinking maybe it could be rotated as a unit vector represents direction.
$endgroup$
– Abhiram V P
Apr 1 at 18:39
$begingroup$
If $ucdot vneq 1$ then $u,v$ are not parallel, so any scalar multiplie of $u$, say, i.e. $lambda u$, can't possibly be equal to $v$.
$endgroup$
– Antinous
Apr 1 at 18:41