Compute transformation matrix in nth dimension

[bohdy]

I have a vector of all ones in n-dimensions. For example (1,1,1) in 3D. I want to find a invertible rotation matrix T that transforms the vector of all ones to the vector (0,0,0,...,0,,1):

Let v be the vector of all ones, and w=(0,0,...,0,1)
Find T such that T.v == wIn low dimension it is easy to find such a vector with the routine below, but computationally intensive when d>5. Is there a better way?

Basic routine: represent the components of T as variables and find a solution to the non-linear problem of finding an invertible matrix T (Det[T]!=0) that satisfies the dot product T.v == w.

Any better ideas?

 

[Lord Crc]

I'm no expert here, but I don't see how this is possible.

Assume that the inverse of T exists, and that it is A = inverse(T).The original problem can then be rewritten:
Tv = w
v = Aw

However, the only way this can hold is if the columns of A are (0, 0, ..., 0, v) (where 0 here is the zero vector). Thus A is obviously not invertible, and thus T cannot be either.

If I'm wrong, I'm sorry. I'd love hear where I'm wrong tho :)

 

[HallsofIvy]

I have a vector of all ones in n-dimensions. For example (1,1,1) in 3D. I want to find a invertible rotation matrix T that transforms the vector of all ones to the vector (0,0,0,...,0,,1):

? Any transformation from R³ to Rⁿ where n> 3 is not square and so is not invertible. Nor can it be a "rotation" matrix.

Let v be the vector of all ones, and w=(0,0,...,0,1)
Find T such that T.v == w


In low dimension it is easy to find such a vector with the routine below, but computationally intensive when d>5. Is there a better way?

Basic routine: represent the components of T as variables and find a solution to the non-linear problem of finding an invertible matrix T (Det[T]!=0) that satisfies the dot product T.v == w.

Any better ideas?

 

[bohdy]

In 2D:
T= {{1, -1}, {1, 0}}
Tinverse = {{0, 1}, {-1, 1}}
T.{1,1}={0,1}

In 3D:
T= {{1, -1, 0}, {-1, 0, 1}, {1, 1, -1}}
Tinverse = {{1, 1, 1}, {0, 1, 1}, {1, 2, 1}}
T.{1,1,1}={0,0,1}

In 4D:
T= {{-1, -2, 0, 3}, {-2, -1, 0, 3}, {-1, 1, -1, 1}, {2, -2, 1, 0}}
Tinverse = {{-(2/3), 1/3, 1, 1}, {-(5/3), 4/3, 1, 1}, {-2, 2, 0, 1}, {-1, 1, 1,
1}}
T.{1,1,1,1}={0,0,0,1}

I'm no expert here, but I don't see how this is possible.

Assume that the inverse of T exists, and that it is A = inverse(T).The original problem can then be rewritten:
Tv = w
v = Aw

However, the only way this can hold is if the columns of A are (0, 0, ..., 0, v) (where 0 here is the zero vector). Thus A is obviously not invertible, and thus T cannot be either.

If I'm wrong, I'm sorry. I'd love hear where I'm wrong tho :)

 

[bohdy]

? Any transformation from R³ to Rⁿ where n> 3 is not square and so is not invertible. Nor can it be a "rotation" matrix.

I'm looking to map Rⁿ to Rⁿ. See above post. I used the 3D vector (1,1,1) as an example.

...so in R⁷ I want to map (1,1,1,1,1,1,1) to (0,0,0,0,0,0,1) etc.

 

[Lord Crc]

Ah, was thinking inside out. Thanks :)

Do you want a proper rotation matrix, or just any invertible matrix that transforms v to w?

Your "basic algorithm" seems to indicate the latter, in which case couldn't you then construct the Tinverse first, which is easy, and then invert that to get T?

 

[bohdy]

Hi,
Can you elaborate on how to construct Tinverse easily? Isn't the computational difficulty the same, in that finding T such that T.v=w is as hard as finding a Tinv such that Tinv.w=v?

Any matrix will do. I will then be applying the transformation to other vectors in the space (I don't want 2 vectors to map onto 1 - hence the invertibility property)

Ah, was thinking inside out. Thanks :)

Do you want a proper rotation matrix, or just any invertible matrix that transforms v to w?

Your "basic algorithm" seems to indicate the latter, in which case couldn't you then construct the Tinverse first, which is easy, and then invert that to get T?

 

[Lord Crc]

Well, again, I might be wrong here but... The requirements of Tinverse is that the last column has to be all 1's (in order to transform w to v), and in order to be invertible the columns must be linearly independent. One way to construct such a matrix would be to use the standard basis vectors e_i for the first n-1 columns, and the use v as the n'th column:

Tinverse = {{1, 0, 0, 1}, {0, 1, 0, 1}, {0, 0, 1, 1}, {0, 0, 0, 1}}

This matrix can then be inverted and will transform v into w:

T = {{1, 0, 0, -1}, {0, 1, 0, -1}, {0, 0, 1, -1}, {0, 0, 0, 1}}

 

[bohdy]

Excellent. That's the kind of solution I was hoping for. Many thanks.

Well, again, I might be wrong here but... The requirements of Tinverse is that the last column has to be all 1's (in order to transform w to v), and in order to be invertible the columns must be linearly independent. One way to construct such a matrix would be to use the standard basis vectors e_i for the first n-1 columns, and the use v as the n'th column:

Tinverse = {{1, 0, 0, 1}, {0, 1, 0, 1}, {0, 0, 1, 1}, {0, 0, 0, 1}}

This matrix can then be inverted and will transform v into w:

T = {{1, 0, 0, -1}, {0, 1, 0, -1}, {0, 0, 1, -1}, {0, 0, 0, 1}}