4 ms·

> Rotation matrices just describe the unit vectors of a new coordinate system More generally, _every_ matrix describes how a change of coordinate system should

> Rotation matrices just describe the unit vectors of a new coordinate system More generally, _every_ matrix describes how a change of coordinate system should happen. For any m x n (m rows, n columns) matrix, each of the the n column vectors represent how the current coordinate system (which may or may not be unit vectors) should be represented in the new coordinate system (for a left multiplied matrix). Whenever a matrix's determinant is zero, it means that you squashed some dimensions. As you can imagine, when m is not equal to n, there will always be dimension squashing. Even when they are equal, that can happen. If you take a 3 x 3 matrix and it transforms all the 3-D vectors into only planes (which are 2-D objects), the determinant will be zero. This would be stated as having a rank of 2. More simply, you'd say the "volume" of the transform to be 0 (because planes have zero volume). EDIT: Make sure to read important clarifications by JadeNB below.

JadeNB 6y ago
> Whenever a matrix's determinant is zero, it means that you squashed some dimensions. As you can imagine, when m is not equal to n, there will always be dimension squashing. Although these two sentences are correct individually, it may be worth it to emphasise that they should not be read together: the determinant is defined only for square matrices. (One can do something like computing the determinant of det(A^{adjoint}A) if you want a general numerical invariant that behaves as you like—and then we're getting in the neighbourhood of the SVD). > If you take a 3 x 3 matrix and it transforms all the 3-D vectors into only planes (which are 2-D objects), the determinant will be zero. Although your meaning seems clear, I think you may have misspoken slightly. Of course 3D vectors are transformed by a 3 x 3 matrix into vectors, not planes. To me, what I think you mean would read better as "If you take a 3 x 3 matrix and there is a plane such that all 3D vectors are transformed into that plane, then …", or, perhaps even better, "… such that the transforms of all 3D vectors lie in that plane …".
- activatedgeek 6y ago
  Those are very important points. I've edited the parent to point out this clarifying comment.
  - JadeNB 6y ago
    Thanks! I should also have mentioned that, assuming we've got an m x n matrix acting by multiplying vectors fed to it on the right, dimension squashing is guaranteed only if n is less than m, not whenever they are unequal. (I guess one could argue that, in the case where n is greater than m, dimension squashing happens if you feed in vectors on the left.)
    - activatedgeek 6y ago
      Yes, I did mean dimension squashing in either direction - left or right.

JadeNB 6y ago

> Whenever a matrix's determinant is zero, it means that you squashed some dimensions. As you can imagine, when m is not equal to n, there will always be dimension squashing. Although these two sentences are correct individually, it may be worth it to emphasise that they should not be read together: the determinant is defined only for square matrices. (One can do something like computing the determinant of det(A^{adjoint}A) if you want a general numerical invariant that behaves as you like—and then we're getting in the neighbourhood of the SVD). > If you take a 3 x 3 matrix and it transforms all the 3-D vectors into only planes (which are 2-D objects), the determinant will be zero. Although your meaning seems clear, I think you may have misspoken slightly. Of course 3D vectors are transformed by a 3 x 3 matrix into vectors, not planes. To me, what I think you mean would read better as "If you take a 3 x 3 matrix and there is a plane such that all 3D vectors are transformed into that plane, then …", or, perhaps even better, "… such that the transforms of all 3D vectors lie in that plane …".

activatedgeek 6y ago
Those are very important points. I've edited the parent to point out this clarifying comment.
- JadeNB 6y ago
  Thanks! I should also have mentioned that, assuming we've got an m x n matrix acting by multiplying vectors fed to it on the right, dimension squashing is guaranteed only if n is less than m, not whenever they are unequal. (I guess one could argue that, in the case where n is greater than m, dimension squashing happens if you feed in vectors on the left.)
  - activatedgeek 6y ago
    Yes, I did mean dimension squashing in either direction - left or right.