755 reputation
311
bio website
location Chemnitz, Germany
age 27
visits member for 2 years, 11 months
seen Apr 15 at 19:03

Feb
8
answered Relationship between nullspace and row/column space
Feb
7
comment matrix equation $(A-B)CA=B$
It may be simple, but why does $Q(CA+I)=A$ make $Q$ invertible?
Feb
3
comment 3D Rotation Matrix Uniqueness
@RahulNarain Hah, indeed!
Feb
3
revised linear interpolation in 3 dimensions
fixed formatting
Feb
3
suggested suggested edit on linear interpolation in 3 dimensions
Feb
3
answered (Graphics Gems IV, Shoemake) From matrix to euler angles explanation
Feb
3
comment 3D Rotation Matrix Uniqueness
@Korchkidu Ok, updated my answer.
Feb
3
revised 3D Rotation Matrix Uniqueness
added 643 characters in body
Feb
3
comment 3D Rotation Matrix Uniqueness
@Korchkidu Well, in fact it's more intuition, since it has to hold for all $\mathbf{v}\in\mathbb{R}^n$, for each and every matrix $M\neq O$ you can always find a $\mathbf{v}$ for which $M\mathbf{v}\neq\mathbf{0}$, but I'm sure there is also some theorem or prove out there for this.
Feb
3
answered Is there a classic Matrix Algebra reference?
Feb
3
answered 3D Rotation Matrix Uniqueness
Feb
3
comment 3D Rotation Matrix Uniqueness
Actually the rotation matrix $R'$ for the negated axis and angle (that you don't consider the same as $R$) is in fact the exact same matrix as the original $R$. So the matrix representation is even more unique than the axis-angle or quaternion representation.
Feb
3
answered How to get a projected 3d line segment, lie on another 3d line parallel to that line segment.
Feb
3
comment What is the conjugate of $\frac{1}{2}+ \frac{3}{2}i$?
@David There is nowhere to arrive from, conjugation is just defined this way. period.
Feb
2
comment Multiplying double-centered matrix to a unit vector
@user506901 But like said, such a translation is not achievable using a matrix-multiplication (and therefore neither with a double-centred matrix). Centering a vector is no translation, since the summands required to center the vector are no constants but depend on the vector itself. Centering a different vector (using the same matrix) would result in different summands, thus the application of the matrix doesn't represent a translation.
Feb
2
comment Multiplying double-centered matrix to a unit vector
@user506901 Or do you mean adding a different constant to each entry? This sounds more like a translation, but again, this cannot be done with a vector (maybe you mean a vector representing the location vector of a point?). And this cannot be achieved by a matrix multiply, because all summands added to the vector entries depend on the vector itself and cannot be constants. Maybe you are mixing things up here and think of the $\mathbb{R}^{n+1}$ projective space usually used in computer graphics and the like. In this space translation can indeed be realized by a matrix-multiply?
Feb
2
comment Multiplying double-centered matrix to a unit vector
@user506901 And when $D$ is orthogonal (or more precisely orthonormal) it doesn't change the length of a vector and thus corresponds to a rotation (if $\det Q=1$) or a reflection (if $\det Q=-1$) around the origin.
Feb
2
comment Multiplying double-centered matrix to a unit vector
@user506901 Ah, that's what you mean by translation (never heard of such a definition for translation, but Ok). But I'm not sure if that is done by a general double-centred matrix. So you mean centering means adding a constant to all its entries? Well in this case, like said, I'm not sure this is achieved by multiplying by a general double-centred matrix. I thought you mean a vector whose element sum is $0$, like in your definition of a double-centred matrix. Maybe you can add this definition to the question (maybe along with what you think is a translation of a vector).
Feb
1
awarded  Critic
Feb
1
comment Finding inverse of a $3\times 4$ or $4\times 3$ matrix
But it is still not an answer to the general question "How would you calculate inverse of such a matrix?", instead if your answer was "by using WolframAlpha". In this case it would just be a bad answer.