Christian Rau
Reputation
805
Top tag
Next privilege 1,000 Rep.
Create tags
 Feb9 revised transformation of 3D coordinate system fixed formatting Feb9 suggested approved edit on transformation of 3D coordinate system Feb9 comment Let $\rho(P)$ be the spectral radius of $P$. Show $\rho( \dfrac{P}{ \rho(P) + \epsilon } ) < 1 \text{ for all } \epsilon >0.$ Can it be you're missing a $c$ in there (first equation)? Otherwise I cannot believe $\lambda$ being an eigenvalue implies $c\lambda$ being an eigenvalue of the same matrix, too. With $\sigma$ you mean the spectrum, right? Feb8 answered Relationship between nullspace and row/column space Feb7 comment matrix equation $(A-B)CA=B$ It may be simple, but why does $Q(CA+I)=A$ make $Q$ invertible? Feb3 comment 3D Rotation Matrix Uniqueness @RahulNarain Hah, indeed! Feb3 revised linear interpolation in 3 dimensions fixed formatting Feb3 suggested approved edit on linear interpolation in 3 dimensions Feb3 answered (Graphics Gems IV, Shoemake) From matrix to euler angles explanation Feb3 comment 3D Rotation Matrix Uniqueness @Korchkidu Ok, updated my answer. Feb3 revised 3D Rotation Matrix Uniqueness added 643 characters in body Feb3 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. Feb3 answered Is there a classic Matrix Algebra reference? Feb3 answered 3D Rotation Matrix Uniqueness Feb3 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. Feb3 answered How to get a projected 3d line segment, lie on another 3d line parallel to that line segment. Feb3 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. Feb2 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. Feb2 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? Feb2 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.