Christian Rau
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 Mar 13 revised I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent fixed formatting and error in last equation Mar 13 suggested approved edit on I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent Mar 13 revised I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent fixed formatting and removed chatter Mar 13 suggested rejected edit on I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent Mar 13 suggested approved edit on I have to show that the matrix $M^TM$ is SPD if and only if the columns of the matrix M are linearly independent Mar 8 comment How to define sparseness of a vector? Sorry if my comments are a bit confusing. It isn't the name sparseness that bothers me, it's the hard fact, that your above function (the one with the $\sqrt{k}$) is 1 for a sparse vector and 0 for a dense vector (no matter how you name it). Mar 8 comment How to define sparseness of a vector? I know, I just wanted to make clear, that your explanation as it stands is wrong, if Sparseness(X) is indeed defined as above. Mar 8 comment How to define sparseness of a vector? Isn't your sparseness function 1 for a sparse vector and 0 for a dense vector? Mar 8 revised How to define sparseness of a vector? improved formatting Mar 8 suggested approved edit on How to define sparseness of a vector? Feb 27 comment Why can we figure out the relationship between a point and a plane by a matrix's determinant? @CChen No, the answer is not accepted, as you forgot to actually accept it. Feb 22 answered Determine the percentage needed to subtract to get the base value Feb 16 comment $\beta_k$ for Conjugate Gradient Method The idea of CG (in my limited understanding of theoretical numerical linear algebra) in contrast to gradient descent is to compute the current solution not only orthogonal to the current error, but also orthogonal to all the previous errors. And by using an appropriate basis (an A-orthogonal one) we get these nice properties of iterative computability and fast convergence. May the mathematicians forgive me if this generalized conceptual understanding is wrong. Feb 16 comment $\beta_k$ for Conjugate Gradient Method When taking the $r_k$ as basis, you don't get a minimal basis, meaning once you get to $r_{n+1}$ it cannot be a basis anymore, as this $r_{n+1}$ has to be linearly dependent on the previous $r$s (as the space is only $n$-dimensional at maximum). But when $A$-orthogonalizing the basis each time, you get a minimal basis of only $n$ vectors at maximum (making CG a direct method in theory). Feb 16 comment $\beta_k$ for Conjugate Gradient Method Of course it can. There are many bases for this sub-space, but only the $p$-basis is $A$-orthogonal, the $r$-basis isn't. Feb 16 comment $\beta_k$ for Conjugate Gradient Method $p_k$ and $r_k$ are not the same for $k>0$, because you don't just use the next orthogonal direction (which would be the gradient $r_k$). You take the gradient direction $r_k$ and orthogonalize this to all the previous $p_k$ (which span the current solution space), but with respect to $A$, making them conjugate with respect to $A$ and not just orthogonal (and this way also different from $r_k$). Feb 11 suggested rejected edit on question in linear algebra, matrices Feb 10 answered Simplest equation for drawing a cube based on its center and/or other vertices Feb 10 comment Simplest equation for drawing a cube based on its center and/or other vertices Well, a mass is not a length, but I guess your cube has a constant density of $1$, in which case the mass $m$ is indeed equal to the volume $V$ and you can determine the side length with a simple $s=\sqrt[3]{V}$. If not, then you need more information than just the mass to gain information about the cube's size. Feb 9 revised transformation of 3D coordinate system fixed formating