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location Chemnitz, Germany
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seen Aug 21 at 11:36

Mar
13
suggested suggested 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 suggested 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 suggested 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 suggested 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
Feb
9
suggested suggested edit on transformation of 3D coordinate system
Feb
9
revised transformation of 3D coordinate system
fixed formatting
Feb
9
suggested suggested edit on transformation of 3D coordinate system