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1d
comment A very detailed book for calculus 1-3.
A very standard textbook is Stewart's Calculus. I don't necessarily recommend it, but you should at least be aware of it.
1d
comment Minimizing convex functions without compatible gradients
So $n$ is not restricted to be an integer? That's unusual notation, it might be more clear to call it $x$.
Apr
27
comment How to prove Lagrange multiplier theorem in a rigorous but intuitive way?
Yes, and it's a beautiful story that's rarely told. We replace the four subspace theorem with Farkas' lemma (the four cone theorem). I added notes about this.
Apr
25
comment Numerical stability of computational results
I should have clarified, my comment was only tangential to the question.
Apr
25
comment Numerical stability of computational results
I think in practice Cramer's rule is never used. It's rare to use determinants also. If you want to numerically solve a linear system of equations, usually you use a method such as Gaussian elimination.
Apr
25
comment Convex optimization qualifying exam
It's better to submit these as separate questions and explain what you have tried for each question (and ideally write the question using Latex rather than linking to image).
Apr
24
comment How to numerically minimize system of equations composed of data and smoothness terms, ensuring minimum solution norm
Hmm, how about adding $\gamma \| g \|^2$ to the objective instead of $\gamma \| g \|$? (Is that what you did already?) I think instead of $A^T A + \Gamma^T \Gamma$ you'll have $A^T A + \Gamma^T \Gamma + \gamma I$, which is still very sparse.
Apr
24
comment What is the motivation for wanting to remove redundant terms that arise from the wedge product of two multilinear functions?
Which terms are you thinking of that could be kept? I think the idea is that $f \otimes g $ is not alternating, but $f \wedge g $ is.
Apr
24
comment How to numerically minimize system of equations composed of data and smoothness terms, ensuring minimum solution norm
You can find a Pareto optimal $g $ by adding $\gamma \| g \|$ to your objective function and solving the resulting optimization problem. There's a section about Pareto optimality in Boyd and Vandenberghe.
Apr
24
comment How does differentiation work?
Actually it's not the same. Taking the limit as $h$ approaches $0$ is not the same thing as plugging in $h = 0$. Here's an example to keep in mind: let $f(x) = \sin(x)/x$. Then $\lim_{x \to 0} f(x) = 1$, and yet $f$ is undefined at $0$, so plugging in $0$ is not an option. Another example: Define $f$ so that $f(x) = x$ when $x \neq 0$, and $f(0) = 1$. Then $\lim_{x \to 0} f(x) = 0$, but $f(0) = 1$, so we can't simply plug in $x = 0$ (in fact the value of $f$ when $x = 0$ is irrelevant when we're taking the limit as $x \to 0$.) Often, there is simply no nice formula for $f$ at all.
Apr
23
comment How many ways are there to prove Cayley-Hamilton Theorem?
What's a $k$-algebra morphism? Can you elaborate on that step?
Apr
23
comment How does differentiation work?
Step 2) won't always work though, so we should have a definition of the derivative that makes sense even if step 2 fails.
Apr
23
comment Is the union of dual cone and polar cone of a convex cone is a vector space?
I think you're right. You could also let $C $ be one of the self-dual cones : the nonnegative orthant, the ice cream cone, or the positive semidefinite cone.
Apr
21
comment learning linear algebra
Since you're interested in tensors, you might also like to learn about integration on manifolds.
Apr
21
comment About the definition of the dimension
@standerQiu Although the word "dimension" is used loosely sometimes, strictly speaking it is only a subspace or a vector space that can have a dimension. If you tell me that a particular subspace has dimension 2, I know exactly what you mean. But, if you tell me that a particular vector has dimension 3, or that a particular vector has dimension 1, then I would have to object that this statement is meaningless, because the "dimension of a vector" has never been defined (and is not a standard term in math). Only a subspace or a vector space can have a dimension, not a vector.
Apr
21
comment About the definition of the dimension
Yes, that works. Since a basis is a set of vectors, I would express your answer using set notation: the set $\{ (1,1,1), (1,1,2) \}$ is a basis for $W$. Since this basis has two elements, $W$ has dimension $2$. (By the way, another basis for $W$ is $\{ (1,1,0), (0,0,1) \}$.)
Apr
21
comment About the definition of the dimension
Yes, assuming I understand you correctly. Here's another example. Let $W = \{ x = (x_1,x_2,x_3) \in \mathbb R^3 \mid x_1 = x_2\}$. (Note that $x_1, x_2$, and $x_3$ are scalars, and $x$ is a vector.) So $W$ is a subspace of $\mathbb R^3$. Can you give me a basis of $W$?
Apr
21
comment About the definition of the dimension
Let's just be very concrete and give one single example of a basis for $U$. The set $\mathcal B = \{ (1,1,1) \}$ is a basis for $U$. And since this basis has one element, the dimension of $U$ is $1$.
Apr
21
comment About the definition of the dimension
@standerQiu What is $x_1$? I don't understand what you mean. I was expecting you to tell me that a basis for $U$ is the set $\mathcal B = \{ (1,1,1) \}$, or something like that.
Apr
21
comment About the definition of the dimension
@standerQiu Before I answer that, maybe forget about "dimension" for a minute, and tell me this: can you give me a basis for the subspace $U$?