22,642 reputation
11441
bio website drexel.edu/math/contact/ta-ra/…
location United States
age 24
visits member for 1 year, 2 months
seen 17 hours ago

I am pursuing a PhD in mathematics at Drexel university in Philadelphia, PA. I started my Teaching assistantship in fall 2013. As of now, I am not sure where my interests lie, though I am leaning towards something along the lines of matrix analysis.

I'm here because I enjoy being a part of the MSE community, and because whether you're asking or answering, you can never get enough practice with math problems.

Some answers I had fun putting together:

Some of my favorite questions/answers:

Useful links:


17h
comment Permutation matrices
Are you sure that should be an "I" (the letter $I$) and not just a "1" (the number $1$)?
17h
answered Permutation matrices
1d
comment Proving result on spectral radius
In fact, there is no need to find the similarity-matrix; it suffices to show that the matrices are similar by considering $\ker(A - \lambda I)$. However, if you prefer, you could use $$ S_n = \pmatrix{ n^{n-1}&&&\\ &n^{n-2}&&\\ &&\ddots&\\ &&& 1 } $$
1d
revised Counting partition of set that $i$ and $i+1$ are not in one part
Grammar
1d
comment Set of measure zero
@DanielFischer ah, cool! That's a neat way to look at it.
1d
comment Set of measure zero
@DanielFischer The set $\{(x,y):(x-y)y = 0\}$ is not an affine subspace of $\Bbb F^n$ (it is the union of two, but not closed under affine combinations), nor is it a submanifold. Though, as the other answers suggest, this does not change the intuitive answer to the problem.
1d
comment Set of measure zero
@DanielFischer are you implying that $S$ is a linear subspace of $\Bbb C^{n \times n}$?
1d
revised Proving result on spectral radius
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1d
revised Proving result on spectral radius
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1d
revised Proving result on spectral radius
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2d
revised Need verification - Given a Hermitian matrix and two eigenvectors corresponding to distinct eigenvalues, show x and y are orthogonal.
edited tags
2d
answered Need verification - Given a Hermitian matrix and two eigenvectors corresponding to distinct eigenvalues, show x and y are orthogonal.
2d
revised Elements in $F=\bigcup \limits_{n=1}^\infty \left(\bigcap \limits_{k=n}^\infty E_k\right)$
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2d
answered Elements in $F=\bigcup \limits_{n=1}^\infty \left(\bigcap \limits_{k=n}^\infty E_k\right)$
2d
revised Connected topological spaces, product is connected
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2d
comment Proving result on spectral radius
Here's what I'm really suggesting. For your favorite operator norm $\|\cdot\|$, we can select a sequence of similar matrices $\{S_n A S_n^{-1}\}_{n=1}^\infty$ whose norm approaches the spectral radius. Importantly, note that you don't have to do this for every operator norm; you just have to choose a single one and show that this works. As a hint towards the choice of similar matrix, note that two matrices are similar if and only if they have the same Jordan canonical form.
2d
comment Proving result on spectral radius
A lot of things to answer here; I'll address them in the order they appear. "Basically what we are saying is that for all ϵ and all induced norms there exists at least an invertible matrix S such that $\|S\|\|S^{-1}\|<\frac{1}{\|A\|}(\rho(A)+\epsilon)$". No, that is not what we are saying. For operator norms, the best we can say for $S$ and $S^{-1}$ is that $1 = \|I\| = \|SS^{-1}\| \leq \|S\| \|S^{-1}\|$. No "hurry", it's just that a lot of askers give absolutely no response, question, comment, or otherwise. Note also that though you've just seen it, it's been up for 2 days.
Aug
27
comment About lemma $\rho(A) \leq \|A^k\|^{1/k}$
@user153012 there's nothing wrong with your more general proof. Not sure why the wiki author decided to stop with that particular lemma. Note: it is also common to consider the matrix $v \mathbf{1}^T$, where $\mathbf 1$ is the column vector of $1$s
Aug
27
comment topology defined on the set $\mathbb{R}^\mathbb{R}$?
@user172061 a lot of the time, with power series in particular, mathematicians define convergence in terms of the $L^\infty([-a,a])$ norm.
Aug
27
comment topology defined on the set $\mathbb{R}^\mathbb{R}$?
@Teri I was addressing the motivation of the question, as given by the comment "I'm asking what topology is being considered on the set of real functions when we talk about, for example, convergence of a power series". In such circumstances, we usually restrict our analysis to "nice" functions. Perhaps it is more fitting to talk about $L^p(X)$ over compact $X \subset \Bbb R$, as this better reflects practice.