James S. Cook
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 2h answered Is this picture of the covariant derivative correct 2h comment Is this picture of the covariant derivative correct Upon further review, I get your picture, maybe choose a simpler $W(p+tv)$ curve to see it and focus on say $t=0$ and $t = \triangle t$ for quasi-small $t$. You should be able to see the covariant derivative as a new vector which describes how $W$ is changing in the $v$-direction at $p$. Perhaps, supermath.info/DifferentialGeometry2015.pdf will be helpful, I started, but they're incomplete, some notes to parallel Oneiil Apr 19 comment How to find basis for orthoogonal complement basis for the following condition? my answer assumes $\langle A, B \rangle = \text{trace}(AB^T)$ aka the usual inner product. Apr 19 answered How to find basis for orthoogonal complement basis for the following condition? Apr 19 comment Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$? @Ian no problem, I have weird ideas sometimes, feel the pain of my students. Apr 19 comment Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$? @Ian But, $IJ=JI$ so, $exp(aI+bJ) = exp(aI)exp(bJ) = exp(aI)exp((b+2\pi k)J) = exp(aI+(b+2\pi k)J)$. Granted, I don't need the $aI$ term, but, I included it to have a bit different looking example. These matrices are diagonalizable over $\mathbb{C}$. Of course, $aI+bJ$ is just the real Jordan form of the matrix $\left[ \begin{array}{cc} a+ib & 0 \\ 0 & a-ib \end{array}\right]$. Apr 18 comment Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$? sure. You can set $a=0$ if you want. Apr 18 revised Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$? added 101 characters in body Apr 18 answered Is there a diagonalizable matrix $A \neq B$ such that $e^A = e^B$? Apr 6 comment Eigenvalues of block matrix related ... Notice $BB^T$ is symmetric and real (I assume, I suppose you can generalize over $\mathbb{C}$ with general spectral theorem) thus diagonalizable via an orthonormal eigenbasis with matrix $P$ such that $P^TP=I$. In particular $D_B = P^T(BB^T)P$ is diagonal. Moreover, $$P^TMP = (\lambda^2I+D_B)-\lambda P^TAP$$ (this is not an answer, but, perhaps someone can make use of it..) Apr 6 comment Eigenvalues of block matrix related Since $B^T$ and $\lambda I$ commute we have the formula below: $$P(\lambda) = \text{det} \left[ \begin{array}{cc} \lambda I - A & -B \\ -B^T & \lambda I \end{array} \right] = \text{det}\left[ (\lambda I - A)(\lambda I)-BB^T \right]$$ thus, $$P(\lambda) = \text{det}\left[ \lambda^2 I - \lambda A -BB^T \right]$$ Let $M = \lambda^2 I - \lambda A -BB^T$... Apr 5 awarded Popular Question Mar 19 comment Quick clarification on the definition of vector field btw, I read this question because of your user name. Nicely done. Mar 19 answered Quick clarification on the definition of vector field Mar 15 comment Are there any instances of significant progress deriving from mathematical 'silliness'? Much of theoretical physics is silly in the sense of your question. Has theoretical physics produced meaningful mathematical progress despite questionable foundations? The answer is clearly yes. Our knowledge of the topology of $4$-manifolds is largely due to what was first wild conjecture (guided by physical intuition). This story is ongoing. Mar 15 comment Are there any instances of significant progress deriving from mathematical 'silliness'? $64/16 = 4$ because you can cancel the $6$'s. Mar 10 comment How to remember hyperbolic functions nice. I should include this in my standard soundbite for these. Mar 10 comment How to remember hyperbolic functions actually, the way to remember these for PDEs is just that $y''-\beta^2y=0$ has solutions $y=\cosh (\beta x)$ and $y=\sinh (\beta x)$ plus the derivative properties of cosh and sinh. Mar 10 answered How to remember hyperbolic functions Feb 29 awarded Enlightened