rcollyer
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 Sep 25 comment Do real matrices always have real eigenvalues? @commenter, no. The equation $A v = \lambda v$ has a non-trivial solution iff $\det(A - \lambda I) = 0$ which is identical to finding the roots of a polynomial in $\lambda$. We know the roots of a polynomial with real coefficients may have complex roots. If we try to restrict ourselves to only real roots, we're not guaranteed to find all of them. Sep 25 awarded Critic Sep 25 comment Do real matrices always have real eigenvalues? No, it doesn't. Finding the eigenvalues of a matrix is identical to finding the zeroes of a polynomial, and polynomials with real coefficients can have complex zeroes. Consider $(a - \lambda)(b - \lambda) - c d = 0$ which has complex roots if $(a+b)^2 < 4 (ab - cd)$. Sep 24 revised Solving an equation with trig functions and two different angles added LaTeX markup Sep 24 suggested approved edit on Solving an equation with trig functions and two different angles Sep 24 answered Pure maths vs applied Sep 22 revised derivatives using chain rule deleted 2 characters in body Sep 22 answered derivatives using chain rule Sep 22 comment Derivative using product rule and chain rule As a second point, I'm not a fan of Newtonian notation to indicate a derivative, e.g. $f'(x)$, and I prefer Leibniz's, e.g. $df/dx$, because it is difficult to tell which variable to use when calculating a derivative like, $\cos(a^3 + x^3)'$. In first semester calculus, this is easier as it is likely to be $x$. But, Leibniz's notation leaves less room for ambiguity, and that can be a life saver when doing these. However, professors insist on using Newton's form, so you have to learn it, also. But, I'd change it to the other form when working these. Sep 22 awarded Citizen Patrol Sep 22 comment Derivative using product rule and chain rule This is just a relatively straightforward application of your previous question. I'd suggest looking at Jonas' answer again. Sep 12 awarded Analytical Sep 8 revised Determinant and power series added align environment: formula exheeded bounding box. Sep 8 suggested approved edit on Determinant and power series Aug 27 comment How to solve a system of dot products You have a $t$ in the denominator of the $b$ term in your known solution. Is that supposed to be a $\tau$? Aug 27 comment Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even Great solution. My only question is what led you to it? Most likely it was the observation that $(2)$ is true, but I'm curious about the thought process. Aug 22 revised Determinant of a special skew-symmetric matrix clarified Aug 22 revised Determinant of a special skew-symmetric matrix added link; rearranged for clarity Aug 22 revised Determinant of a special skew-symmetric matrix fixed typo Aug 22 revised Determinant of a special skew-symmetric matrix deleted 1 characters in body