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August 24

Real cube roots of the identity matrix

Does a solution for ${𝐀}^3=𝐈$ exist, where $𝐈\ne 𝐀\in {ℝ}^{3\times 3}$? I have been trying to find a solution for hours. Thanks, 85.250.176.130 (talk) 08:18, 24 August 2011 (UTC)

$𝐀=\left(\begin{array}{ccc}-\frac{1}{2}& -\frac{\sqrt{3}}{2}& 0\\ \frac{\sqrt{3}}{2}& -\frac{1}{2}& 0\\ 0& 0& 1\end{array}\right)$

Bo Jacoby (talk) 08:36, 24 August 2011 (UTC).

This is a perfect example of where it's easier to think of composition of linear transformations, rather than multiplication of matrices. What linear transformation composed with itself 3 times will return the identity matrix? Well, clearly, if you rotate around an arbitrary axis with angle 120°, then you get such a function, and then you only need to confirm that the function is, in fact, a linear transformation. --COVIZAPIBETEFOKY (talk) 13:09, 24 August 2011 (UTC)

A simpler solution is

$𝐀=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right)$

Bo Jacoby (talk) 16:46, 24 August 2011 (UTC).

... which is a rotation about the line x = y = z. Gandalf61 (talk) 17:04, 24 August 2011 (UTC)

Problem from Spivak's Calculus

Let f be two-times differentiable with f(0) = 0, f(1) = 1, and fTemplate:'(0) = fTemplate:'(1) = 0. Prove that |fTemplate:'Template:'(a)| ≥ 4 for some a in (0,1). Spivak hints that we should try to prove that either fTemplate:'Template:'(a) ≥ 4 for some a in (0,1/2), or fTemplate:'Template:'(a) ≤ −4 for some a in (1/2, 1), but I'm not sure how to do that. I managed to prove the claim for the case when f has a maximum point over (0,1) in (0, 1/2), but I'm not convinced this is the approach he's suggesting. Thanks for the help. —Anonymous Dissident^Talk 21:44, 24 August 2011 (UTC)

Try a proof by contradiction. Suppose $f^{″}<4$ on the interval $(0,1/2)$. Using $f^{\prime }(0)=0$ and $f(0)=0$, find a bound for $f(1/2)$. Repeat using $f^{″}>-4$ on $(1/2,1)$, $f^{\prime }(1)=0$, $f(1)=1$ to get a different bound. Conclude a contradiction.--Antendren (talk) 22:04, 24 August 2011 (UTC)

Success! Cheers. Maybe I should stop considering contradiction as a last resort... —Anonymous Dissident^Talk 08:11, 25 August 2011 (UTC)

Actually no. I thought I had it but now things went wrong. I began with the false premises. Then I used the mean value theorem on the interval (0, x) to show that fTemplate:'(x) < 4x for all x in (0, 1/2], and then the MVT again to show f(x) < 4x², whence f(1/2) < 1. Similar methods on [1/2, 1) lead merely to f(1/2) > 0, which isn't helpful. What have I done wrong? —Anonymous Dissident^Talk 12:00, 25 August 2011 (UTC)

The bound $f(x)<4x^2$ is too weak. Hint: $(2x^2{)}^{\prime }=4x$. -- Meni Rosenfeld (talk) 13:38, 25 August 2011 (UTC)

Let x=f(t) be the position of a car standing still on the position x=0, then speeding up at t=0 at maximum acceleration x ' '=4 until t=1/2, when it brakes at maximum deceleration x ' '=−4, until t=1. Then x=1 and x '=0. This is the only motion with |x ' '|≤4 that satifies the boundary conditions. But f(t) is not two times differentiable at the time t=1/2 when speeding up is suddenly changed to braking. A two times differentiable motion satisfies the sharp inequality f ' '(a)>4 for some a. Bo Jacoby (talk) 13:52, 25 August 2011 (UTC).

Given that the problem as stated is even easier, calling only for proof that |fTemplate:'Template:'(a)| ≥ 4 somewhere (not strictly greater), how far can the differentiability condition be relaxed? Would it be sufficient that f be a differentiable function, twice differentiable almost everywhere, or would f need to be twice differentiable at all by finitely many locations? -- 110.49.248.74 (talk) 00:11, 26 August 2011 (UTC)

yes. If f is unbounded, we're done. Otherwise f is in L^1, so the FTC holds. Sławomir Biały (talk) 00:30, 26 August 2011 (UTC)