Wikipedia:Reference desk/Archives/Mathematics/2011 August 24

= August 24 =

Real cube roots of the identity matrix
Does a solution for $$\mathbf A^3=\mathbf I$$ exist, where $$\mathbf I\ne\mathbf A\in\mathbb R^{3\times3}$$? I have been trying to find a solution for hours. Thanks, 85.250.176.130 (talk) 08:18, 24 August 2011 (UTC)
 * $$\mathbf A=\begin{pmatrix}

-\frac 1 2       & -\frac{\sqrt 3}2 & 0 \\ \frac{\sqrt 3}2 & -\frac 1 2       & 0 \\ 0              &      0           & 1 \end{pmatrix}$$ 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
 * $$\mathbf A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$$

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 f(0) = f(1) = 0. Prove that |f(a)| ≥ 4 for some a'' in (0,1). Spivak hints that we should try to prove that either f'(a) ≥ 4 for some a in (0,1/2), or f'(a) ≤ &minus;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 $$f'(0) = 0$$ and $$f(0) = 0$$, find a bound for $$f(1/2)$$.  Repeat using $$f > -4$$ on $$(1/2, 1)$$, $$f'(1) = 0$$, $$f(1) = 1$$ 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 f'(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: $$(2x^2)'=4x$$. -- 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 ' '=&minus;4, until t=1. Then x=1 and x '=0. This is the only motion with |x ' '|&le;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 |f(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)