Linear Algebra/Tensor Algebra: Symmetry of a (1,1) tensor.

B L
Messages
7
Reaction score
0

Homework Statement


Let M be a differentiable manifold, p \in M.
Suppose A \in T_{1,p}^1(M) is symmetric with respect to its indices (i.e. A^i_j = A^j_i) with respect to every basis.
Show that A^i_j = \lambda \delta^i_j, where \lambda \in \mathbb{R}.

Homework Equations



The Attempt at a Solution


I've tried various ways of using the change of basis formula to arrive at the desired result, but I can't make it work. I imagine I need to use something else that I'm not thinking of.
 
Last edited:
Physics news on Phys.org
Do you know that a symmetric real matrix has a complete set of orthogonal eigenvectors? That means A is diagonal in some basis. Now can you show if A is diagonal in some basis with unequal diagonal elements, then it is not symmetric in some basis? Construct that basis from the original diagonal basis. It would be quite enough to do this for a 2x2 matrix.
 
I don't think we're supposed to use eigenvectors, but I'll give that a shot, thanks!

Any other ideas? I'm way too stumped given how seemingly simple this thing is (especially compared to the rest of the assignment).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top