Proving A Must Be of Rank 2: The 2x2 Matrix Dilemma

LukasMont
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Homework Statement
4.1 Show that one may express any second rank matrix as the sum of a symmetric
and an antisymmetric matrix.
Relevant Equations
I was able to proof that any matrix could be constructed by adding a symmetric and antisymmetric matrix:

A= A/2 + A/2 + A'/2 - A'/2,
A= (A/2 + A'/2) + (A/2 - A'/2), where A' is the transposed matrix. Now,

A/2 + A'/2 is symmetric, since (A/2 +A'/2)' = A'/2 + A/2 (equal) and
A/2 - A'/2 is antisymmetric, since (A/2 - A'/2)' = - A'/2 + A/2= -(A/2 - A'/2).
My trouble is being to show A must be of rank 2. Any ideas?
 
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You don't have to show that A has rank 2. You are given that as a premise.
I don't know why they give you that as a premise though, because the theorem is true for any matrix, not just any rank-2 matrix, as your proof shows.
Anyway, you have proven what they asked you to.
 
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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...

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