A question about findind an operator that

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Operator
transgalactic
Messages
1,386
Reaction score
0
Physics news on Phys.org
You should notice the fine print. The question doesn't say dim(Ker(S))+dim(Im(S))=dim(M2x2). That's the LAW. It says Ker(S)+Im(S)=M2x2. Suppose the matrix m is [[0,1],[0,0]] and S(x)=x*m. Show Ker(S)+Im(S) is 2 dimensional and doesn't span M2x2. I'd almost consider this a trick question.
 
Last edited:
thanks
 
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