I have a general question regarding eigenvalues/vectors. Say you are

  • Thread starter Thread starter kr0z3n
  • Start date Start date
  • Tags Tags
    General
kr0z3n
Messages
8
Reaction score
0
I have a general question regarding eigenvalues/vectors. Say you are working on a matrix
[ 1, 1; 4, 1] and you find the eigenvalues to be 3 and -1.
Does it matter which eigenvalue is first...meaning does it matter when it is λ1= 3 and λ2=-1 or the other way around?
When you write out the general solution in the form of c1*e^-t[1, -2] + c2*e^3t[1, 2] , how do you know which eigenvalue is written first? In other words why can the general solution not be written as c1*e^3t[1, 2] + c2*e^-t[1, -2]?
 
Physics news on Phys.org


No. It doesn't matter at all.
 


Thank you
 
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