Determining Eigenvalue 4 Invertibility in Linear Algebra

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To find out if 4 in an eigenvalue of A, decide if A-4I is invertible...

So, if A-4I is invertible, then its cols are lin ind by IMT, and also there is only the trivial solution to A-4I=0, so thus 4 is not an eigenvalue of A

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? The definition of "eigenvalue" is that \lambda is an eigenvalue if and only if Av= \lambda v has non-trivial solutions. That is the same as saying that Av- \lambda v= (A- \lambda I)v= 0 has non-trivial solutions. Since v=0 is obviously a solution, saying it has non-trivial solutions means it does NOT have a "unique" solution. If the matrix M has an inverse, then the equation Mv= u has the unique solution [math]v= M^{-1}u[/math].
 
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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