Proving Matrix Equality: A^3+4A^2-2A+7i=0 Implies A^T Also Satisfies It?

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ahmed dawod
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1. show that if a square matrix A satisfies A^3+4A^2 -2A+7i=0 then so does A^T

i is the identity matrix





The Attempt at a Solution



I tried to multiply the equation by A^T but all in vain
 
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Use:
(A+B)^T = A^T + B^T \qquad (AB)^T = A^TB^T \qquad (cA)^T = cA^T
to compute the transpose by using:
0 = 0^T = (A^3 + 4A^2 - 2A + 7I)^T
 
thanks
this is useful

problem solved:biggrin:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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