Is T[kx(t)] equal to kT[x(t)] in a linear system?

[magnifik]

y(t) = x(t) + $$\int$$ (t - $$\tau$$)x($$\tau$$)d$$\tau$$

for it to be linear, T[kx(t)] = kT[x(t)] so i have
T[kx(t)] = kx(t) + $$\int$$ (t - $$\tau$$)x($$\tau$$)d$$\tau$$
and
kT[x(t)] = k[x(t) + $$\int$$ (t - $$\tau$$)x($$\tau$$)d$$\tau$$] = kx(t) + k$$\int$$ (t - $$\tau$$)x($$\tau$$)d$$\tau$$
so they aren't equal and aren't linear. however, I'm not sure about this answer because for the first part, T[kx(t)], I'm not sure if x($$\tau$$) should also be multiplied by k, making it a linear system

any help would be appreciated. thx.

 

[ystael]

You should, in fact, be multiplying $$x(\tau)$$ by $$k$$ inside the integral -- simple substitution is how you see this. If $$(Tx)(t) = x(t) + \int (t - \tau) x(\tau) \,d\tau$$ and $$x_1(t) = kx(t)$$ then $$(Tx_1)(t) = x_1(t) + \int (t - \tau) x_1(\tau) \,d\tau = kx(t) + \int (t - \tau) kx(\tau) \,d\tau$$.