# How is $x = 0$ a solution to x' = Ax?

This may seem obvious, but how would you explain something like this? This is referring to the homogeneous system of differential equations, originally from

$$x'(t) = A(t)x(t) + b(t)$$

where $b(t) = 0$ to bring us the homogeneous system.

I ask about $x=0$ being a solution because it is necessary to demonstrate that the set of all solutions $x$ is a subspace of $V_n(I)$.

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If you just plug it in, it works doesn't it?: if $x_0(t)=0$ then $x'_0(t)=0$ and $Ax_0(t)=0$ for all $t$ so $x'_0(t)=Ax_0(t)=0$ for all $t$ so it is indeed a solution. –  tibL Dec 2 '12 at 10:42
Oh okay - thanks for spelling that out for me! –  dmonopoly Dec 2 '12 at 11:02

if $x_0(t)=0$ then $x'_0(t)=0$ and $Ax_0(t)=0$ for all $t$ so $x'_0(t)=Ax_0(t)=0$ for all $t$