Vector in 9 dimensions always has a solution?

Think of choosing two random vectors in the plane. The probability that they lie on the same line (so one is a multiple of the other) is $0$ - lines are very thin - so the probability that they are independent and thus span the plane is $1$.

In $9$ dimensions the probability that $9$ random vectors all happen to lie in some subspace of dimension less than $9$ is $0$.

(To say this rigorously you need to be precise about what it means to choose a random vector, but geometric intuition strongly supports Strang's assertion. You should think about it that way, not in terms of equations.)

Well, let's first start in $2$ dimensions.

If we pick the vectors, $(1,1)$, and $(2,0)$, it is apparent that they point in different directions and are linearly independent. Linear combinations of these two vectors span the entire plane.

However, if we were to pick the vectors $(1,0)$, and $(-1,0$), they are not linearly independent. No matter which way we make linear combinations of these two vectors, they can only span the $x$-axis.

In this way, if and only if every vector is linearly independent from the other $n-1$ vectors, then the $n$ vectors can span $R^n$.