# Vector in 9 dimensions always has a solution?

I was watching Professor Gilbert Strang's lecture on Linear Algebra, and in the video, he talks about when we can or cannot say that linear combinations of vectors span the entire region in that dimension.

He said that if we had nine vectors, each of dimension 9, and if the vectors were all random, then the answer is probably yes. But if it is the case that two of the nine vectors are the same, then they don't add anything new to the equation, and what we will get is probably an 8-dimensional plane.

I don't understand how this is the case.

Also, could someone explain what this means in terms of the number of solutions we will get?

• You don't understand how what is the case? The statement has two parts. Which of them do you understand or not understand? – saulspatz Sep 14 at 22:33
• I cannot understand why if two vectors are the same, then we don't have a solution. – Oishika Chaudhury Sep 14 at 22:36
• I takes $n$ linearly independent vectors to span $n$-space. If two of the are are the same, you only really have $n-1$ vectors. Think about the natural basis vectors. – saulspatz Sep 14 at 22:40

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 $$R^n$$.

• I think you meant 'linearly independent' in the first sentence. – Nick Guerrero Sep 14 at 23:01
• @NickGuerrero good catch, i changed it – Saketh Malyala Sep 14 at 23:22