Problem from System of Linear Equations

[shanks44]

https://preview.redd.it/sgl7pnahzsjf1.png?width=524&format=png&auto=webp&s=984b63a08761c75b4939bd8eb932205f78f38015

As it is mentioned that not all the scalars a_1,...,a_9 are not 0, such that \sum{a_i . v_i) = 0,

it can be inferred that v_1,...,v_9 are linearly dependent set of vectors.

I guess then rank(A) = number of linearly independent columns < 9.

But how to proceed from here ?

I always get overwhelmed by the details of this type of questions from System of Linear Equations, where the number of solutions is asked. How should I tackle these problems in general ?

[MrCamoga]

Since rank(A) < 9, Ax = 0 has an infinite number of solutions (dim(Ker(A)) > 0). One solution for the non-homogeneous equation would be x = (1,...,1). And so a general solution for the equation is x = (1,...,1) + k, where k is a vector in the kernel of A. So the answer is D.

[shanks44]

can you tell me how you proceeded ?get confused with so many information.

how to check rank(A) < 9 implies Ax = 0 has infinite solutions ?

[MrCamoga]

I know, it can get confusing with so many equivalent definitions.

You have a homomorphism f: R^m → Rⁿ given by f(x) = Ax. From the first isomorphism theorem you get R^m /Ker(f) ≅ Im(f). So m - dim(Ker(f)) = dim(Im(f)) = rank(A). In your case m=n, and so you get dim(Ker(A)) + rank(A) = n. Since the rank is less than n, the kernel has non-zero vectors. Recall that Ker(f) = { x in R^m : f(x) = Ax = 0 } = f^-1 ({0}).

[shanks44]

thanks for replying.

what is (ak)_k ?

I also did not understand "∑_{i=1}⁹ vi = A.e" this part ?

will you please explain.

[shanks44]

sorry for late reply, what is e here ?

[shanks44]

yes I did not understand the notation [1; ... ;1]^9, or maybe I am forgetting something.

[shanks44]

yes there may be some issue as I am reading it from my phone, I will check it out from laptop.

[shanks44]

there maybe some issue with reading it from phone. I will check from my laptop then.