# Prove that the maximal number of linearly independent vectors for the given vectors in the $k$ -vector space $k^3$ is 0 or 2.

I have a question that has me completely stumped in my linear algebra homework this week. I'm not looking for a full blown solution here. I just need a hint on where to begin. It asks the following:

"Prove that the maximal number of linearly independent vectors among $$x=(a_2,-a_1,0),\quad y=(a_3,0,a_1),\quad z=(0,a_3,a_2)$$ in the $k$-vector space $k^3$ is $0$ or $2$."

What is this even asking? Are $x,y$ and $z$ vectors? And what is a $k$-vector space in this context? I swear half of the battle in doing problems at this level is knowing the notation and definitions, and knowing them well.

As I mentioned above, I'm not sure how to go about even starting this one. I am pretty awful with this material, so you can safely assume I have no clue what I am doing here. The book we are using is Abstract Linear Algebra by Morton L. Curtis. We are only in the first chapter (topics include everything vector spaces, linear maps, isomorphisms, basis, dimension, rank, etc.), so we haven't discussed anything with matrices yet, so any tips on how to start this one using non-matrix related methods would be immensely appreciated. And yes, I know what a matrix is and how they function, but we aren't supposed to use them until we get to that section.

• Yes, x,y,z are Vectors. K is a field and $K^3$ a three dimensional vectorspace over K. Do you know how to check if vectors are linearly independent ? – jonask Feb 10 '17 at 8:44
• Thank you for that clarification. I think I know how. Don't you try and solve the system for zero, and if the scalars are all zero, then the vectors are all linearly independent? – Thy Art is Math Feb 10 '17 at 9:23
• Exactly. The homework question already hints that there will be two seperate cases to examine, so try finding those two first and then solve the two systems of linear equations. – jonask Feb 10 '17 at 10:12
• Thank you so much for your help! – Thy Art is Math Feb 10 '17 at 14:39

$k$ is a field and $k^3=\{(a,b,c): a,b,c \in k\}$. $k^3$ is a $k$ - vector space.
$x=(a_2,-a_1,0),\quad y=(a_3,0,a_1)\quad$ and $z=(0,a_3,a_2)$ ar elements of $k^3$.
• Thank you for defining $k^3$ in that way, it seems to make more sense now. However, even knowing this information, I am not sure how to go about proving what this question is actually asking me to prove. – Thy Art is Math Feb 10 '17 at 9:26